Free access until 31 December 2026 · Physics and Maths questions and notes, updated continuously · Build a test · Get an email before anything changes

ExaminerPrep › Question bank › Cambridge IGCSE Physics › Topic 2 Thermal physics

Cambridge IGCSE Physics 0625 · Core and Extended · Topic 2

Thermal physics

327 free, high-quality past-paper-style questions on Thermal physics for Cambridge IGCSE Physics, each with a full mark scheme. Covers: Conduction, convection and radiation; Particle model and gas pressure; Thermal expansion, specific heat capacity and changes of state. For when you have finished the past papers, or want to practise one topic at a time.Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,126 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1IGPC-036-02-01Conduction, convection and radiation Paper CEasy1 mark

Which method of thermal energy transfer can take place through a vacuum?

Show mark scheme
Marking pointMarkNotes
Radiation (infrared) is an electromagnetic wave and needs no medium, so it can cross a vacuum. Conduction needs particles to pass on vibrations, and convection needs a fluid that can flow. A vacuum has no particles, so neither of these can happen in it.✓ 1Answer C

Answer: C

Common error: Conduction and convection both need particles. Only radiation can cross empty space, which is how energy from the Sun reaches the Earth.

2IGPC-036-02-02Conduction, convection and radiation Paper CEasy1 mark

Air directly above a hot heater rises. Why does this happen?

Show mark scheme
Marking pointMarkNotes
When the air is heated, its particles move faster and spread further apart, so the air expands. The same mass now takes up a larger volume, so its density falls. The cooler, denser air around it sinks and pushes the warm air upwards. The particles themselves do not change mass, and they speed up rather than slow down.✓ 1Answer A

Answer: A

Exam tip: In convection answers, always give the chain: heated, expands, less dense, rises. The cooler, denser fluid sinks to take its place.

3IGPC-036-02-03Conduction, convection and radiation Paper CEasy1 mark

Four metal plates are all at the same temperature. They have different surface finishes. Which plate emits infrared radiation at the greatest rate?

Show mark scheme
Marking pointMarkNotes
Dark, matt (dull) surfaces are the best emitters and absorbers of infrared radiation. Shiny and light-coloured surfaces are poor emitters. A shiny silver surface is the worst emitter of the four.✓ 1Answer A

Answer: A

Common error: The colour only matters for infrared because dull black is also a good emitter. Good absorbers are good emitters, and shiny surfaces are poor at both.

4IGPC-036-02-04Conduction, convection and radiation Paper CEasy1 mark

Rods of copper, glass, wood and steel have the same length and thickness. A small bead of wax is stuck to one end of each rod. The other ends are placed in boiling water at the same time. The wax on the copper rod melts first. What does this show?

boiling watercopperglasswoodsteelwax bead
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The energy travels along each rod by conduction. The wax on the copper rod reaches its melting point first, so energy was conducted along copper fastest. Copper is therefore the best thermal conductor. Convection cannot happen in solids, and the test does not measure emission or melting point.✓ 1Answer A

Answer: A

Exam tip: For a fair test, the rods must have the same length and thickness and start heating at the same time. Only the material should change.

5IGPC-036-00-01Particle model and gas pressure Paper CEasy1 mark

Which statement describes the particles in a liquid?

Show mark scheme
Marking pointMarkNotes
In a liquid the particles are almost as close together as in a solid. They have no regular pattern and can slide past one another, which is why a liquid flows and takes the shape of its container. Option A describes a gas and option B describes a solid. Option D describes no state of matter, because particles are always moving.✓ 1Answer C

Answer: C

Common error: Liquid particles are not far apart. The separation in a liquid is similar to that in a solid, which is why liquids are very hard to compress.

6IGPC-036-00-02Particle model and gas pressure Paper CEasy1 mark

The air in a classroom is at a temperature of 27 °C.
What is this temperature in kelvin?

Show mark scheme
Marking pointMarkNotes
T (in K) = θ (in °C) + 273, so T = 27 + 273 = 300 K. Option C comes from subtracting 273 instead of adding it. Option A comes from subtracting the wrong way round. Option B assumes the two scales have the same zero.✓ 1Answer D

Answer: D

Exam tip: A kelvin temperature can never be negative. If your conversion gives a negative kelvin value, you have subtracted instead of adding 273.

7IGPC-036-00-03Particle model and gas pressure Paper CEasy1 mark

A sealed metal can of air is heated. The volume of the can does not change.
Why does the pressure of the air in the can increase?

Show mark scheme
Marking pointMarkNotes
Heating raises the average kinetic energy of the molecules, so they move faster. They collide with the walls more frequently and each collision exerts a larger force, so the pressure rises. Molecules do not expand (A). The can is sealed, so the number of molecules is fixed (C). Gas pressure comes from collisions, not from attraction to the walls (D).✓ 1Answer B

Answer: B

Common error: Molecules themselves do not expand when heated. It is the spacing and speed of the particles that change, not their size.

8IGPC-036-00-09Particle model and gas pressure Paper CEasy1 mark

A bicycle pump is sealed at the end. The piston is pushed in slowly, so the temperature of the air inside stays the same.
Why does the air pressure increase?

Show mark scheme
Marking pointMarkNotes
The temperature is constant, so the average speed of the molecules is unchanged. In a smaller volume the same molecules are closer together, so they strike each unit area of wall more often. This raises the pressure. Options A and D would only be true if the temperature increased. Molecules do not change size (B).✓ 1Answer C

Answer: C

Exam tip: Link each change to the right cause. A change in temperature changes the speed of the molecules. A change in volume at constant temperature changes only how often they hit the walls.

9IGPC-036-01-01Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

A concrete road is laid in sections. Small gaps are left between the sections and filled with a soft, flexible material.
Why are the gaps left?

Show mark scheme
Marking pointMarkNotes
When the temperature rises, the concrete expands. Without gaps, neighbouring sections would push against each other and crack or buckle. The soft filler is squashed as the sections expand. Materials contract when cooled, not when heated. The gaps are not drainage channels, and they are not there to reduce conduction.✓ 1Answer A

Answer: A

Exam tip: Gaps in bridges, railway tracks and roads are all explained by expansion on heating. Say clearly that the material expands when hot.

10IGPC-036-01-04Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

Pure water is boiling in an open pan at standard atmospheric pressure. The pan continues to be heated.
What happens to the temperature of the water while it is boiling?

Show mark scheme
Marking pointMarkNotes
At standard atmospheric pressure, pure water boils at 100 °C. While it is boiling, the energy supplied is used to change liquid water into steam, so the temperature stays at 100 °C. 0 °C is the melting point of ice, not the boiling point of water.✓ 1Answer B

Answer: B

Common error: More heating while water boils makes it boil faster. It does not make the water hotter.

11IGPC-036-01-10Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

Hot liquid wax is left to cool. Its temperature is measured every minute. For several minutes, the temperature stays constant at 52 °C, even though the wax is still losing energy to the surroundings.
What is happening to the wax during this time?

Show mark scheme
Marking pointMarkNotes
A constant temperature while the wax is cooling shows a change of state. The liquid wax is solidifying (freezing). As its particles settle into fixed positions, energy is released, and this balances the energy lost to the surroundings. Boiling would need energy to be supplied, not lost. The question says energy is still being lost, so option D is wrong.✓ 1Answer C

Answer: C

Exam tip: A flat section on a heating or cooling curve always means a change of state is taking place.

12IGPC-036-01-11Thermal expansion, specific heat capacity and changes of state Paper CEasy4 marks

Overhead electricity cables are hung between tall pylons. Engineers fit the cables on a warm summer day.

(a)

State what happens to the length of a cable when its temperature falls.

(1)
(b)

Explain why the cables are left sagging, not pulled tight, when they are fitted in summer.

(2)
(c)

Suggest one other example where a gap or space is left to allow for thermal expansion.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The length decreases (the cable contracts).✓ 1
Part (b)
In cold weather the cables contract (become shorter).✓ 1
If they were tight, they could snap or pull on the pylons, so slack allows for this contraction.✓ 1
Part (c)
Any one, e.g. gaps between sections of railway track / expansion joints in bridges / gaps between concrete road sections.✓ 1

Exam tip: Link the season to the change. Cold weather means contraction, so a cable fitted tight in summer would be under tension in winter.

13IGPC-036-02-12Conduction, convection and radiation Paper CEasy5 marks

A student has two identical metal cans. One is painted dull black and the other is polished shiny silver. She fills each can with hot water at 80 °C and records the temperature of the water every minute as it cools in a room at 20 °C.

dull blackshiny silvereach can filled with water at 80 °C, in a room at 20 °C
Diagram not to scale
(a)

State two variables, other than the starting temperature, that the student must keep the same to make the test fair.

(2)
(b)

Predict which can cools faster. Explain your answer.

(2)
(c)

The water in the black can cools from 80 °C to 62 °C in 10 minutes. Calculate the average rate of fall of temperature, in °C per minute.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Acceptable variables (any two): the volume or mass of water in each can; whether a lid is used (the same for both cans); the position in the room or exposure to draughts; the temperature of the surroundings.—
First valid variable✓ 1
Second valid variable✓ 1
Part (b)
The dull black can cools faster .✓ 1
A dull black surface is a better emitter of infrared (thermal) radiation than a shiny silver surface. Also accept: it emits infrared radiation at a greater rate .✓ 1
Part (c)
(80 − 62) ÷ 10 = 1.8 °C per minute✓ 1

Common error: Do not just say the black can is 'better at radiating heat'. Say that a dull black surface is a better emitter of infrared radiation.

14IGPC-036-00-11Particle model and gas pressure Paper CEasy4 marks

A block of copper, some liquid cooking oil and the air above them are all at room temperature.

(a)

Describe the arrangement and the motion of the particles in the solid copper.

(2)
(b)

State one difference between the separation of the particles in the oil and the separation of the particles in the air.

(1)
(c)

Explain, in terms of particles, why air is much easier to compress than oil.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Arrangement: the particles are close together (touching) in a regular, fixed pattern . Motion: they vibrate about fixed positions .✓ 2
Part (b)
Oil particles are close together (touching), while air particles are far apart, or much further apart .✓ 1
Part (c)
Either air has large spaces between its particles, so they can be pushed closer together, or oil particles are already touching, so there is almost no space to push them into .✓ 1

Exam tip: When asked about arrangement and motion, give one point about each. Two points about arrangement alone will not earn both marks.

15IGPC-036-00-07Particle model and gas pressure Paper CEasy1 mark

What happens to the particles of a substance at absolute zero (−273 °C)?

Show mark scheme
Marking pointMarkNotes
Temperature is related to the average kinetic energy of the particles. Absolute zero is the lowest possible temperature. At absolute zero the particles have the least possible kinetic energy, so the substance cannot be cooled any further. That makes B correct.
A reverses the link between temperature and particle speed. C wrongly assumes that the particles themselves change size when cooled. Only the spacing and motion of particles change, not their size. D confuses 0 °C with 0 K: at 0 °C (273 K) the particles still have plenty of kinetic energy.
✓ 1Answer B

Answer: B

Common error: Absolute zero is −273 °C, which is 0 K. It is not 0 °C, the melting point of ice.

16IGPC-036-01-12Thermal expansion, specific heat capacity and changes of state Paper CEasy4 marks

This question is about the changes of state of water.

(a)

Name the change of state when steam turns into liquid water on a cold window.

(1)
(b)

Describe how the arrangement and the motion of the particles change when ice melts to form liquid water.

(2)
(c)

State the boiling temperature of pure water at standard atmospheric pressure.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Condensation.✓ 1
Part (b)
Arrangement: the particles change from a regular, fixed pattern to an irregular (random) one, but stay close together.✓ 1
Motion: the particles change from vibrating about fixed positions to moving around and sliding past each other.✓ 1
Part (c)
100 °C.✓ 1

Common error: When ice melts, the particles do not move far apart. Liquid particles are still close together. Only in a gas are the particles far apart.

17IGPC3-2-02c-01Particle model and gas pressure Paper CEasy1 mark

A sealed rigid steel canister contains air. The canister is left in the sun and the air inside warms up. Which statement explains the increase in gas pressure?

Show mark scheme
Marking pointMarkNotes
Correct: B. Higher temperature means a greater average speed, so there are more collisions per second and each collision exerts a larger force. A is wrong: the molecules themselves do not expand; and the volume of the rigid canister is fixed. C is wrong: heating does not add molecules; the canister is sealed, so the number is fixed. D is wrong: the mass of each molecule is unchanged by heating.✓ 1Answer B

Answer: B

Exam tip: Pressure comes from many molecular collisions with the container walls. Describe both how often and how hard they collide.

18IGPC3-2-02c-02Conduction, convection and radiation Paper CEasy1 mark

Why is a metal spoon in hot soup much hotter at its handle after a minute than a plastic spoon of the same size?

Show mark scheme
Marking pointMarkNotes
Correct: D. free electrons carry energy through the metal much faster than vibrations alone, which is why metals are good conductors. A is wrong: convection cannot occur in a solid because the particles are fixed in position. B is wrong: particles in many plastics are also fixed in place; the difference is the free electrons in metals. C is wrong: radiation does not need contact and is not the main process inside a solid metal spoon.✓ 1Answer D

Answer: D

Exam tip: Mention free electrons when explaining why metals conduct better than non-metals.

19IGPC3-2-02c-03Particle model and gas pressure Paper CEasy1 mark

A steel bridge is slightly longer on a hot afternoon than on a cold morning. Which statement explains this thermal expansion?

Show mark scheme
Marking pointMarkNotes
Correct: C. Heating gives particles more kinetic energy, so they vibrate more and the average separation grows, making the solid larger. A is wrong: the particles do not change size; the spacing between them increases. B is wrong: the number of particles is unchanged by heating. D is wrong: that happens in a liquid, not in a solid that has only warmed up.✓ 1Answer C

Answer: C

Common error: Thermal expansion is about the spacing between particles, never the particles getting bigger.

20IGPC3-2-02c-09Particle model and gas pressure Paper CEasy6 marks

The particle model is used to explain the properties of solids, liquids and gases.

(a)

Describe the arrangement of the particles in a solid.

(2)
(b)

State how the motion of the particles in a liquid differs from the motion of the particles in a solid.

(1)
(c)

Explain how the particles of a gas produce pressure on the walls of a container.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles are close together✓ 1
in a regular pattern / fixed positions✓ 1
Part (b)
particles in a liquid can move past each other / slide over one another, but in a solid they only vibrate about fixed positions✓ 1
Part (c)
gas particles move randomly at high speed✓ 1
they collide with the walls✓ 1
each collision exerts a force on the wall, and the force per unit area is the pressure✓ 1

Exam tip: In a gas explanation include movement, collisions with the walls and force on an area.

21IGPC3-2-02c-13Conduction, convection and radiation Paper CEasy6 marks

A saucepan of water is heated on an electric hob. The base of the saucepan is made of metal. The handle is made of plastic.

plastic handleelectric hobwatermetal base
Side view of the saucepan
(a)

Name the process by which thermal energy passes through the metal base.

(1)
(b)

Name the process by which the water at the top of the saucepan becomes warm.

(1)
(c)

Explain how the water at the top of the saucepan becomes warm.

(3)
(d)

Suggest why the handle is made of plastic.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
conduction✓ 1
Part (b)
convection✓ 1
Part (c)
water at the bottom is heated, expands and becomes less dense✓ 1
the warm water rises✓ 1
cooler, denser water sinks to take its place, setting up a circulation✓ 1
Part (d)
plastic is a poor conductor (an insulator) of thermal energy, so the handle stays cool enough to hold✓ 1

Exam tip: Use the words density, rises and sinks when you explain convection.

22IGPC3-2-07r7-01Particle model and gas pressure Paper CEasy1 mark

The air in a sealed brake reservoir on a train is warmed by strong sunshine. The volume of the reservoir does not change. Which statement explains why the air pressure rises?

Show mark scheme
Marking pointMarkNotes
Heating increases the average kinetic energy of the particles, so they collide with the walls more frequently and with greater force, which raises the pressure. Option A is wrong: particles themselves do not expand; it is the average speed that changes. Option C is wrong: the reservoir is sealed, so the number of particles is fixed. Option D is wrong: the mass of each particle does not change on heating.✓ 1Answer B

Answer: B

Common error: Pressure rises on heating because particles move faster, not because the particles themselves grow or gain mass.

23IGPC3-2-07r7-02Conduction, convection and radiation Paper CEasy1 mark

On a frosty morning a metal handrail on a station platform feels much colder to touch than a wooden bench, although both are at the same temperature. What is the reason?

Show mark scheme
Marking pointMarkNotes
Metal is a good conductor, so energy flows from the warm hand into it quickly and the hand cools; wood is a poor conductor. Option B is wrong: both are at the same temperature as the air. Option C is wrong: wood is an insulator, so it transfers energy to or from the hand more slowly. Option D is wrong: radiation is not the main process in touching; the difference is the rate of conduction.✓ 1Answer A

Answer: A

Common error: Touch tells you how fast energy leaves your hand, not the temperature of the object.

24IGPC3-2-07r7-07Particle model and gas pressure Paper CEasy1 mark

Liquid water has been detected beneath the surface of a distant moon. Which row describes the particles in a liquid?

Show mark scheme
Marking pointMarkNotes
Liquid particles are close and touching but not in a fixed pattern, so they can move around each other. Option A is wrong: this describes a solid. Option B is wrong: this describes a gas. Option C is wrong: a regular pattern belongs to a solid; liquid particles are irregular.✓ 1Answer D

Answer: D

Exam tip: Compare three things for each state: spacing, arrangement and type of motion.

25IGPC3-2-07r7-09Thermal expansion, specific heat capacity and changes of state Paper CEasy4 marks

Steel railway lines are laid in long sections with small gaps between the ends on a hot day.

(a)

State what happens to the length of a steel rail when its temperature increases.

(1)
(b)

Explain, using the particle model, why a solid expands when heated.

(2)
(c)

Suggest why engineers leave a small gap between rail sections.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
it increases (the rail expands)✓ 1
Part (b)
particles gain energy / vibrate with greater amplitude✓ 1
so the average separation between particles increases✓ 1
Part (c)
so the rails can expand in hot weather without buckling / bending✓ 1

Exam tip: In particle-model answers mention the vibration of the particles and the change in their separation.

26IGPC3-2-07r7-13Conduction, convection and radiation Paper CEasy4 marks

A marathon runner is given a thin shiny foil blanket at the finish line.

(a)

State the name of the process by which thermal energy can travel from the runner's skin to the surroundings through a vacuum.

(1)
(b)

Explain why the shiny surface of the foil helps to keep the runner warm.

(2)
(c)

The blanket traps a layer of air next to the runner. Explain why this also reduces energy loss by conduction.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
(thermal / infrared) radiation✓ 1
Part (b)
a shiny surface is a poor emitter and a good reflector of infrared radiation, so it reflects energy back towards the runner✓ 1
so less thermal energy is lost from the runner✓ 1
Part (c)
air is a poor conductor (insulator)✓ 1

Exam tip: Shiny surfaces are poor emitters and poor absorbers of radiation; dull black surfaces are good at both.

27IGPC3-2-07r9-01Particle model and gas pressure Paper CEasy1 mark

A gas is sealed in a syringe. Which statement explains why the gas exerts a pressure on the walls of the syringe?

Show mark scheme
Marking pointMarkNotes
C is correct: random collisions of molecules with the walls each exert a small force, and the total force per unit area is the pressure. A is wrong because pressure acts equally in all directions, not just downwards, so weight is not the cause. B is wrong because the pressure comes from collisions with the walls, not from repulsion between molecules. D is wrong because molecules bounce off the walls rather than sticking to them.✓ 1Answer C

Answer: C

Common error: Do not explain gas pressure using the weight of the molecules; gas pressure acts equally on all walls because of collisions.

28IGPC3-2-07r9-02Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

Hand sanitiser gel feels cold on the skin as the alcohol in it evaporates. Which statement explains why evaporation makes the remaining liquid colder?

Show mark scheme
Marking pointMarkNotes
B is correct: only the fastest, most energetic molecules can leave the surface, so the average kinetic energy of the molecules left behind falls and the liquid cools. A is wrong because it is the fastest, not the slowest, molecules that escape, and this would cause warming. C is wrong because evaporation occurs below the boiling point. D is wrong because molecules do not change size when they change state.✓ 1Answer B

Answer: B

Common error: Evaporation happens at any temperature below the boiling point and only at the surface; do not confuse it with boiling.

29IGPC3-2-07r9-08Conduction, convection and radiation Paper CEasy1 mark

In a hospital ward, a patient's warm breath rises in the cooler air of the room. Why does warm air rise?

Show mark scheme
Marking pointMarkNotes
B is correct: warm air expands, so its density is lower than that of the surrounding cooler air and it floats upwards. A is wrong because molecules keep the same mass when heated; it is the spacing between them that increases. C is wrong because warm air is less dense, not more dense. D is wrong because heat is energy, not a substance, and energy can move in any direction.✓ 1Answer B

Answer: B

Common error: Say 'less dense', not 'lighter'. The mass of each molecule does not change when the air is heated.

30IGPC3-2-07r9-09Particle model and gas pressure Paper CEasy5 marks

A sealed glass vial contains air. A technician in a hospital laboratory places the vial in a beaker of warm water.

(a)

State two differences between the arrangement and movement of molecules in a gas and in a liquid.

(2)
(b)

Explain, in terms of molecules, why the pressure of the air in the vial increases as the vial warms.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
In a gas the molecules are far apart, but in a liquid they are close together✓ 1
Gas molecules move randomly and freely at high speed, but liquid molecules only slide past one another✓ 1
Part (b)
The molecules gain kinetic energy and move faster✓ 1
They hit the walls of the vial harder and more often✓ 1
The force per unit area on the walls is greater, so the pressure increases✓ 1

Exam tip: In 'explain' answers about gas pressure, link speed, then collisions, then force per unit area.

31IGPC3-2-07r9-10Conduction, convection and radiation Paper CEasy6 marks

A solar hot-water panel on a roof has a flat metal absorber plate with pipes attached. Water flows through the pipes to a storage tank above the panel.

(a)

State the name of the process by which energy travels from the Sun to the panel through space.

(1)
(b)

The absorber plate is painted matt black. Explain why.

(2)
(c)

Hot water from the panel rises to the tank. Explain, in terms of particles, why the hot water rises.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
(thermal) radiation✓ 1
Part (b)
Matt black surfaces are good absorbers of radiation✓ 1
so more energy is absorbed and the plate warms up faster✓ 1
Part (c)
The water particles move faster and spread further apart when heated✓ 1
so the hot water is less dense than the cold water✓ 1
Hot water rises and cooler, denser water sinks to take its place, forming a convection current✓ 1

Common error: Do not say that 'heat rises'; say that hot fluid is less dense, so it rises.

32IGPC3-2-07r12-01Particle model and gas pressure Paper CEasy1 mark

Before a flight, a ground engineer pumps more air into an aircraft tyre. The volume of the tyre and the temperature of the air stay the same. Which statement explains why the pressure of the air in the tyre increases?

Show mark scheme
Marking pointMarkNotes
The correct option is the one about more particles: the temperature is constant, so the particles' speed and the force of each collision are unchanged; more particles in the same volume means more collisions per second, giving a greater force per unit area. The option about faster particles is wrong because the temperature has not risen. The option about larger particles is wrong because particles do not change size. The option about a greater force per particle is wrong because the average speed is unchanged.✓ 1Answer B

Answer: B

Common error: Pressure depends on both how often and how hard particles collide with the wall. Check which of these quantities actually changes.

33IGPC3-2-07r12-02Conduction, convection and radiation Paper CEasy1 mark

A farmer fits a solar water heater to a barn roof. The tank must absorb as much of the Sun's thermal radiation as possible. Which surface finish should the outside of the tank have?

Show mark scheme
Marking pointMarkNotes
Matt black is the best absorber (and best emitter) of thermal radiation. Matt white and shiny silver reflect most of the radiation, so they absorb poorly. Shiny black is better than white but still reflects more than a matt surface, so it is not the best choice.✓ 1Answer D

Answer: D

Exam tip: Dull, dark surfaces are the best absorbers and best emitters; light, shiny surfaces are the best reflectors.

34IGPC3-2-07r12-06Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

Steel beams in an aircraft hangar roof are fitted with small expansion gaps. Which statement explains why a steel beam gets longer when it is heated?

Show mark scheme
Marking pointMarkNotes
Heating gives atoms in a solid more kinetic energy, so they vibrate more widely and the average separation grows. The atoms do not change size, no atoms are created, and atoms in a solid stay in fixed positions; moving in straight lines between collisions describes a gas.✓ 1Answer D

Answer: D

Common error: Expansion is caused by a change in the spacing between particles, not by a change in the particles themselves.

35IGPC3-2-07r12-09Particle model and gas pressure Paper CEasy7 marks

An aircraft carries a sealed cylinder of oxygen gas. The volume of the cylinder is fixed.

(a)

State two differences between the arrangement or motion of particles in a gas and in a solid.

(2)
(b)

Explain how the gas particles cause a pressure on the wall of the cylinder.

(2)
(c)

The cylinder is moved from a cold cargo hold at 5 °C to a warm cabin at 25 °C. Explain why the pressure of the gas increases.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Gas particles are far apart but particles in a solid are close together✓ 1
Gas particles move randomly and freely but particles in a solid only vibrate about fixed positions (or are in a regular pattern)✓ 1
Part (b)
Particles collide with the wall (and rebound)✓ 1
Each collision exerts a force on the wall; many collisions every second give a force per unit area, which is the pressure✓ 1
Part (c)
The particles gain kinetic energy and move faster✓ 1
They hit the wall more often✓ 1
and hit with a greater force each time, so the pressure is greater✓ 1

Common error: Do not say that the particles 'expand' when heated: they speed up, and it is the collisions with the wall that change.

36IGPC3-2-07r12-14Thermal expansion, specific heat capacity and changes of state Paper CEasy5 marks

A beekeeper melts beeswax and lets it cool. The temperature of a sample is recorded every 2 minutes.
Time / min: 0, 2, 4, 6, 8, 10, 12, 14
Temperature / °C: 78, 70, 64, 62, 62, 62, 57, 49

(a)

State the freezing point of the beeswax.

(1)
(b)

State the time interval during which the beeswax is freezing.

(1)
(c)

Explain, using the particle model, why the temperature stays constant while the wax freezes.

(2)
(d)

Describe the change in the movement of the wax particles when it changes from liquid to solid.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
62 °C✓ 1
Part (b)
from about 6 min to about 10 min (while the temperature stays at 62 °C)✓ 1
Part (c)
Energy is still being lost to the surroundings, but the particles are attracting and forming bonds/fixed positions✓ 1
so the energy transferred out comes from the potential energy of the particles, not their kinetic energy, so the temperature does not fall✓ 1
Part (d)
Particles move freely past each other in the liquid but only vibrate about fixed positions in the solid✓ 1

Common error: A flat section on a cooling table means a change of state, not that cooling has stopped.

37IGPC3-2-05r13-01Particle model and gas pressure Paper CEasy1 mark

A cyclist leaves a sealed bicycle tyre in strong sunshine and the temperature of the air inside rises. Which statement correctly explains why the air pressure increases?

Show mark scheme
Marking pointMarkNotes
Correct: higher temperature means greater average kinetic energy, so more frequent and harder collisions with the wall. Particle size does not change on heating; no particles are added to a sealed tyre; gas particles do not exert significant repulsive forces on the wall.✓ 1Answer B

Answer: B

Common error: Particles themselves do not expand when a gas is heated; it is their speed that changes.

38IGPC3-2-05r13-02Conduction, convection and radiation Paper CEasy1 mark

An archaeologist warms one end of a bronze rod in a flame. Which statement best describes how thermal energy is conducted along the rod?

Show mark scheme
Marking pointMarkNotes
Correct: free electrons in a metal carry energy rapidly by collisions. Atoms in a solid only vibrate about fixed positions; convection needs a fluid; infrared is emitted at surfaces and does not pass through metal.✓ 1Answer C

Answer: C

Exam tip: Name the particles responsible when you explain conduction in a metal: free electrons as well as vibrating atoms.

39IGPC3-2-05r13-04Particle model and gas pressure Paper CEasy1 mark

Warm air rises above the servers in a data hall. What is the reason that warm air rises?

Show mark scheme
Marking pointMarkNotes
Correct: heating makes the air expand, so the same mass occupies more volume and the density falls below that of the surroundings. The mass of each particle does not change; there is no attraction to the ceiling; thermal energy is not a substance with weight.✓ 1Answer A

Answer: A

Common error: Warm air rises because its density decreases, not because the particles themselves get lighter.

40IGPC3-2-05r13-09Particle model and gas pressure Paper CEasy5 marks

Replica bronze axe heads are made by pouring molten bronze into moulds, where it cools and solidifies.

(a)

Describe the arrangement and motion of the particles in molten bronze.

(2)
(b)

State the arrangement and motion of the particles in solid bronze.

(2)
(c)

Explain why the axe head shrinks slightly as it cools to room temperature.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles close together but in an irregular arrangement✓ 1
particles move around and slide past each other✓ 1
Part (b)
particles in a regular pattern / fixed positions✓ 1
particles vibrate about fixed positions✓ 1
Part (c)
particles vibrate less / with smaller amplitude so average separation decreases✓ 1

Exam tip: 'Describe' answers for liquids need both arrangement and motion; write at least one comment on each.

41IGPC3-2-05r13-11Particle model and gas pressure Paper CEasy4 marks

A cyclist checks the air pressure in a sealed bicycle tyre.

(a)

Explain, in terms of particles, how the air inside the tyre exerts a pressure on the inner wall.

(2)
(b)

The tyre is warmed on a hot day. Explain why the pressure rises. The volume of the tyre stays constant.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles collide with the wall and rebound, each collision exerting a force✓ 1
very large number of collisions gives a (steady) force per unit area✓ 1
Part (b)
particles move faster (greater average kinetic energy) so collide more frequently with the wall✓ 1
each collision exerts a greater force✓ 1

Exam tip: Pressure explanations earn marks for both the frequency and the force of collisions.

42IGPC3-2-07r13-01Particle model and gas pressure Paper CEasy1 mark

A football is inflated by pumping extra air into it. The temperature of the air inside stays the same and the volume of the ball does not change. Why does the air pressure inside the ball increase?

Show mark scheme
Marking pointMarkNotes
B is correct: with the same temperature and volume, more molecules means more collisions per second per unit area, so a larger average force per unit area. A is wrong because the temperature is unchanged, so the average speed is unchanged. C is wrong because molecules do not change size. D is wrong because gas pressure acts equally in all directions and is not due to the weight of the air.✓ 1Answer B

Answer: B

Common error: Gas pressure does not come from molecules pushing down under gravity. It comes from molecules colliding with the container walls in all directions.

43IGPC3-2-07r13-02Particle model and gas pressure Paper CEasy1 mark

On a hot day a steel railway track becomes slightly longer. Which statement explains this in terms of particles?

Show mark scheme
Marking pointMarkNotes
C is correct: heating makes the particles vibrate more strongly about fixed positions, so the average separation, and so the length, increases. A is wrong because the particles do not change size. B is wrong because the number of particles stays the same. D describes a liquid, not a solid that is still solid.✓ 1Answer C

Answer: C

Common error: When a solid expands the particles do not get bigger. It is the spaces between them that grow.

44IGPC3-2-07r13-05Conduction, convection and radiation Paper CEasy1 mark

A railway company wants the roof of a carriage to absorb as little thermal radiation from the Sun as possible. Which roof surface is best?

Show mark scheme
Marking pointMarkNotes
A is correct: shiny, light surfaces are the poorest absorbers (and the best reflectors) of thermal radiation. B is the best absorber. C absorbs less than black but still more than a shiny surface, because a dull surface absorbs more than a shiny one. D absorbs more than white or silver.✓ 1Answer A

Answer: A

Exam tip: Compare both the colour and the finish: dull black surfaces are the best absorbers and emitters, shiny light surfaces are the best reflectors.

45IGPC3-2-07r13-09Particle model and gas pressure Paper CEasy6 marks

A football is inflated by a pump at a sports club.

(a)

Describe how the air molecules inside the football produce a pressure on the inside surface of the ball.

(2)
(b)

The football is inflated at constant temperature. Explain why pumping in more air increases the pressure.

(2)
(c)

The football is left in a hot car and its temperature rises. Explain, in terms of molecules, why the pressure increases.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The molecules move randomly and collide with the surface of the ball✓ 1
Each collision exerts a force on the surface; the total force per unit area is the pressure✓ 1
Part (b)
More molecules in the same volume✓ 1
so more collisions each second per unit area, so greater force per unit area✓ 1
Part (c)
The molecules move faster / have greater average kinetic energy✓ 1
so they collide with the surface more often and with greater force✓ 1

Common error: Do not say that molecules 'expand' or 'get heavier' when gas is heated. Refer to speed, number of collisions and force of collisions.

46IGPC3-2-07r13-10Thermal expansion, specific heat capacity and changes of state Paper CEasy5 marks

Long sections of steel railway track are laid in a country with hot summers and cold winters.

(a)

State why small gaps are left between sections of track.

(1)
(b)

Explain, in terms of particles, why a steel rail gets longer when its temperature rises.

(2)
(c)

In very hot countries some rails are painted white. Suggest why this helps.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
To allow the track to expand when it gets hot (so it does not buckle)✓ 1
Part (b)
The particles vibrate with greater amplitude✓ 1
so the average separation between the particles increases✓ 1
Part (c)
White (light) surfaces absorb less thermal radiation from the Sun / reflect more✓ 1
so the temperature rise is smaller and there is less expansion✓ 1

Exam tip: In 'explain' questions about expansion, link the change in temperature to the particle motion, and then to the change in size.

47IGPC3-2-05r14-01Particle model and gas pressure Paper CEasy1 mark

A sealed steel cylinder contains air. Which statement correctly explains how the air exerts a pressure on the walls of the cylinder?

Show mark scheme
Marking pointMarkNotes
A is correct: gas pressure is the force per unit area from huge numbers of particle collisions with the wall. B is wrong because there is no repulsion between the particles and the wall that causes pressure. C is wrong because weight is negligible and pressure acts equally on all walls, not just downwards. D is wrong because collisions between particles do not push the walls; only collisions with the wall do.✓ 1Answer A

Answer: A

Common error: Pressure in a gas acts equally in all directions, so it cannot be explained by the weight of the particles.

48IGPC3-2-05r14-03Conduction, convection and radiation Paper CEasy1 mark

A cyclist notices that the metal handlebar feels much colder to the touch than the plastic saddle, although both have been outside overnight and are at the same temperature. What is the reason?

Show mark scheme
Marking pointMarkNotes
A is correct: the metal conducts thermal energy away from the warm hand faster, so the hand cools more quickly. B is wrong because both are at the same temperature. C is wrong because the effect is caused by conduction on contact, not radiation. D is wrong because 'cold' is not something stored in a material.✓ 1Answer A

Answer: A

Common error: What the skin senses is how fast thermal energy leaves it, not the temperature of the object touched.

49IGPC3-2-05r14-08Conduction, convection and radiation Paper CEasy1 mark

Four identical metal cans are filled with the same volume of hot water at 80 °C and left in a cool room. The cans have different outer surfaces. Which can's water cools fastest?

Show mark scheme
Marking pointMarkNotes
Dull black surfaces are the best emitters of infrared radiation, so the water in the dull black can loses thermal energy fastest. Shiny surfaces are poor emitters, and a dull white surface emits less than dull black.✓ 1Answer D

Answer: D

Exam tip: Dull black surfaces are good emitters and good absorbers; shiny light surfaces are poor at both.

50IGPC3-2-05r14-09Conduction, convection and radiation Paper CEasy6 marks

Heat sinks in a data centre are made of copper and have thin vertical fins. A processor transfers thermal energy to the copper base of its heat sink, and warm air then leaves the fins.

(a)

State the name of the process by which thermal energy passes through the solid copper base.

(1)
(b)

Explain, in terms of particles, why copper is a better thermal conductor than plastic.

(2)
(c)

Explain why the air next to the fins rises.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
conduction✓ 1
Part (b)
copper has free (delocalised) electrons✓ 1
these electrons move through the metal and transfer energy to other particles more quickly than vibrations alone✓ 1
Part (c)
the air is heated by the fins and expands (particles move further apart)✓ 1
so its density decreases✓ 1
it is less dense than the surrounding cooler air, so it is pushed upwards✓ 1

Exam tip: For convection, always link the temperature change to the change in density before saying that the fluid rises.

51IGPC3-2-05r14-15Thermal expansion, specific heat capacity and changes of state Paper CEasy4 marks

Steel rails on a cycle-track bridge are laid in sections with small gaps between the ends of neighbouring sections.

(a)

State what happens to the length of a steel rail when its temperature rises.

(1)
(b)

Explain, in terms of particles, why this happens.

(2)
(c)

Suggest why gaps are left between sections of rail.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
it increases (expands)✓ 1
Part (b)
the particles vibrate with greater amplitude✓ 1
so the average separation of the particles increases✓ 1
Part (c)
to allow room for the rails to expand in hot weather so they do not buckle✓ 1

Exam tip: Explain expansion using the particle spacing; the particles themselves stay the same size.

52IGPC3-2-07r14-01Particle model and gas pressure Paper CEasy1 mark

A long steel rail on a library's rolling-ladder track is warmed by sunlight through a window and becomes very slightly longer. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
B is correct: heating increases the vibrational energy of the particles, so they vibrate with larger amplitude and the average spacing grows, making the rail longer. A is wrong because the particles do not change size. C is wrong because no particles are added. D is wrong because the rail is still a solid; particles only vibrate about fixed positions.✓ 1Answer B

Answer: B

Common error: Expansion is caused by a change in particle spacing, not by a change in the size or number of particles.

53IGPC3-2-07r14-02Conduction, convection and radiation Paper CEasy1 mark

At the edge of a swimming pool a metal handrail and a wooden bench have both been in the shade for hours and are at the same temperature, 20 °C. The metal feels much colder when touched. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
D is correct: metal is a good thermal conductor, so energy flows quickly from the warmer hand into the metal and the hand cools. A is wrong because both are at 20 °C. B is wrong because wood is a poor conductor, so little energy leaves the hand. C is wrong because cold is not radiated; energy flows from hot to cold.✓ 1Answer D

Answer: D

Common error: Different materials at the same temperature can feel different because they conduct thermal energy at different rates.

54IGPC3-2-07r14-06Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

A conservator in a library uses a wax that has a melting point of 54 °C and a boiling point of 280 °C. Which description applies to the particles of the wax at 120 °C?

Show mark scheme
Marking pointMarkNotes
C is correct: 120 °C is between the melting point and the boiling point, so the wax is a liquid, whose particles are close together and free to move past each other. A and D describe a solid, which the wax is only below 54 °C. B describes a gas, which the wax becomes only above 280 °C.✓ 1Answer C

Answer: C

Exam tip: Compare the temperature with both the melting point and the boiling point to decide the state.

55IGPC3-2-07r14-09Thermal expansion, specific heat capacity and changes of state Paper CEasy4 marks

In a library, a steel hot-water heating pipe runs along a wall for 30 m. When hot water is first pumped through, the pipe becomes longer.

(a)

Explain, in terms of particles, why the pipe becomes longer.

(2)
(b)

Suggest one way in which the design of the pipe system allows for this change.

(1)
(c)

State what happens to the mass of the pipe when it is heated.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The particles vibrate with larger amplitude as they gain energy✓ 1
so the average separation of the particles increases✓ 1
Part (b)
Include an expansion loop / bend / sliding supports so the pipe can lengthen without being damaged✓ 1
Part (c)
It stays the same / does not change✓ 1

Exam tip: In an expansion explanation, mention vibration amplitude and average separation; do not say the particles get bigger.

56IGPC3-2-07r14-16Particle model and gas pressure Paper CEasy4 marks

An inflatable swimming ring is filled with air and sealed.

(a)

Explain, using molecules, how the air produces a pressure on the inside wall of the ring.

(2)
(b)

The ring is left in the Sun and its temperature rises. State what happens to the pressure of the air and explain your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Air molecules move randomly and collide with the wall✓ 1
Each collision exerts a force on the wall; the many collisions give an average force per unit area, which is the pressure✓ 1
Part (b)
The pressure increases✓ 1
The molecules move faster, so they hit the wall more often and with greater force✓ 1

Common error: Do not say the molecules expand; their speed increases, not their size.

57IGPC3-2-07r15-01Particle model and gas pressure Paper CEasy1 mark

A sealed steel container of fixed volume holds a gas. The gas is heated. Which statement correctly explains why the gas pressure increases?

Show mark scheme
Marking pointMarkNotes
B is correct: a higher temperature means a higher average speed, giving more frequent and harder collisions with the walls. A is wrong because the particles themselves do not expand. C is wrong because the mass of each particle does not change. D is wrong because the volume is fixed, so the particles cannot move further apart on average.✓ 1Answer B

Answer: B

Exam tip: For gas pressure, mention both how often the particles hit the walls and how hard they hit.

58IGPC3-2-07r15-02Conduction, convection and radiation Paper CEasy1 mark

A box of fresh fish is packed in crushed ice. While the ice is melting, its temperature stays at 0 °C. Why does the temperature not rise?

Show mark scheme
Marking pointMarkNotes
B is correct: during melting the energy goes into overcoming the attractions between particles, so temperature does not change. A is wrong because the fish are warmer than the ice and energy does flow. C is wrong because particles in ice still vibrate. D is wrong because the energy melts the ice and is not simply passed to the box.✓ 1Answer B

Answer: B

Common error: A flat part of a heating curve does not mean that no energy is being supplied.

59IGPC3-2-07r15-09Particle model and gas pressure Paper CEasy4 marks

A fish-processing plant stores compressed air in a sealed steel tank.

(a)

Describe how the particles of a gas produce pressure on the walls of the tank.

(2)
(b)

The tank is left in sunshine and its temperature rises. The volume does not change. State what happens to the gas pressure and explain your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The particles move randomly and collide with the walls✓ 1
Each collision exerts a force on the wall, and the force per unit area is the pressure✓ 1
Part (b)
The pressure increases✓ 1
The particles move faster on average, so they hit the walls more often and with greater force✓ 1

Exam tip: Pressure is caused by collisions of particles with the walls; always refer to force and area or to the frequency of collisions.

60IGPC3-2-07r15-10Conduction, convection and radiation Paper CEasy4 marks

A winter jacket is filled with soft fibres that trap air.

(a)

Explain why the trapped air makes the jacket a good insulator.

(3)
(b)

The inside lining of the jacket is shiny silver. Suggest why this helps to keep the wearer warm.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Air is a poor conductor✓ 1
The air is trapped in small spaces, so it cannot circulate freely✓ 1
So convection currents are reduced or prevented✓ 1
Part (b)
It reflects infrared radiation back towards the body, and it is a poor emitter of radiation✓ 1

Common error: Do not say that the jacket 'produces' or 'keeps in' cold or heat; explain which method of thermal transfer is reduced.

61IGPC3-2-07r15-11Conduction, convection and radiation Paper CEasy4 marks

In a fish-smoking oven, a heater at the bottom warms the air.

airheateroven
Diagram not to scale
(a)

Name the process by which the warm air carries thermal energy upwards through the oven.

(1)
(b)

Explain why the warm air rises.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Convection✓ 1
Part (b)
The particles move faster and spread further apart✓ 1
So the density of the warm air decreases✓ 1
The warm air is less dense than the cooler air around it, so it rises✓ 1

Exam tip: Explain rising warm air with particle spacing, then density, then the comparison with the surroundings.

62IGPC3-2-06r16-01Conduction, convection and radiation Paper CEasy1 mark

A cyclist grips a metal handlebar and a plastic grip. Both have been outside overnight at the same temperature of 5 °C, but the metal feels much colder. What is the reason?

Show mark scheme
Marking pointMarkNotes
B is correct: metal is a good conductor, so energy leaves the warm hand quickly and the surface feels cold. A is wrong because both are at 5 °C. C reverses the conductivities. D is wrong because cold is not a substance that flows; energy flows from the hand to the metal.✓ 1Answer B

Answer: B

Common error: Cold does not travel; energy always transfers from the hotter object to the colder one.

63IGPC3-2-06r16-02Particle model and gas pressure Paper CEasy1 mark

The air in a bicycle tyre is heated by the sun on a hot day, and its pressure rises. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
D is correct: higher temperature means a greater average speed, so collisions with the wall are more frequent and each exerts a larger force. A is wrong because the particles themselves do not expand. B is wrong because their mass does not change. C is wrong because the tyre is sealed.✓ 1Answer D

Answer: D

Exam tip: Link pressure to the force and frequency of particle collisions with the wall.

64IGPC3-2-06r16-04Particle model and gas pressure Paper CEasy1 mark

Archaeologists find a sealed ancient jar containing air. Which row describes the air particles in the jar?

Show mark scheme
Marking pointMarkNotes
C is correct for a gas. A describes a solid and B describes a liquid. D mixes the large spacing of a gas with the fixed positions of a solid, which is not a state of matter.✓ 1Answer C

Answer: C

Common error: Gas particles are far apart and move freely; they do not vibrate about fixed positions.

65IGPC3-2-06r16-07Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

Small gaps are left between sections of steel railway track. Why?

Show mark scheme
Marking pointMarkNotes
A is correct: steel expands on heating, and the gaps allow room for that expansion. B is not the engineering reason. C is false, as solids expand when heated. D is wrong because the gaps are too small for air cooling.✓ 1Answer A

Answer: A

Exam tip: Thermal expansion in solids is small but large forces result if it is prevented.

66IGPC3-2-06r16-11Thermal expansion, specific heat capacity and changes of state Paper CEasy5 marks

Archaeologists heat a sample of ancient beeswax in a laboratory.

(a)

Describe the arrangement and motion of the particles in solid wax.

(2)
(b)

The wax melts at 62 °C. State the temperature at which the wax changes from liquid to solid when cooled slowly.

(1)
(c)

The wax is heated from 20 °C to 50 °C. State and explain what happens to the speed of vibration of the particles.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles are close together in a regular pattern✓ 1
particles vibrate about fixed positions✓ 1
Part (b)
62 °C (the freezing point equals the melting point)✓ 1
Part (c)
the vibrations become faster (or larger)✓ 1
because the average kinetic energy of the particles increases as the temperature rises✓ 1

Common error: Melting point and freezing point of a pure substance are the same temperature.

67IGPC3-2-07r17-01Particle model and gas pressure Paper CEasy1 mark

A sealed syringe contains air. The plunger is pushed in so that the volume of the air is halved. The temperature of the air stays the same.
Which statement explains why the pressure of the air increases?

Show mark scheme
Marking pointMarkNotes
B is correct: the same particles, moving at the same average speed, now strike each unit area of wall more often, so the force per unit area rises. A is wrong because the temperature is unchanged, so the average speed is unchanged. C is wrong because the force in each collision depends on the speed, which has not changed. D is wrong because the syringe is sealed, so the number of particles is fixed.✓ 1Answer B

Answer: B

Common error: Pressure changes in a gas come from the rate of collisions with the walls and the size of each collision; identify which of these changes before choosing.

68IGPC3-2-07r17-02Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

Steel railway lines are laid with small gaps between the sections. In hot weather the sections get longer.
Which statement describes what happens to the particles in the steel when its temperature rises?

Show mark scheme
Marking pointMarkNotes
B is correct: a higher temperature gives the particles in the solid more energy, so they vibrate more and push slightly further apart, so the solid expands. A is wrong because the particles do not change size. C is wrong because the steel remains a solid and the particles stay in fixed positions. D is wrong because the number of particles is unchanged.✓ 1Answer B

Answer: B

Common error: Expansion is a change in the spacing of the particles, not in the particles themselves.

69IGPC3-2-07r17-05Conduction, convection and radiation Paper CEasy1 mark

A tank of water is heated by a heater at the bottom. The water at the bottom warms up and rises through the cooler water above it.
Why does the warmed water rise?

waterheater
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
B is correct: the warmed water expands, so the same mass occupies a larger volume and its density is less than that of the cooler water; it is pushed up and a convection current forms. A is wrong because the particles do not change mass. C is wrong because the particles do not themselves expand. D is wrong because the question is about the movement of the water, which is convection, not conduction.✓ 1Answer B

Answer: B

Common error: State that the fluid becomes less dense; never say that the particles get lighter.

70IGPC3-2-07r17-09Thermal expansion, specific heat capacity and changes of state Paper CEasy5 marks

Steel railway lines are made in sections and laid end to end with small gaps between the sections.

(a)

Explain why the gaps are left.

(2)
(b)

Describe what happens to the particles in the steel when it is heated.

(2)
(c)

State one other structure where a gap or joint is needed to allow for expansion.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The steel expands when its temperature rises (in hot weather)✓ 1
Without gaps the rails would push against each other and buckle / bend✓ 1
Part (b)
The particles vibrate with greater energy / larger amplitude✓ 1
The average separation of the particles increases (the particles themselves do not change size)✓ 1
Part (c)
Any suitable example, e.g. bridge, concrete road, long metal pipe✓ 1

Exam tip: When explaining expansion, say that the particles move further apart on average; do not say that the particles get bigger.

71IGPC3-2-07r17-10Conduction, convection and radiation Paper CEasy6 marks

A space telescope orbits the Sun in a vacuum. It has a shiny silver sunshield facing the Sun. The side of the telescope that faces away from the Sun is painted matt black.

(a)

Explain why the energy from the Sun reaches the telescope by radiation and not by conduction or convection.

(2)
(b)

Explain why the sunshield is shiny.

(2)
(c)

Suggest why the side facing away from the Sun is matt black.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Space is a vacuum / there are no particles between the Sun and the telescope✓ 1
Conduction and convection need particles, but radiation (infrared and other electromagnetic waves) can travel through a vacuum✓ 1
Part (b)
A shiny surface is a poor absorber (a good reflector) of radiation✓ 1
so less energy is absorbed and the telescope warms up less✓ 1
Part (c)
A matt black surface is a good emitter of (infrared) radiation✓ 1
so energy is radiated away quickly and the detectors are kept cold✓ 1

Exam tip: Link the surface to both absorption and emission: shiny surfaces are poor at both, matt black surfaces are good at both.

72IGPC3-2-06r25-01Particle model and gas pressure Paper CEasy1 mark

A sealed steel container holds a sample of gas collected from a volcanic vent. The container is heated and its volume does not change. Which statement explains why the gas pressure increases?

Show mark scheme
Marking pointMarkNotes
Heating makes the particles move faster, so they collide with the walls more frequently and each collision exerts a larger force, increasing the pressure. The particles themselves do not expand, the number of particles in a sealed container cannot change, and heating does not change the mass of a particle.✓ 1Answer C

Answer: C

Common error: Gas pressure depends on the frequency and force of collisions with the walls; do not say that particles 'expand' or 'become heavier' when heated.

73IGPC3-2-06r25-02Conduction, convection and radiation Paper CEasy1 mark

A theatre lamp has a glass bulb containing an almost perfect vacuum. Energy from the hot filament reaches the glass wall mainly by which process?

Show mark scheme
Marking pointMarkNotes
Radiation (infrared waves) needs no medium, so it can cross a vacuum. Conduction and convection both need particles to carry energy, and there are almost none in the vacuum. Evaporation is a change of state, not a way of crossing a gap.✓ 1Answer A

Answer: A

Exam tip: Radiation is the only method of thermal energy transfer that can cross a vacuum.

74IGPC3-2-06r25-08Conduction, convection and radiation Paper CEasy1 mark

In a theatre, warm air from the stage lights collects near the ceiling. Which statement gives the reason?

Show mark scheme
Marking pointMarkNotes
Warm air expands, so a given volume contains fewer particles and the air is less dense than the surrounding cooler air. It is pushed upwards by the denser air, which is convection. The particles themselves are not lighter, there is no attraction to the ceiling, and heating does not remove particles from the room; it is the expansion that lowers the density.✓ 1Answer B

Answer: B

Common error: Hot air rises because it is less dense, not because the individual particles have less mass.

75IGPC3-2-06r25-09Conduction, convection and radiation Paper CEasy4 marks

A technician is adjusting a theatre stage lamp that has been switched on for some time. The metal handle of the lamp is attached to a hot metal housing.

(a)

State the name of the process by which energy is transferred along the metal handle to the technician's hand.

(1)
(b)

The air above the lamp becomes warm. Describe how the warm air transfers energy to the ceiling.

(2)
(c)

The lamp casing can be painted either dull black or shiny silver. State which paint causes the casing to emit more thermal radiation when both casings are at the same temperature.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
conduction✓ 1
Part (b)
the warm air expands and becomes less dense than the surrounding air✓ 1
so it rises, while cooler, denser air moves in to replace it (convection current)✓ 1
Part (c)
dull black✓ 1

Exam tip: Name the process first, then describe what the particles or the air do; 'heat rises' alone does not earn marks.

76IGPC3-2-06r25-10Particle model and gas pressure Paper CEasy5 marks

A volcanologist collects a sample of gas from a vent in a rigid sealed metal flask.

(a)

State how the particles of a gas move.

(1)
(b)

Explain how the gas in the flask produces a pressure on its walls.

(2)
(c)

The flask is placed in sunlight and its temperature rises. Explain why the pressure of the gas increases.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
randomly / in all directions, at high speed✓ 1
Part (b)
gas particles collide with the walls of the flask✓ 1
each collision exerts a force on the wall; the force per unit area is the pressure✓ 1
Part (c)
the particles move faster (greater average speed)✓ 1
so collisions with the walls are more frequent and each exerts a greater force✓ 1

Common error: Link both frequency and force of collisions to the pressure change; one of the two alone gives only part of the explanation.

77IGPC3-2-09r33-01Particle model and gas pressure Paper CEasy1 mark

A sealed glass sampling flask of fixed volume holds air. It is left in strong sunshine on the deck of a research vessel and the air temperature rises. Which statement explains why the gas pressure rises?

Show mark scheme
Marking pointMarkNotes
Faster particles collide with the walls more frequently and each collision exerts a larger force, so the pressure rises. The particles do not expand when heated, the sealed flask lets no extra particles in, and the mass of each particle does not change with temperature.✓ 1Answer B

Answer: B

Common error: Particles themselves do not expand or change mass when a gas is heated; it is their speed and spacing that change.

78IGPC3-2-09r33-02Conduction, convection and radiation Paper CEasy1 mark

On a cold morning a steel handrail and a wooden handrail on a footbridge are at the same temperature. The steel feels much colder to the touch. What is the reason?

Show mark scheme
Marking pointMarkNotes
Steel is a much better thermal conductor, so energy leaves the warm hand quickly and the skin cools. Both rails are at the same temperature, cold is not a substance that flows, and radiation is not the cause of the different feel.✓ 1Answer D

Answer: D

Common error: Objects at the same temperature can feel different because of how fast they conduct energy, not because of different temperatures.

79IGPC3-2-09r33-07Conduction, convection and radiation Paper CEasy1 mark

Which finish on a flat roof would keep a building coolest in a hot, sunny climate?

Show mark scheme
Marking pointMarkNotes
A shiny white surface is a poor absorber and a good reflector of infrared and visible radiation from the Sun, so the roof heats up least. Matt black, matt dark grey and shiny black all absorb far more of the incident radiation.✓ 1Answer A

Answer: A

Common error: Black surfaces are good emitters, but in strong sunshine they absorb far more energy than they lose, so they end up hotter.

80IGPC3-2-09r33-09Particle model and gas pressure Paper CEasy5 marks

A marine biology laboratory keeps tropical seawater in a tank. Some of the water evaporates and the vapour spreads through the room. The laboratory also stores air in a sealed steel cylinder of fixed volume.

(a)

State two differences between the particles in liquid water and the particles in water vapour.

(2)
(b)

The sealed cylinder is moved from a cold store to a warm room. Explain, in terms of particles, why the pressure of the air in the cylinder increases.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any one difference: particles in vapour are much further apart than in the liquid✓ 1
Any second different difference: particles in the vapour move freely in random directions, but in the liquid they slide past each other / forces between particles are negligible in the vapour but strong in the liquid / the vapour fills its container but the liquid has a surface✓ 1
Part (b)
The air particles gain kinetic energy / move faster✓ 1
They hit the walls of the cylinder more often✓ 1
Each collision exerts a greater force (greater change of momentum), so the force per unit area is larger✓ 1

Common error: Do not say that the particles expand or that the volume changes; the cylinder is sealed and rigid.

81IGPC3-2-09r33-12Conduction, convection and radiation Paper CEasy6 marks

An architect designs a house for a cold climate. The windows are double glazed, with a sealed layer of air between two sheets of glass.

glassglassairsealed layer
Diagram not to scale
(a)

Explain why the layer of air between the glass sheets reduces the transfer of energy by conduction.

(2)
(b)

A radiator warms the air near the floor of a room. Explain how a convection current is set up in the room.

(3)
(c)

State how thermal energy from the Sun reaches the roof of the house.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Air is a poor conductor (an insulator)✓ 1
The particles in a gas are far apart, so energy is passed on slowly by collisions✓ 1
Part (b)
The air near the radiator is heated, expands and becomes less dense✓ 1
The warm air rises✓ 1
Cooler, denser air sinks to take its place, creating a circulating current✓ 1
Part (c)
By infrared radiation (electromagnetic waves), which do not need a medium✓ 1

Common error: Do not say that 'heat rises'; say that warm fluid rises because it is less dense.

82IGPC3-2-08r36-01Particle model and gas pressure Paper CEasy1 mark

A sealed glass vial of medicine contains only air. The vial is moved from a cold fridge to a warm room. Its volume does not change. Which statement explains why the pressure of the air in the vial increases?

Show mark scheme
Marking pointMarkNotes
The warmer particles have more kinetic energy and move faster, so they collide with the walls more often and with greater force, which raises the pressure. Particles do not change in size when heated. The vial is sealed, so the number of particles stays the same. The mass of each particle does not change either.✓ 1Answer B

Answer: B

Common error: Heating a gas changes how fast its particles move, not their size, mass or number.

83IGPC3-2-08r36-02Conduction, convection and radiation Paper CEasy1 mark

An engineer touches a metal water pipe and a plastic water pipe. Both are at 10 °C but the metal pipe feels colder. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
Metal is a good conductor, so thermal energy flows from the warm hand into the metal quickly and the hand cools, which feels cold. Both pipes are at the same temperature. Cold is not a substance that flows; thermal energy flows from hot to cold. Plastic is a poor conductor, which is why it feels warmer.✓ 1Answer D

Answer: D

Exam tip: Things feel cold when they take thermal energy from your hand quickly; describe the energy flowing from the hand, not cold flowing in.

84IGPC3-2-08r36-04Conduction, convection and radiation Paper CEasy1 mark

A long steel water pipe is fitted with expansion joints so that it does not buckle on hot days. Which statement describes what happens in the steel when the pipe is heated?

Show mark scheme
Marking pointMarkNotes
Heating gives the atoms more energy so they vibrate more and the average spacing increases, which makes the pipe longer. The atoms themselves stay the same size. No atoms are added to the steel. The bonds do not become stronger on heating.✓ 1Answer A

Answer: A

Common error: Solids expand because the particles move further apart, not because the particles themselves get bigger.

85IGPC3-2-08r36-06Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

Which change would increase the rate at which water evaporates from the surface of an open reservoir?

Show mark scheme
Marking pointMarkNotes
Wind carries away water vapour from just above the surface, so more particles can escape and the rate increases. Lower temperature means fewer particles have enough energy to escape. A smaller surface area gives fewer escape routes. Higher humidity means the air already holds a lot of vapour, so evaporation is slower.✓ 1Answer C

Answer: C

Exam tip: Evaporation is faster when the liquid is warmer, has a bigger surface area, or when air movement removes vapour.

86IGPC3-2-08r36-09Particle model and gas pressure Paper CEasy7 marks

A pharmacist handles medicines in three different states: a solid tablet, a liquid syrup and a gas in an inhaler can.

(a)

Describe the arrangement and motion of the particles in the solid tablet.

(2)
(b)

Describe the arrangement and motion of the particles in the liquid syrup.

(2)
(c)

State the name of the change of state that happens when a liquid at its boiling point becomes a gas.

(1)
(d)

Explain why the gas in the inhaler can be compressed easily but the liquid syrup cannot.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles are close together in a fixed, regular pattern✓ 1
particles vibrate about fixed positions✓ 1
Part (b)
particles are close together or touching, but in no regular pattern✓ 1
particles move around and slide past each other✓ 1
Part (c)
boiling✓ 1
Part (d)
particles in a gas are far apart, so there is a lot of empty space to squash them into✓ 1
particles in a liquid are already touching, so there is almost no space to move them closer✓ 1

Exam tip: When asked about arrangement and motion, give both: how close the particles are and how they move.

87IGPC3-2-09r41-01Conduction, convection and radiation Paper CEasy1 mark

A volcanologist stands 6 m to one side of a glowing lava channel and feels her face grow warm. Which process mainly transfers thermal energy from the lava to her face?

Show mark scheme
Marking pointMarkNotes
Hot lava emits infrared radiation in all directions, and radiation needs no medium, so it reaches her face directly. Convection carries hot air upwards, not sideways towards her. Air is a very poor conductor, so conduction through it is tiny. Conduction through the ground would not reach her face.✓ 1Answer C

Answer: C

Exam tip: Warmth felt beside (not above) a hot object is mainly radiation, because convection currents rise.

88IGPC3-2-09r41-02Particle model and gas pressure Paper CEasy1 mark

A sealed canister of gas inside a theatre fog machine is warmed. Which statement explains why the gas pressure increases?

Show mark scheme
Marking pointMarkNotes
Warming gives the particles more kinetic energy, so each collision with the wall exerts a larger force and collisions happen more often. The particles themselves do not expand, their mass does not change, and the canister is sealed so the number of particles stays the same.✓ 1Answer B

Answer: B

Common error: Write that the particles move faster, not that the particles expand or get bigger.

89IGPC3-2-09r41-06Particle model and gas pressure Paper CEasy1 mark

The mercury in a ship's engine-room thermometer rises up the tube as it is warmed. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
Heating gives the particles more kinetic energy so they move further apart on average and the liquid takes up more volume. The particles themselves do not change size, the number of particles is fixed, and the mass of each particle does not change.✓ 1Answer D

Answer: D

Common error: Expansion is an increase in the spacing between particles, not in the size of the particles.

90IGPC3-2-09r41-09Conduction, convection and radiation Paper CEasy6 marks

A volcanologist studies a hot vent in a rocky crater floor.

(a)

State the name of the process by which thermal energy passes through the solid rock beneath her boots.

(1)
(b)

Hot gas rises from the vent. Explain why the gas rises.

(3)
(c)

Her protective suit has a shiny silver outer surface. Suggest why this helps keep her safe near the vent.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
conduction✓ 1
Part (b)
the gas particles gain energy, move faster and spread out, so the gas expands✓ 1
the hot gas has a lower density than the cooler air around it✓ 1
the denser cooler air pushes the less dense hot gas upwards✓ 1
Part (c)
shiny silver surfaces are poor absorbers of infrared radiation✓ 1
so most of the radiation from the vent is reflected and her suit stays cooler✓ 1

Exam tip: 'Explain' questions on convection need three links: expands, less dense, rises.

91IGPC3-2-09r41-13Conduction, convection and radiation Paper CEasy5 marks

Long steel pipes carry hot fluid along the deck of a ship.

(a)

State what happens to the length of a steel pipe when its temperature rises.

(1)
(b)

Explain this change in terms of the particles in the steel.

(2)
(c)

The pipes are fitted with U-shaped loops. Suggest why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the length increases (the pipe expands)✓ 1
Part (b)
the particles vibrate with greater amplitude (more energy)✓ 1
so the average separation between the particles increases✓ 1
Part (c)
the pipes expand when they become hot✓ 1
the loops take up the extra length so the pipes do not buckle or break✓ 1

Exam tip: Solids expand because the particles vibrate more and push slightly further apart, not because the particles get bigger.

92IGPC3-2-07r42-01Conduction, convection and radiation Paper CEasy1 mark

On a cold morning, a person rests a bare hand on a metal handrail and on a wooden handrail of the same pier. Both rails are at the same temperature as the air. The metal rail feels much colder. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
Both rails are at the same temperature, so the difference is in how quickly energy is conducted away from the skin. Metal is a good thermal conductor, so energy leaves the hand rapidly and the metal feels colder. The suggestion that the metal is colder is wrong because both are at air temperature. Cold is not a substance that flows; energy flows from hot to cold. Convection to the air does not explain the sensation at the point of contact.✓ 1Answer B

Answer: B

Common error: Cold is not a substance: say that energy flows out of the warmer object, not that cold flows in.

93IGPC3-2-07r42-02Conduction, convection and radiation Paper CEasy1 mark

A warehouse in a hot climate needs a roof surface that keeps the building as cool as possible by absorbing the least thermal radiation from the Sun. Which finish is best?

Show mark scheme
Marking pointMarkNotes
Shiny white surfaces reflect most of the incident radiation and are the poorest absorbers, so the roof heats least. Dull black and matt dark grey are good absorbers. Shiny black reflects better than dull black but still absorbs far more than white because dark colours absorb strongly.✓ 1Answer D

Answer: D

Exam tip: Link colour and finish to both absorption and emission: dull black is best at both, shiny white is worst at both.

94IGPC3-2-07r42-09Thermal expansion, specific heat capacity and changes of state Paper CEasy4 marks

A research vessel uses a liquid-in-glass thermometer to measure the temperature of seawater at different depths. The thermometer contains alcohol in a narrow glass tube.

alcoholnarrow glass tube
Simplified diagram
(a)

State what happens to the volume of the alcohol when the thermometer is placed in warm water.

(1)
(b)

Explain, in terms of particles, why the alcohol behaves in this way.

(2)
(c)

Suggest why the tube is made very narrow.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The volume increases / the alcohol expands✓ 1
Part (b)
The particles gain kinetic energy and vibrate or move faster✓ 1
so the average separation of the particles increases / the particles move further apart✓ 1
Part (c)
A small increase in volume gives a large change in the length of the liquid column, so the thermometer is more sensitive✓ 1

Common error: Do not write that the particles themselves expand; it is the spaces between them that grow.

95IGPC3-2-07r42-14Conduction, convection and radiation Paper CEasy4 marks

A research station on a polar coast uses a solar panel to heat water. The panel is a flat metal plate with water pipes attached, facing the Sun.

(a)

State the colour and finish of the surface of the plate that is best at absorbing radiation.

(1)
(b)

Name the method by which thermal energy reaches the panel from the Sun across space.

(1)
(c)

A shiny metal sheet is fixed behind the plate, on the side away from the Sun. Explain how this reduces energy loss from the panel.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Dull black (matt black)✓ 1
Part (b)
Radiation✓ 1
Part (c)
A shiny surface is a poor emitter of radiation✓ 1
and it reflects radiation back towards the plate✓ 1

Exam tip: Remember that good absorbers are also good emitters, so shiny surfaces are used to reduce losses.

96IGPC3-2-08r47-01Conduction, convection and radiation Paper CEasy1 mark

A small paddling pool is heated by an electric heater placed at the bottom. Which statement explains why the warmed water moves upwards?

Show mark scheme
Marking pointMarkNotes
Heating makes the water expand, so the same mass occupies a larger volume and the density falls; the denser cold water sinks and pushes the warm water up (convection). The particles do not change mass, so the idea that they become lighter is wrong. 'Heat rises' is not an explanation, since it is the less dense fluid that rises. Water is a poor conductor, and conduction is not what moves the water.✓ 1Answer B

Answer: B

Common error: Particles do not get lighter or bigger when heated; the spacing between them increases, so density falls.

97IGPC3-2-08r47-02Particle model and gas pressure Paper CEasy1 mark

A rigid, sealed box of air in a library store is moved from a cool room into a warm room. The pressure of the air in the box increases. What is the reason?

Show mark scheme
Marking pointMarkNotes
A higher temperature gives the particles more kinetic energy, so they travel faster and hit the walls more frequently, and each collision exerts a bigger force. The particles themselves do not expand. The box is sealed, so the number of particles cannot increase. Collisions with the walls are what cause pressure, so collisions between particles alone cannot explain it.✓ 1Answer D

Answer: D

Common error: Pressure comes from particle collisions with the container walls, not from collisions between particles.

98IGPC3-2-08r47-07Conduction, convection and radiation Paper CEasy1 mark

On a cold day a technician touches a steel wind-turbine tower and a wooden door at the base. Both are at the same temperature, but the steel feels colder. Why?

Show mark scheme
Marking pointMarkNotes
Metals are good thermal conductors, so energy flows rapidly from the warm hand into the steel and the skin cools quickly. Wood is a poor conductor, so the skin stays warmer. The two objects are at the same temperature, so the steel is not colder in that sense. The amount of thermal energy stored is not what the hand senses. Radiation is not the main transfer in contact between the hand and the surface.✓ 1Answer C

Answer: C

Common error: Things 'feel' hot or cold because of how quickly energy moves to or from your skin, not only because of temperature.

99IGPC3-2-08r47-09Particle model and gas pressure Paper CEasy6 marks

A library keeps rare documents in a sealed, rigid metal box that contains air.

(a)

Describe the arrangement and the movement of the particles in a gas.

(2)
(b)

Explain how the air particles produce a pressure on the walls of the box.

(2)
(c)

The box is moved from a store at 10 °C to a room at 25 °C. Explain why the pressure of the air in the box increases.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles are far apart and arranged randomly✓ 1
particles move randomly and at high speed✓ 1
Part (b)
particles collide with the walls✓ 1
each collision exerts a force on the wall, and the force acting on each unit area of the wall is the pressure✓ 1
Part (c)
the particles gain kinetic energy and move faster✓ 1
they hit the walls more often and with a greater force✓ 1

Exam tip: Link every pressure explanation to collisions with the walls: how often, and how hard.

100IGPC3-2-08r47-11Conduction, convection and radiation Paper CEasy6 marks

A heated hut at a wind farm has metal walls and a roof. A wind-turbine nacelle outside is painted white.

(a)

Name the process by which thermal energy passes through the metal wall of the hut.

(1)
(b)

A heater stands on the floor of the hut. Explain, in terms of density, why the warm air next to the heater rises.

(3)
(c)

Explain how the white paint helps to keep the inside of the nacelle cool in sunshine.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
conduction✓ 1
Part (b)
the air next to the heater is heated and expands✓ 1
its density becomes less than that of the surrounding air✓ 1
the cooler, denser air sinks and pushes the warm air upwards (convection)✓ 1
Part (c)
white surfaces are poor absorbers of infrared radiation✓ 1
they reflect much of the radiation from the Sun✓ 1

Exam tip: Convection needs a fluid (liquid or gas); conduction in solids and radiation do not.

101IGPC3-2-09r47-01Particle model and gas pressure Paper CEasy1 mark

A sealed rigid metal flask of air is left in strong sunshine on a mountain ledge and its temperature rises. Which statement explains why the gas pressure inside the flask increases?

Show mark scheme
Marking pointMarkNotes
Warmer gas has particles with greater average kinetic energy. They hit the walls more often and each collision exerts a larger force, so the pressure rises. The particles do not expand, the number of particles in a sealed flask stays the same, and their mass does not change with temperature.✓ 1Answer A

Answer: A

Common error: Pressure comes from particle collisions with the walls; the particles themselves do not change size, number or mass when a sealed gas is heated.

102IGPC3-2-09r47-03Conduction, convection and radiation Paper CEasy1 mark

Which statement explains why metals are good conductors of thermal energy?

Show mark scheme
Marking pointMarkNotes
Free electrons in a metal move through the lattice and transfer energy quickly, in addition to the transfer by vibrating atoms. Trapped air is a poor conductor, convection does not occur in solids, and shininess affects emission of radiation from a surface rather than conduction inside the metal.✓ 1Answer C

Answer: C

Common error: Convection only happens in fluids (liquids and gases), so it can never be the explanation for conduction in a solid metal.

103IGPC3-2-09r47-04Thermal expansion, specific heat capacity and changes of state Paper CEasy1 mark

A chef heats 0.40 kg of oil from 20 °C to 160 °C. The specific heat capacity of the oil is 2000 J / (kg °C). How much thermal energy does the oil gain?

Show mark scheme
Marking pointMarkNotes
Energy = mass × specific heat capacity × temperature change = 0.40 × 2000 × (160 − 20) = 112000 J = 112 kJ. The value 16 kJ uses the starting temperature as the temperature change, 128 kJ uses the final temperature, and 280 kJ leaves out the mass.✓ 1Answer D

Answer: D

Exam tip: The temperature change is the difference between the final and initial temperatures, not either temperature on its own.

104IGPC3-2-09r47-09Particle model and gas pressure Paper CEasy6 marks

A mountaineer carries a sealed metal canister of gas for a camping stove.

(a)

Explain, in terms of particles, how the gas exerts a pressure on the walls of the canister.

(2)
(b)

The canister is left in strong sunshine and the gas warms up. State what happens to the pressure of the gas and explain why.

(2)
(c)

The gas exerts a force of 480 N on a section of the canister wall of area 0.0030 m2. Calculate the pressure on this section of wall.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 particles of the gas collide with the walls of the canister✓ 1
B1 each collision exerts a force on the wall, and the force per unit area is the pressure✓ 1
Part (b)
B1 the pressure increases✓ 1
B1 particles move faster, so they hit the walls more often and with greater force✓ 1
Part (c)
M1 pressure = force / area = 480 / 0.0030✓ 1
A1 160000 Pa (1.6 × 105 Pa)✓ 1

Exam tip: In 'explain' questions about pressure, mention both how often the particles hit the wall and how hard they hit it.

105IGPC3-2-09r47-11Thermal expansion, specific heat capacity and changes of state Paper CEasy5 marks

A cook heats 0.25 kg of water in a pan. The water is heated from 18 °C to 98 °C. The specific heat capacity of water is 4200 J / (kg °C).

(a)

Calculate the thermal energy gained by the water.

(2)
(b)

The hob supplies energy to the water at a rate of 700 W. Assume all of this energy is gained by the water. Calculate the time taken to heat the water. Use your answer to (a).

(2)
(c)

Suggest why the actual time taken is longer than your answer to (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 energy = mass × specific heat capacity × temperature change = 0.25 × 4200 × (98 − 18)✓ 1
A1 84000 J✓ 1
Part (b)
M1 time = energy / power = 84000 / 700✓ 1
A1 120 s (ECF from (a))✓ 1
Part (c)
B1 some energy is transferred to the pan or to the surroundings, so less than 700 W heats the water✓ 1

Common error: Use the temperature change (80 °C here), not the final temperature, in the specific heat capacity equation.

106IGPC-036-02-06Conduction, convection and radiation Paper CMedium1 mark

A vacuum flask keeps hot drinks hot. Which row correctly links a feature of the flask with the main energy transfer it reduces?

stopperhot drinkvacuumsilvered surfaces
Cross-section of a vacuum flask
Show mark scheme
Marking pointMarkNotes
A vacuum contains no particles, so energy cannot cross it by conduction or convection. The silvered surfaces are poor emitters of infrared. They also reflect infrared back towards the drink, which reduces transfer by radiation. The vacuum cannot stop radiation, because infrared can travel through a vacuum.✓ 1Answer A

Answer: A

Exam tip: Learn each flask feature together with the transfer it reduces: vacuum for conduction and convection, silvering for radiation, and the stopper for convection and evaporation.

107IGPC-036-02-07Conduction, convection and radiation Paper CMedium1 mark

The cooling unit of an old-style refrigerator is placed at the top of the food compartment. Why is this the best position?

cooling unitfoodfoodfoodold-style refrigerator
Side view of the food compartment
Show mark scheme
Marking pointMarkNotes
Air next to the cooling unit loses energy, contracts and becomes denser, so it sinks. Warmer, less dense air from lower down rises to replace it and is cooled in turn. This convection current circulates air through the whole compartment. Cold air sinks, and 'cold' is not something that is transferred: energy always moves from the warmer food and air to the cooler unit.✓ 1Answer A

Answer: A

Common error: Avoid phrases like 'cold travels' or 'cold is conducted'. Thermal energy always moves from hotter to colder regions.

108IGPC-036-02-10Conduction, convection and radiation Paper CMedium1 mark

On a sunny day, the land near the coast warms up faster than the sea. Which statement describes the movement of air near the beach during the day?

sealandSunsunny day: land warms faster than sea
Side view of the coast
Show mark scheme
Marking pointMarkNotes
The air above the warmer land is heated, expands, becomes less dense and rises. Cooler, denser air from above the sea moves in to take its place, producing a breeze blowing from the sea onto the land. Convection happens in all fluids, both liquids and gases.✓ 1Answer A

Answer: A

Exam tip: In any convection question, first find the warmest region. The fluid rises above it, and cooler fluid flows in at a lower level to replace it.

109IGPC-036-02-14Conduction, convection and radiation Paper CMedium6 marks

A room is heated by a hot-water radiator fixed to the wall, just below a window. The house has cavity walls: two layers of brick with a gap between them.

(a)

Describe how the radiator heats the air throughout the room.

(3)
(b)

The back of the radiator has thin metal fins attached to it. Suggest why.

(1)
(c)

The gap in the cavity wall is filled with plastic foam that contains many small trapped pockets of air. Explain how this reduces the rate of energy loss through the wall.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Air next to the radiator is heated, expands and becomes less dense . This warm air rises towards the ceiling and spreads across the room . Cooler, denser air sinks and moves along the floor back to the radiator, forming a convection current .✓ 3
Part (b)
The fins increase the surface area, so energy is transferred to the air at a greater rate .✓ 1
Part (c)
Air is a poor thermal conductor, so little energy is conducted across the foam . The air is trapped in small pockets, so it cannot move to form convection currents .✓ 2

Common error: Foam insulation works because it traps air. It does not 'block cold'. It reduces both conduction and convection.

110IGPC-036-00-04Particle model and gas pressure Paper CMedium1 mark

A student looks through a microscope at smoke particles in a small glass cell filled with air. The smoke particles are lit from the side.
What does the student see, and what causes it?

Show mark scheme
Marking pointMarkNotes
This is Brownian motion. The visible smoke particles move in a random, zig-zag way. The cause is collisions from the invisible, fast-moving air molecules, which strike each smoke particle unevenly from different directions. Convection (A) would give a steady drift, not random motion. Collisions between smoke particles (D) are far too rare to explain the continuous jerky movement.✓ 1Answer B

Answer: B

Exam tip: Be clear about which particles you can see. The smoke particles are visible, but the air molecules that hit them are not.

111IGPC-036-00-15Particle model and gas pressure Paper CMedium5 marks

A small glass cell is filled with air and a little smoke. The cell is lit brightly from the side and viewed through a microscope. The smoke particles appear as tiny bright specks.

(a)

Describe the motion of the bright specks.

(2)
(b)

Explain what causes this motion.

(2)
(c)

Suggest how the motion of the specks changes if the air in the cell is warmed.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
They move in a random direction . The motion is jerky or zig-zag, with sudden changes of direction .✓ 2
Part (b)
Air molecules, which are too small to be seen, collide with the smoke particles . The collisions are uneven or random from different sides, so each smoke particle is pushed in a changing direction .✓ 2
Part (c)
The specks move more vigorously or faster, because the air molecules move faster .✓ 1

Exam tip: Say clearly that the air molecules hit the smoke particles. Writing just 'the particles collide' is too vague to earn the mark.

112IGPC-036-01-03Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

A straight bimetallic strip is made by joining a strip of brass to a strip of iron. For the same temperature rise, brass expands more than iron.
What happens when the strip is heated?

brassironclampstraight strip before heating
Strip before heating (side view)
Show mark scheme
Marking pointMarkNotes
The brass becomes longer than the iron. Because the two strips are fixed together, the strip has to curve. The longer metal (brass) ends up on the outside of the curve, where the path is longer. It cannot stay straight, because the two metals expand by different amounts.✓ 1Answer B

Answer: B

Exam tip: The metal that expands more always ends up on the outside, the longer side, of the curve.

113IGPC-036-01-05Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

Which statement is true for evaporation but not for boiling?

Show mark scheme
Marking pointMarkNotes
Evaporation happens only at the liquid surface. It can happen at any temperature, because some of the more-energetic particles near the surface escape. Boiling happens throughout the liquid, with bubbles forming, and only at the boiling point. Both processes change a liquid into a gas, so option D does not tell them apart.✓ 1Answer C

Answer: C

Exam tip: Learn two differences: where it happens (surface or throughout) and when it happens (any temperature or a fixed temperature).

114IGPC-036-00-12Particle model and gas pressure Paper CMedium6 marks

A car tyre is pumped up in the morning when the air temperature is 15 °C. By the afternoon, after driving on a hot road, the air inside the tyre is at 45 °C. No air leaks out of the tyre, and its volume stays almost constant.

(a)

Describe the motion of the air molecules inside the tyre.

(1)
(b)

Explain how the air molecules exert a pressure on the inside wall of the tyre.

(2)
(c)

Convert 45 °C into kelvin.

(1)
(d)

Explain, in terms of molecules, why the pressure in the tyre is higher in the afternoon.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
They move randomly, in all directions, at high speed .✓ 1
Part (b)
The molecules collide with the wall and rebound, so each collision exerts a force on the wall . Pressure is the total force from many collisions acting on each unit area of the wall .✓ 2
Part (c)
45 + 273 = 318 K✓ 1
Part (d)
At a higher temperature the molecules have a greater average kinetic energy, so they move faster . They hit the wall more often and with greater force, so the force per unit area, and so the pressure, increases .✓ 2

Common error: Do not say that the molecules 'expand' or that there are 'more molecules'. The tyre holds the same molecules, and they simply move faster on average.

115IGPC3-2-02c-05Conduction, convection and radiation Paper CMedium1 mark

Four identical metal cans each contain 200 g of water at 90 °C. Each can has a different outer surface. Which can cools fastest by thermal radiation?

Show mark scheme
Marking pointMarkNotes
Correct: C. A matt black surface is the best emitter of thermal radiation, so energy leaves it fastest. A is wrong: a shiny surface is a poor emitter as well as a poor absorber, so it cools slowest. B is wrong: shiny surfaces emit poorly; colour alone does not make a good emitter, the surface must also be matt. D is wrong: a matt white surface emits less well than matt black, because light-coloured surfaces are poorer emitters.✓ 1Answer C

Answer: C

Exam tip: Good absorbers of radiation are also good emitters. Dull, dark surfaces are the best of both.

116IGPC3-2-02c-08Conduction, convection and radiation Paper CMedium1 mark

A room heater is placed near the floor. Which statement correctly describes how it warms the air in the room by convection?

Show mark scheme
Marking pointMarkNotes
Correct: A. Convection currents are driven by density differences in a fluid: warm air rises and cool air sinks, setting up a circulation. B is wrong: heat is energy, not a substance with weight; it is the warm air that rises because it is less dense. C is wrong: the particles do not change mass; the air expands so its density falls. D is wrong: radiation transfers energy by waves; it does not push air upward.✓ 1Answer A

Answer: A

Common error: Explain convection with density: warm fluid expands, becomes less dense and rises.

117IGPC3-2-02c-10Thermal expansion, specific heat capacity and changes of state Paper CMedium6 marks

A bimetallic strip is made from a strip of brass joined firmly to a strip of iron of the same length. It is used as a switch in a fire alarm circuit.

(a)

Explain why the strip bends when it is heated.

(3)
(b)

Describe how the bending of the strip is used to switch on the alarm.

(1)
(c)

A liquid-in-glass thermometer has a very narrow tube. Explain why this makes the thermometer more sensitive.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
both metals expand when heated✓ 1
brass expands more than iron for the same temperature rise✓ 1
the strip bends with the brass on the outside of the curve✓ 1
Part (b)
the strip bends until it touches a contact✓ 1
Part (c)
the liquid expands by a small volume when heated✓ 1
in a narrow tube this volume produces a large change in length of the liquid thread✓ 1

Exam tip: When explaining the bimetallic strip, state that the two metals expand by different amounts.

118IGPC3-2-02c-14Conduction, convection and radiation Paper CMedium7 marks

Thermal radiation affects how buildings and containers gain and lose energy.

(a)

In hot countries many roofs are painted white. Explain how this keeps the building cooler.

(2)
(b)

The pipes of a solar water heater are painted matt black. Explain why.

(2)
(c)

Describe how a vacuum flask keeps a hot drink hot.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
white (light, shiny) surfaces are poor absorbers of infrared radiation✓ 1
they reflect most of the radiation from the Sun✓ 1
Part (b)
matt black surfaces are good absorbers of infrared radiation✓ 1
so more energy from the Sun is absorbed and transferred to the water✓ 1
Part (c)
the vacuum between the walls stops conduction and convection✓ 1
the inner surfaces are shiny/silvered so little radiation is emitted and the radiation is reflected back✓ 1
the stopper (made of an insulator) reduces conduction and convection from the top✓ 1

Exam tip: A good absorber of radiation is also a good emitter. Match the surface (matt black or shiny) to the job.

119IGPC3-2-02c-16Thermal expansion, specific heat capacity and changes of state Paper CMedium5 marks

A solid sample is heated steadily by a heater and its temperature is recorded every minute. At 0, 1, 2, 3, 4, 5, 6, 7 and 8 minutes the temperatures are 30, 44, 58, 72, 72, 72, 72, 84 and 96 °C respectively.

012345678time / mintemperature / °C
Graph plotted from the data in the question
(a)

State the melting point of the sample.

(1)
(b)

Explain why the temperature stays constant between 3 and 6 minutes although the heater is still on.

(2)
(c)

Calculate the rate of temperature rise of the solid during the first 3 minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
72 °C✓ 1
Part (b)
the thermal energy supplied is used to break the bonds / overcome the forces between particles as the solid melts✓ 1
so the energy does not increase the kinetic energy of the particles (the temperature does not rise)✓ 1
Part (c)
temperature rise = 72 − 30 = 42 °C✓ 1
42 ÷ 3 = 14 °C per minute✓ 1

Exam tip: During a change of state the temperature stays constant while energy is still being supplied.

120IGPC3-2-07r7-03Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

Two blocks of equal mass receive the same quantity of thermal energy. Block X has a specific heat capacity of 400 J/(kg °C) and block Y has 1200 J/(kg °C). Which statement about their temperature rises is correct?

Show mark scheme
Marking pointMarkNotes
Temperature rise = energy ÷ (mass × specific heat capacity), so for equal energy and mass the rise is inversely proportional to c; X has one third of Y's capacity, so rises three times as much. Option A is wrong: the rise depends on the specific heat capacity as well as the energy. Option B is wrong: the rise is inversely proportional to specific heat capacity, so the material with the larger value rises less. Option D is wrong: the capacities cannot simply be subtracted; the ratio of rises is 1200 : 400.✓ 1Answer C

Answer: C

Exam tip: Large specific heat capacity means a small temperature change for a given energy input.

121IGPC3-2-07r7-05Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

An athlete sweats on a warm day and feels cooler as the sweat evaporates. Which statement explains the cooling?

Show mark scheme
Marking pointMarkNotes
Fast molecules overcome attractions and leave the surface, so the average kinetic energy, and hence the temperature, of the remaining liquid decreases and it takes energy from the skin. Option B is wrong: only the fastest molecules at the surface escape, so molecules do not all lose equal energy. Option C is wrong: evaporation occurs at any temperature; boiling needs 100 °C for water. Option D is wrong: the remaining molecules move more slowly on average, which lowers the temperature.✓ 1Answer A

Answer: A

Common error: Evaporation happens only at the surface and at any temperature; boiling happens throughout the liquid at the boiling point.

122IGPC3-2-07r7-08Particle model and gas pressure Paper CMedium1 mark

A sealed sports ball filled with gas is heated by the sun. When the gas is heated by the same amount, why does it expand much more than a similar-sized solid ball would?

Show mark scheme
Marking pointMarkNotes
In solids strong forces hold particles in place, so expansion is small; in gases the forces are negligible, so faster particles spread out much more. Option A is wrong: particle size is similar and the particles do not grow on heating. Option B is wrong: solid particles also gain energy, but they stay bound together and only vibrate more. Option D is wrong: the number of particles does not change on heating; the difference is in the forces between them.✓ 1Answer C

Answer: C

Common error: Expansion is caused by particles moving further apart, not by the particles becoming bigger.

123IGPC3-2-07r7-11Particle model and gas pressure Paper CMedium6 marks

A sealed capsule on a space probe contains helium gas. The volume of the capsule is fixed.

(a)

Describe, in terms of particles, how the helium produces a pressure on the walls of the capsule.

(2)
(b)

The capsule warms up as the probe approaches the Sun. Explain why the pressure of the gas rises.

(2)
(c)

In a laboratory test the same gas at constant temperature is squeezed into a smaller volume. State what happens to the pressure and explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles move randomly and collide with the walls✓ 1
each collision exerts a force on the wall; the many collisions together give a pressure✓ 1
Part (b)
particles move faster / have more kinetic energy✓ 1
so they collide with the walls more often and with greater force✓ 1
Part (c)
pressure increases✓ 1
particles hit the walls more often because they have less distance to travel between collisions✓ 1

Exam tip: Link pressure to both frequency and force of the collisions.

124IGPC3-2-07r7-14Conduction, convection and radiation Paper CMedium6 marks

A sports hall has a heater mounted low on one wall. The walls are lined with foam containing trapped pockets of air.

(a)

Describe how convection currents spread the warm air through the hall.

(3)
(b)

Explain why the heater is placed low in the hall rather than near the ceiling.

(1)
(c)

Explain why the foam lining reduces energy loss through the walls.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
air next to the heater is warmed and expands, so its density decreases✓ 1
warm air rises and cooler, denser air sinks to take its place✓ 1
this circulation carries energy around the hall✓ 1
Part (b)
warm air rises, so a heater near the ceiling would not warm the lower air by convection✓ 1
Part (c)
air is a poor conductor and the foam traps air in pockets✓ 1
trapped air cannot circulate, so convection is prevented✓ 1

Exam tip: Name the cause of convection: a change in density, not simply 'heat rises'.

125IGPC3-2-07r9-03Conduction, convection and radiation Paper CMedium1 mark

A shop sells copper-based and glass saucepans. When both are placed on the same hot ring, the copper base warms much faster. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
D is correct: metals contain free electrons which gain kinetic energy, move through the metal and collide with other particles, transferring energy quickly. A is wrong because atoms in a solid only vibrate about fixed positions and do not travel. B is wrong because shiny surfaces are poor absorbers of radiation, and radiation is not the main process in a pan base in contact with the ring. C is wrong because density does not determine how quickly energy is conducted.✓ 1Answer D

Answer: D

Exam tip: When explaining conduction in a metal, mention free electrons moving through the metal, not just 'particles vibrating'.

126IGPC3-2-07r9-04Conduction, convection and radiation Paper CMedium1 mark

An engineer tests four identical metal plates in strong sunlight for a solar water heater. Each plate has a different surface finish. Which finish gives the greatest temperature rise in ten minutes?

Show mark scheme
Marking pointMarkNotes
A is correct: matt black surfaces are the best absorbers of infrared and visible radiation, so the plate gains the most energy. B is wrong because shiny silver reflects most of the radiation. C is wrong because white surfaces reflect much of the radiation. D is wrong because, although black, a shiny surface reflects part of the radiation, so it absorbs less than a matt surface.✓ 1Answer A

Answer: A

Common error: Both colour and texture matter: the best absorber (and the best emitter) is a dull, matt black surface.

127IGPC3-2-07r9-11Thermal expansion, specific heat capacity and changes of state Paper CMedium6 marks

A shop has a long steel rail fixed above its entrance. The rail is hotter on a summer afternoon than on a winter morning.

(a)

State what happens to the length of the rail as it warms.

(1)
(b)

Explain this change in terms of the particles in the rail.

(2)
(c)

The shop's heater is controlled by a thermostat containing a strip made of brass joined to iron. Brass expands more than iron for the same temperature rise. Explain how the strip switches the heater off when the room becomes too warm.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The rail gets (slightly) longer / expands✓ 1
Part (b)
The particles vibrate with larger amplitude / gain kinetic energy✓ 1
so the average distance between neighbouring particles increases and the rail expands✓ 1
Part (c)
Both metals expand, but the brass expands more than the iron✓ 1
so the strip bends (with the brass on the outside of the curve)✓ 1
The bending separates the contacts / breaks the circuit, so the heater switches off✓ 1

Exam tip: For a bimetallic strip, always say which metal expands more and which way the strip bends.

128IGPC3-2-07r9-12Conduction, convection and radiation Paper CMedium5 marks

A cafe serves hot coffee in metal cups with a metal spoon. Customers hold the cups using a cardboard sleeve.

(a)

The bowl of a metal spoon is placed in the hot coffee. Explain why the handle of the spoon soon feels warm.

(3)
(b)

The cardboard sleeve contains trapped air. Suggest why the sleeve reduces the energy transferred to the customer's hand.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Free electrons in the hot end gain kinetic energy✓ 1
They move through the metal and collide with other particles✓ 1
and transfer energy along the spoon to the cooler handle✓ 1
Part (b)
Air is a poor conductor (an insulator), so conduction through the sleeve is slow✓ 1
The air is trapped so it cannot circulate / convection currents cannot form✓ 1

Common error: Do not write that insulators 'trap heat'; explain that they slow the transfer of energy.

129IGPC3-2-07r9-14Conduction, convection and radiation Paper CMedium6 marks

A cool box carries vaccines to a rural clinic. It has thick walls of polystyrene foam. Ice packs are placed inside the box.

(a)

State the changes in the arrangement and the movement of the particles when the ice in an ice pack melts.

(2)
(b)

Explain why the polystyrene foam walls reduce the energy transferred into the box by conduction.

(2)
(c)

Suggest why the ice packs are placed on top of the vaccines rather than underneath them.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Particles change from a regular fixed pattern to a random arrangement✓ 1
and they gain freedom to move / slide past one another instead of only vibrating✓ 1
Part (b)
Foam contains many pockets of trapped air✓ 1
Air is a poor conductor, so energy is transferred slowly through the walls✓ 1
Part (c)
Air next to the ice is cooled and becomes denser, so it sinks✓ 1
Warmer air rises, creating a convection current that cools the whole contents✓ 1

Exam tip: Link density to movement: cold fluid is denser and sinks, warm fluid is less dense and rises.

130IGPC3-2-07r12-03Conduction, convection and radiation Paper CMedium1 mark

A vacuum flask used by a farm worker keeps tea hot. There is a vacuum between the two glass walls. Why does the vacuum reduce thermal energy transfer by conduction and convection?

stopperhot teavacuumglass walls
Cross-section of a vacuum flask
Show mark scheme
Marking pointMarkNotes
A vacuum has no particles, so energy cannot be passed from particle to particle (conduction) and there is no fluid to flow (convection). Reflection of radiation is done by the silvered surfaces, not by the vacuum. A vacuum contains no trapped air. Particles in a vacuum do not exist, so they cannot be colder.✓ 1Answer B

Answer: B

Common error: Do not confuse the vacuum, which stops conduction and convection, with the silvered surfaces, which reduce radiation.

131IGPC3-2-07r12-08Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

A musician touches a brass key and a wooden panel on a stage. Both are at 18 °C and the musician's hand is at 33 °C. The brass key feels colder. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
Both objects are at the same temperature; brass feels colder because as a good conductor it removes energy from the skin faster. Saying the brass is colder contradicts the data. Wood is an insulator, not a better conductor. Specific heat capacity does not explain the difference in sensation.✓ 1Answer A

Answer: A

Common error: Feeling hot or cold depends on how quickly energy flows to or from your skin, not only on the temperature of the object.

132IGPC3-2-07r12-10Conduction, convection and radiation Paper CMedium6 marks

A farmer grows tomatoes in a greenhouse. The Sun's energy warms the inside of the greenhouse.

(a)

Name the process by which thermal energy from the Sun reaches the greenhouse across space.

(1)
(b)

The farmer paints the inside of the back wall matt black. State and explain how this helps to keep the greenhouse warm.

(2)
(c)

An electric heater is placed on the floor. Explain how the warm air spreads through the greenhouse.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
radiation (infrared radiation / electromagnetic waves)✓ 1
Part (b)
A matt black surface is a good absorber of radiation✓ 1
so more of the radiation is absorbed, and the wall warms up and warms the air (instead of the radiation being reflected back out)✓ 1
Part (c)
Air near the heater gains energy and expands, so it becomes less dense✓ 1
The warm air rises✓ 1
Cooler, denser air moves in / sinks to take its place, setting up a convection current✓ 1

Exam tip: In convection explanations, always link a change in density to the movement of the fluid.

133IGPC3-2-07r12-11Thermal expansion, specific heat capacity and changes of state Paper CMedium5 marks

A temperature switch in the cargo-hold heating system of an aircraft contains a bimetallic strip. The strip is made of a brass layer and an iron layer joined together. For the same rise in temperature, brass expands more than iron.

(a)

Explain why the strip bends when its temperature rises.

(2)
(b)

The strip bends away from an electrical contact when the hold becomes warm. Describe how this switches the heater off.

(2)
(c)

Suggest one change to the strip that would make the switch respond to a smaller temperature change.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Brass expands more than iron for the same temperature rise✓ 1
The layers are joined, so the strip must curve, with the brass on the outside (longer side) of the curve✓ 1
Part (b)
The strip moves away from the contact, so the contact separates✓ 1
The circuit is broken, so the heater is switched off✓ 1
Part (c)
Use a longer (or thinner) strip, or use two metals whose expansions differ by more✓ 1

Common error: Both metals expand; the bending happens because one expands more than the other while they are fixed together.

134IGPC3-2-07r12-16Conduction, convection and radiation Paper CMedium7 marks

A technician in a concert hall tests two identical metal cans, one painted matt black and one polished silver. Each can is filled with 200 g of water at 85 °C and left in the same room. After 10 minutes the black can is at 52 °C and the silver can is at 63 °C.

dull blackshiny silvereach can: 200 g of water at 85 °C, same room
Diagram not to scale
(a)

State which can emits thermal radiation more quickly.

(1)
(b)

Explain why the water in the black can cools more.

(2)
(c)

State two variables the technician must keep the same for a fair test.

(2)
(d)

Calculate how much more the temperature of the black can's water fell than the temperature of the silver can's water.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the matt black can✓ 1
Part (b)
A matt black surface is a better emitter of radiation than a polished silver surface✓ 1
so the black can transfers energy to the surroundings faster✓ 1
Part (c)
One correct variable, e.g. mass or volume of water, starting temperature, time allowed or position in the room✓ 1
A second, different correct variable✓ 1
Part (d)
black fell 33 °C and silver fell 22 °C✓ 1
difference = 11 °C✓ 1

Exam tip: In a fair test, only the variable being investigated (here, the surface finish) is allowed to change.

135IGPC3-2-05r13-03Conduction, convection and radiation Paper CMedium1 mark

The roof of a data-centre container is exposed to sunshine. Which finish on the roof would keep the container coolest?

Show mark scheme
Marking pointMarkNotes
Correct: shiny silver is the best reflector and poorest absorber of infrared radiation. Matt black is the best absorber, and matt dark grey and dull dark red also absorb much of the radiation.✓ 1Answer D

Answer: D

Common error: A good absorber is also a good emitter; for keeping something cool in sunshine you want a poor absorber, which is a shiny, light surface.

136IGPC3-2-05r13-06Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

An archaeologist heats a sample of beeswax steadily and records its temperature. The temperature rises, then stays constant at 62 °C for several minutes while the wax melts, and then rises again. Which statement explains the constant temperature?

Show mark scheme
Marking pointMarkNotes
Correct: during a change of state the energy goes into separating the particles, so average kinetic energy and temperature stay constant. The heater is still supplying energy; particles keep moving; if all energy were lost the wax would not melt.✓ 1Answer B

Answer: B

Common error: Temperature measures average kinetic energy, so a flat section on a heating graph means energy is changing the arrangement of particles, not their speed.

137IGPC3-2-05r13-08Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

A liquid-in-glass thermometer is placed in the hot aisle of a data centre and the liquid thread rises. Why does the liquid level rise?

Show mark scheme
Marking pointMarkNotes
Correct: both liquid and glass expand, but the liquid expands more, so it is forced up the narrow tube. Glass expands rather than contracts; particles do not change size; the thermometer is sealed so no particles enter.✓ 1Answer B

Answer: B

Common error: Solids and liquids both expand on heating; the thermometer works because the liquid expands more than the glass.

138IGPC3-2-05r13-10Conduction, convection and radiation Paper CMedium6 marks

A server in a data centre has a metal heat sink with fins painted matt black. A fan is switched off.

(a)

Name the process by which thermal energy passes along the metal fins and explain it in terms of particles.

(3)
(b)

Explain why warm air rises above the heat sink.

(2)
(c)

Suggest why the fins are painted matt black.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
conduction✓ 1
free electrons gain kinetic energy and move through the metal, colliding with other particles to pass on energy✓ 1
atoms/ions vibrate more and pass vibrations to neighbouring atoms✓ 1
Part (b)
warm air expands so its density is lower than the surrounding air✓ 1
cooler denser air sinks and pushes the warm air upwards✓ 1
Part (c)
matt black surface is a good emitter of infrared radiation so energy is lost faster✓ 1

Common error: Do not say that 'heat rises'; say that warm fluid rises because its density is lower.

139IGPC3-2-07r13-06Conduction, convection and radiation Paper CMedium1 mark

The immersion heater in a hot-water tank is placed near the bottom of the tank. Why is this the best position?

waterimmersionheater
Hot-water tank (side view)
Show mark scheme
Marking pointMarkNotes
C is correct: the water near the heater expands and becomes less dense, so it rises, and cooler, denser water moves in to replace it, setting up a convection current that heats all the water. A is wrong because water is a poor conductor. B is wrong because heated water is less dense, not more. D is wrong because water absorbs radiation within a short distance, so it is not the main process.✓ 1Answer C

Answer: C

Common error: Hot fluids rise because they are less dense, not because heat itself 'rises'.

140IGPC3-2-07r13-08Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

A thermostat in a railway carriage heater uses a strip made of brass joined to a strip of iron. Brass expands more than iron for the same rise in temperature. What happens when the joined strip is heated?

brassironclampstraight strip before heating
Strip before heating (side view)
Show mark scheme
Marking pointMarkNotes
D is correct: the brass becomes longer than the iron, so it must be on the outside of the curve. A is the wrong way round. B is wrong because the metals expand by different amounts. C is wrong because the strip that expands more has to be on the outside.✓ 1Answer D

Answer: D

Common error: A bimetallic strip bends towards the metal that expands less, so the metal that expands more is on the outside of the bend.

141IGPC3-2-07r13-13Conduction, convection and radiation Paper CMedium6 marks

The metal walls of a train carriage are heated by a heater near the floor.

(a)

State the name of the process by which thermal energy is transferred through the metal wall.

(1)
(b)

Describe, in terms of particles, how thermal energy is conducted from the hot side of the wall to the cold side.

(2)
(c)

Explain how the heater near the floor warms the air in the whole carriage.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
Conduction✓ 1
Part (b)
Particles at the hot side vibrate more strongly / gain kinetic energy✓ 1
and pass energy on to neighbouring particles through collisions, so energy travels along the wall✓ 1
Part (c)
Air near the heater is heated and expands, so it becomes less dense✓ 1
The warm air rises✓ 1
Cooler, denser air moves in to take its place, setting up a convection current that circulates through the carriage✓ 1

Exam tip: For convection, always include the density change: the fluid expands, becomes less dense and rises.

142IGPC3-2-05r14-02Particle model and gas pressure Paper CMedium1 mark

A fixed mass of gas in a cylinder is kept at constant temperature. The piston is pushed in so that the volume of the gas is halved, and the pressure doubles. Which statement explains the increase in pressure?

Show mark scheme
Marking pointMarkNotes
B is correct: the temperature is constant so the average speed and the force per collision are unchanged, but in a smaller volume the particles collide with the walls more frequently per unit area. A is wrong because speed depends on temperature, which is constant. C is wrong because the force of each collision depends on speed, which has not changed. D is wrong because the mass of gas is fixed.✓ 1Answer B

Answer: B

Common error: If the temperature is constant, the particles' average speed does not change, so a pressure change must come from the number of collisions per second.

143IGPC3-2-05r14-06Particle model and gas pressure Paper CMedium1 mark

A solid metal rod is heated and its length increases slightly. What happens at the particle level?

Show mark scheme
Marking pointMarkNotes
A is correct: heating increases the vibration of the particles and the average spacing grows, so the rod expands. B is wrong because the particles do not change size. C is wrong because the particles in a solid stay in fixed positions and only vibrate. D is wrong because the number of particles is unchanged.✓ 1Answer A

Answer: A

Common error: Expansion is caused by the increased spacing between particles, not by the particles growing.

144IGPC3-2-05r14-10Particle model and gas pressure Paper CMedium5 marks

A cyclist's tyre contains air at constant volume.

(a)

Explain, in terms of particles, how the air produces a pressure on the inside wall of the tyre.

(2)
(b)

After a long descent with the brakes on, the temperature of the air in the tyre rises. Explain why the pressure of the air rises.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
air particles move randomly and collide with the wall✓ 1
each collision exerts a force on the wall, and the total force on each unit area of the wall is the pressure✓ 1
Part (b)
particles gain kinetic energy and move faster on average✓ 1
they hit the wall more often✓ 1
and with a greater force in each collision, so the force per unit area is greater✓ 1

Common error: Do not say that the particles 'expand' or 'get hotter' when a gas is heated; say that they move faster on average.

145IGPC3-2-05r14-13Conduction, convection and radiation Paper CMedium5 marks

A student compares how well two metal cans emit thermal radiation. One can is painted dull black and the other is shiny silver. Each can is filled with hot water and left to cool in the same room.

dull blackshiny silvereach can filled with hot water, same room
Diagram not to scale
(a)

State two variables that the student should keep the same.

(2)
(b)

Describe how the student can tell which can emits thermal radiation faster.

(2)
(c)

The temperature of the water in the dull black can fell by 18 °C and in the shiny silver can by 7 °C. State the conclusion.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
any two from: initial temperature of the water; volume (or mass) of water; size and shape of the cans; room temperature or position in the room; time allowed to cool✓ 2
Part (b)
measure the temperature of the water in each can at the start and after the same time using a thermometer✓ 1
the can whose water has the larger temperature fall emits faster✓ 1
Part (c)
the dull black surface is a better emitter of thermal radiation than the shiny silver surface✓ 1

Exam tip: In a fair test only the variable being investigated may change; list the others explicitly.

146IGPC3-2-07r14-03Conduction, convection and radiation Paper CMedium1 mark

A heater is fixed near the bottom of an indoor swimming pool. The water next to the heater rises. Which statement explains why?

heaterwater
Side view of the swimming pool
Show mark scheme
Marking pointMarkNotes
A is correct: warmed water expands, its density falls and it floats upwards, while denser cooler water sinks to replace it (convection). B is wrong because the mass of each particle does not change. C is wrong because liquids are poor conductors and conduction does not depend on direction. D is wrong because there is no such repulsion.✓ 1Answer A

Answer: A

Exam tip: For convection, link the temperature change to a change in density, then to the movement of the fluid.

147IGPC3-2-07r14-04Conduction, convection and radiation Paper CMedium1 mark

The metal cabinet at the base of a wind turbine in a hot, sunny region is painted matt white to reduce heating by the Sun. Which statement explains why this helps?

Show mark scheme
Marking pointMarkNotes
C is correct: light-coloured, smooth surfaces are poor absorbers of infrared radiation, so less energy from the Sun is absorbed. A is wrong because paint does not act as a conductor of energy through the cabinet. B is wrong because colour has no effect on convection. D is wrong because black surfaces are the better emitters.✓ 1Answer C

Answer: C

Common error: Dull black surfaces are the best absorbers and best emitters of radiation; shiny white surfaces are the worst.

148IGPC3-2-07r14-08Particle model and gas pressure Paper CMedium1 mark

A sealed, rigid air tank beside a swimming-pool pump is warmed by the Sun. The mass of air and the volume of the tank stay constant. Which statement explains why the gas pressure increases?

Show mark scheme
Marking pointMarkNotes
D is correct: a higher temperature means a higher average molecular speed, giving more frequent and harder collisions with the walls. A is wrong because molecules do not expand. B is wrong because the tank is sealed. C is wrong because pressure on the walls comes from collisions with the walls, not between molecules.✓ 1Answer D

Answer: D

Common error: Gas pressure comes from molecules colliding with the container walls; say 'with the walls' in your explanation.

149IGPC3-2-07r14-10Conduction, convection and radiation Paper CMedium5 marks

An indoor swimming pool is warmed by a heater fixed near the bottom of the tank. At night a foam cover, which contains many trapped pockets of air, is placed over the water.

heaterwater
Side view of the swimming pool
(a)

Explain how the heater causes a convection current in the water.

(3)
(b)

Explain why the foam cover is a good thermal insulator.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The water next to the heater is warmed, expands and becomes less dense✓ 1
The warm water rises✓ 1
Cooler, denser water from the sides or top sinks to take its place✓ 1
Part (b)
Air is a poor conductor of thermal energy✓ 1
The air is trapped in small pockets, so it cannot circulate and there is little convection✓ 1

Exam tip: Name the process (convection) and describe the density change, the rise and the sinking: three linked steps.

150IGPC3-2-07r14-13Conduction, convection and radiation Paper CMedium5 marks

A wind-farm engineer wants to compare how well two surfaces emit thermal radiation. She has two identical metal cans, one painted matt black and one painted shiny white, a kettle of hot water, two thermometers and a timer.

(a)

Describe how she could use this apparatus to compare the emission of thermal radiation.

(3)
(b)

State the result she should find.

(1)
(c)

State the type of electromagnetic wave that carries most thermal radiation.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Put equal volumes of hot water at the same starting temperature into the two cans✓ 1
Measure the temperature of each can's water at regular time intervals, or after the same time✓ 1
Keep other conditions the same (same position, same surroundings, both with or without lids) and compare the temperature falls✓ 1
Part (b)
The water in the black can cools faster / has a larger temperature fall✓ 1
Part (c)
Infrared✓ 1

Exam tip: In a comparison experiment, state what is kept the same, what is changed and what is measured.

151IGPC3-2-07r15-03Conduction, convection and radiation Paper CMedium1 mark

Why is a metal spoon a better thermal conductor than a wooden spoon of the same size?

Show mark scheme
Marking pointMarkNotes
C is correct: free electrons transfer energy rapidly through a metal. A is wrong because both spoons are at the same temperature before heating, so particles are not hotter. B is wrong because particle size is not the reason. D is wrong because the atoms of a solid stay in fixed positions and only vibrate.✓ 1Answer C

Answer: C

Common error: In conduction the particles do not travel along the solid; only the energy is passed on.

152IGPC3-2-07r15-04Conduction, convection and radiation Paper CMedium1 mark

A designer leaves four fabric patches of the same size in strong sunshine. Which patch will warm up fastest?

Show mark scheme
Marking pointMarkNotes
C is correct: dull black surfaces are the best absorbers of infrared radiation. A and D are wrong because shiny surfaces reflect most of the radiation. B is wrong because white surfaces, even dull ones, reflect more radiation than black ones.✓ 1Answer C

Answer: C

Exam tip: Dark, dull surfaces are good absorbers and good emitters; light, shiny surfaces are poor at both.

153IGPC3-2-07r15-05Particle model and gas pressure Paper CMedium1 mark

Which statement about the particles in a liquid is correct?

Show mark scheme
Marking pointMarkNotes
B is correct for a liquid. A describes a solid. C describes a gas. D is wrong because a regular arrangement is a feature of solids, and liquid particles are not in a regular pattern.✓ 1Answer B

Answer: B

Exam tip: Describe particle arrangement, spacing and movement separately when comparing states.

154IGPC3-2-07r15-13Thermal expansion, specific heat capacity and changes of state Paper CMedium5 marks

A batik artist heats a sample of wax steadily. The temperature is recorded every 2 minutes.
Time / min: 0, 2, 4, 6, 8, 10, 12, 14
Temperature / °C: 20, 34, 48, 60, 60, 60, 71, 82

02468101214020406080100time / mintemperature / °C
Graph plotted from the data in the question
(a)

State the melting point of the wax.

(1)
(b)

State the physical state of the wax at 3 minutes and at 13 minutes.

(2)
(c)

Explain, in terms of particles, why the temperature stays constant between 6 and 10 minutes even though energy is still being supplied.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
60 °C✓ 1
Part (b)
Solid at 3 minutes✓ 1
Liquid at 13 minutes✓ 1
Part (c)
The energy is used to overcome the forces between particles (to melt the wax)✓ 1
So the potential energy of the particles increases while their average kinetic energy stays the same✓ 1

Common error: Do not say the particles stop moving at the melting point; the average kinetic energy simply stops increasing.

155IGPC3-2-06r16-03Conduction, convection and radiation Paper CMedium1 mark

In a data centre, the air just above a group of hot servers rises. Why does this happen?

Show mark scheme
Marking pointMarkNotes
A is correct: heating makes the air expand, lowering its density, so it floats upward through cooler, denser air. B is wrong because particle mass is constant. C is wrong because expansion means fewer particles per unit volume. D is wrong because heated air expands rather than contracts.✓ 1Answer A

Answer: A

Exam tip: For convection, always state the order: heated, expands, less dense, rises.

156IGPC3-2-06r16-09Particle model and gas pressure Paper CMedium6 marks

A cyclist uses a pump to inflate a bicycle tyre.

(a)

Explain, in terms of particles, how the air in the tyre exerts a pressure on the tyre wall.

(2)
(b)

The cyclist pumps more air into the tyre. The temperature stays the same. Explain why the pressure increases.

(2)
(c)

Air trapped in the pump is squeezed into a smaller volume at constant temperature. Explain why its pressure rises.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
air particles move randomly at high speed✓ 1
they collide with the wall and exert a force, over an area, which is the pressure✓ 1
Part (b)
more particles in the same volume✓ 1
more collisions per second with the wall✓ 1
Part (c)
particles are closer together in a smaller volume✓ 1
so collisions with each unit area of the walls happen more often✓ 1

Exam tip: Name the cause (more particles, faster particles or less volume) and then the effect (more frequent or harder collisions).

157IGPC3-2-06r16-10Conduction, convection and radiation Paper CMedium7 marks

A processor in a data centre is attached to an aluminium heat sink with many thin fins. A fan blows air across the fins.

(a)

Explain why a metal such as aluminium is a good conductor of thermal energy.

(2)
(b)

The fins are painted matt black. Explain how this helps to cool the processor.

(2)
(c)

State why the heat sink has many fins.

(1)
(d)

Explain why blowing air across the fins increases the rate of cooling.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
free (delocalised) electrons move through the metal✓ 1
carrying energy quickly, and colliding with other particles to pass energy on✓ 1
Part (b)
matt black surfaces are good emitters of infrared radiation✓ 1
so more energy is radiated per second✓ 1
Part (c)
large surface area to transfer energy to the air✓ 1
Part (d)
warm air next to the fins is replaced by cooler air✓ 1
so the temperature difference stays large and energy transfer is faster✓ 1

Exam tip: Pair every material or surface property with the energy transfer process it affects.

158IGPC3-2-06r16-15Conduction, convection and radiation Paper CMedium6 marks

A cyclist carries a hot drink in a vacuum flask on a winter ride.

hot drinkvacuumshiny silver surfacesplastic foam stopper
Diagram not to scale
(a)

Explain why the vacuum between the walls of the flask reduces energy transfer by conduction and convection.

(2)
(b)

The inner surfaces of the flask are shiny silver. Explain why.

(2)
(c)

The stopper is made of plastic foam. Explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
there are no particles in a vacuum to conduct energy✓ 1
convection needs a fluid (moving particles), which is absent✓ 1
Part (b)
shiny surfaces are poor emitters of radiation✓ 1
and good reflectors, so radiation is reflected back to the drink✓ 1
Part (c)
plastic is a poor conductor (insulator)✓ 1
trapped air in the foam reduces convection✓ 1

Exam tip: For insulation questions, address each of the three processes separately.

159IGPC3-2-07r17-03Conduction, convection and radiation Paper CMedium1 mark

A satellite orbiting the Earth has its fuel tank wrapped in shiny metallic foil. The satellite is in the vacuum of space.
Why is the foil useful?

Show mark scheme
Marking pointMarkNotes
A is correct: a shiny surface reflects most of the radiation from the Sun, so the tank warms up less. B is wrong because spreading energy around does not reduce the total absorbed. C is wrong because the foil does not control convection inside the fuel, and the energy reaches the satellite by radiation anyway. D is wrong because metallic foil is a conductor, and no energy reaches the satellite by conduction through the vacuum.✓ 1Answer A

Answer: A

Exam tip: In a vacuum only radiation transfers energy between the Sun and an object, so the surface finish is what matters.

160IGPC3-2-07r17-04Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

On a hot day an athlete sweats, and the sweat evaporates from the skin. The athlete's skin feels cooler.
Which statement explains this?

Show mark scheme
Marking pointMarkNotes
A is correct: the fastest, most energetic particles escape from the surface, so the particles that remain have a lower average kinetic energy and a lower temperature, which cools the skin. B is wrong because evaporation removes particles selectively. C is wrong because it is the most energetic particles that escape, not the least. D is wrong because evaporation occurs at all temperatures; only boiling happens at a fixed temperature.✓ 1Answer A

Answer: A

Common error: Do not confuse evaporation, which happens at any temperature from the surface only, with boiling.

161IGPC3-2-07r17-06Particle model and gas pressure Paper CMedium1 mark

A strong, sealed steel cylinder of fixed volume contains a gas. The cylinder is heated and the pressure of the gas rises.
Which statement explains the rise in pressure?

Show mark scheme
Marking pointMarkNotes
B is correct: a higher temperature gives a higher average particle speed, so there are more collisions per second with the walls and each collision exerts a larger force. A is wrong because the particles do not change size. C is wrong because the particles do not gain mass or weight. D is wrong because the cylinder has fixed volume and is strong, and any tiny expansion would lower the pressure rather than raise it.✓ 1Answer B

Answer: B

Exam tip: For a gas at fixed volume, link the temperature rise to faster particles, then to more frequent and harder collisions.

162IGPC3-2-07r17-12Particle model and gas pressure Paper CMedium7 marks

A football is pumped up with air. The air inside the ball is at a constant volume.

(a)

Explain, in terms of particles, how the air inside the ball produces a pressure on the inside surface of the ball.

(3)
(b)

More air is pumped into the ball. Explain why the pressure inside the ball increases.

(2)
(c)

The ball is left in a hot car and the pressure rises further. Explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The air particles move randomly at high speed✓ 1
They collide with the inside surface of the ball and exert a force on it✓ 1
Pressure is the force per unit area on the surface from many collisions✓ 1
Part (b)
There are more particles in the same volume✓ 1
so more collisions per second with each unit area of the surface✓ 1
Part (c)
The particles move faster at the higher temperature✓ 1
so they collide with the surface more often and with greater force✓ 1

Exam tip: Always mention both the number of collisions per second and the force of each collision when explaining pressure changes.

163IGPC3-2-07r17-15Conduction, convection and radiation Paper CMedium6 marks

On a bench in a sports hall there is a metal seat and a wooden seat. Both seats are at the same temperature of 15 °C.

(a)

A student touches both seats. The metal seat feels colder. Explain why.

(2)
(b)

Describe how thermal energy is conducted along a solid rod when one end is heated.

(2)
(c)

A heater is to be fitted in the hall. Explain why it is better to place it near the floor than near the ceiling.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Metal is a better conductor (of thermal energy) than wood✓ 1
so energy is conducted away from the student's hand faster by the metal✓ 1
Part (b)
The particles at the hot end vibrate with more energy✓ 1
They pass energy to neighbouring particles through collisions / vibrations, and so energy passes along the rod✓ 1
Part (c)
Air heated by the heater expands and becomes less dense, so it rises✓ 1
This sets up a convection current that circulates warm air through the whole room (a heater at the ceiling would not warm the lower air)✓ 1

Exam tip: Name the process (conduction, convection or radiation) before explaining it, and keep to the particle ideas for that process.

164IGPC3-2-06r25-03Particle model and gas pressure Paper CMedium1 mark

Steel deck plates on a ship are laid with small gaps between them. On a hot day the gaps become narrower. Which statement describes the change in the steel particles?

Show mark scheme
Marking pointMarkNotes
On heating, the particles vibrate with a larger amplitude and their average separation increases, so each plate expands and the gaps narrow. The particles themselves do not get larger, they move faster not more slowly, and no particles are added to the plate.✓ 1Answer D

Answer: D

Common error: Expansion is caused by particles moving further apart on average, not by the particles themselves growing in size.

165IGPC3-2-06r25-04Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

On a beach, equal masses of sand (specific heat capacity 800 J/(kg °C)) and seawater (specific heat capacity 4200 J/(kg °C)) absorb the same amount of energy from the Sun. What happens to their temperatures?

Show mark scheme
Marking pointMarkNotes
Temperature rise = energy ÷ (mass × specific heat capacity). With equal energy and mass the rise is inversely proportional to the specific heat capacity, so the sand rises by 4200 ÷ 800 ≈ 5 times as much. A larger specific heat capacity means a smaller rise, so water does not rise more; equal energy does not mean equal rise unless the specific heat capacities are equal; and the ratio is the wrong way round in the last option.✓ 1Answer B

Answer: B

Common error: A large specific heat capacity means a small temperature change for a given energy input, not a large one.

166IGPC3-2-06r25-06Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

A geologist heats a sample of sulfur powder steadily. While the sulfur melts at 115 °C its temperature stays constant. Why does the temperature not rise during melting?

Show mark scheme
Marking pointMarkNotes
During melting the energy supplied goes into separating the particles against the forces holding them in the solid, so the average kinetic energy, and hence the temperature, stays constant. The heater keeps supplying energy, the particles still vibrate, and losing exactly all the energy to the surroundings would not be a general explanation.✓ 1Answer D

Answer: D

Common error: Energy supplied during a change of state changes the potential energy of the particles, not their average kinetic energy.

167IGPC3-2-06r25-13Thermal expansion, specific heat capacity and changes of state Paper CMedium7 marks

A steel pipeline on a ship is fixed at both ends. Part of the pipeline is made as a U-shaped loop.

U-shaped loopsteel pipelinefixedfixed
Diagram not to scale
(a)

State what happens to the length of a steel pipe when its temperature rises.

(1)
(b)

Describe how the movement of the steel particles changes when the pipe is heated.

(2)
(c)

Suggest why the U-shaped loop is included in the pipeline.

(2)
(d)

A thermostat in a stage-lighting cabinet uses a strip made of two different metals joined together. Explain why the strip bends when it is heated.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
it increases (expands)✓ 1
Part (b)
the particles vibrate more (with larger amplitude / faster)✓ 1
the average separation of the particles increases✓ 1
Part (c)
the loop can flex when the pipe expands or contracts✓ 1
this stops the pipe buckling or putting stress on the fixed ends✓ 1
Part (d)
the two metals expand by different amounts✓ 1
the metal that expands more is on the outside of the curve, so the strip bends✓ 1

Common error: Say that the particles vibrate more and move further apart on average; do not say that the particles themselves expand.

168IGPC3-2-06r25-16Thermal expansion, specific heat capacity and changes of state Paper CMedium5 marks

A student heats a sample of solid sulfur steadily and records its temperature every 2 minutes.
Time / min: 0, 2, 4, 6, 8, 10, 12, 14
Temperature / °C: 85, 100, 115, 115, 115, 115, 127, 139

02468101214020406080100120140160time / mintemperature / °C
Graph plotted from the data in the question
(a)

State the melting point of sulfur from the data.

(1)
(b)

State the state or states of sulfur at 7 minutes.

(1)
(c)

State what happens to the average kinetic energy of the sulfur particles between 4 and 10 minutes.

(1)
(d)

Describe the arrangement and movement of the particles in the sulfur at 14 minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
115 °C✓ 1
Part (b)
solid and liquid (a mixture of both)✓ 1
Part (c)
it stays the same (constant)✓ 1
Part (d)
particles are close together in an irregular arrangement✓ 1
particles move around and slide past each other✓ 1

Common error: A flat section of the data means the temperature is constant, but energy is still being supplied, so it is not 'no heating'.

169IGPC3-2-09r33-03Conduction, convection and radiation Paper CMedium1 mark

An aquarium heater in a marine laboratory is fixed near the bottom of a tank of seawater rather than near the top. Why is this the better position?

Show mark scheme
Marking pointMarkNotes
Warmed water expands, becomes less dense and rises while cooler, denser water sinks to replace it, so convection currents distribute the energy. Water is a poor conductor, 'heat' does not itself rise by conduction, and radiation is not the main process in water.✓ 1Answer A

Answer: A

Exam tip: Explain convection using density: the warmed fluid expands, becomes less dense and rises.

170IGPC3-2-09r33-04Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

Lumps of ice at 0 °C are heated steadily in a beaker in a marine laboratory. While the ice is melting, the thermometer stays at 0 °C even though energy is still being supplied. What happens to the energy?

Show mark scheme
Marking pointMarkNotes
During melting the supplied energy overcomes the attractions between particles, so the average speed of the particles (and hence the temperature) stays the same. It is not all lost to the surroundings, it does not raise the particle speed, and mass does not increase.✓ 1Answer C

Answer: C

Common error: A flat section on a heating graph does not mean energy has stopped being supplied; it is being used to change the state.

171IGPC3-2-09r33-08Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

Water evaporates from the surface of a shallow tide pool and the remaining water becomes cooler. Why does the remaining water cool?

Show mark scheme
Marking pointMarkNotes
Only molecules with enough energy can escape the surface, so the molecules left behind have a lower average kinetic energy and the temperature falls. If the slowest escaped, the pool would warm. The slowing is not caused by pushing against the air, and the process is not radiation only.✓ 1Answer C

Answer: C

Exam tip: In evaporation explanations say that the more energetic molecules escape, lowering the average energy of those left.

172IGPC3-2-09r33-14Thermal expansion, specific heat capacity and changes of state Paper CMedium6 marks

Engineers design a steel bridge. Small gaps are left between sections of the road deck.

brasssteelstraight at room temperature
Diagram not to scale
(a)

Explain why the gaps are needed.

(2)
(b)

Explain, in terms of particles, why a solid steel beam expands when heated.

(2)
(c)

A strip made of brass bonded to steel is heated. Brass expands more than steel for the same rise in temperature. State which metal is on the outside of the curve when the strip bends, and explain why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Steel expands when its temperature rises✓ 1
The gaps give room for expansion, preventing buckling or cracking✓ 1
Part (b)
The particles vibrate with larger amplitude✓ 1
The average separation of the particles increases (the particles themselves do not get bigger)✓ 1
Part (c)
Brass is on the outside of the curve✓ 1
Brass expands more, so the brass layer becomes longer than the steel layer✓ 1

Common error: Do not say that the atoms themselves expand; the spaces between them increase because they vibrate more.

173IGPC3-2-09r33-16Conduction, convection and radiation Paper CMedium6 marks

Two identical tanks at a port are filled with the same mass of fuel at the same temperature. One tank is painted matt black and the other is painted shiny white. Both stand outdoors.

(a)

At night, state which tank cools faster and explain why.

(2)
(b)

The tanks are placed side by side to compare their rates of cooling. Suggest two variables that must be controlled to make this a fair test.

(2)
(c)

In steady sunshine the temperature of the matt black tank rises, then stops rising. Explain why the temperature stops rising.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The matt black tank cools faster✓ 1
Matt black surfaces are better emitters of infrared radiation than shiny white surfaces✓ 1
Part (b)
Any one: same mass or volume of fuel, same initial temperature, same size and shape of tank, same wall thickness, same wind or surroundings✓ 1
Any second different variable from the list above✓ 1
Part (c)
As the tank gets hotter it emits radiation at a greater rate✓ 1
When the rate of emitting energy equals the rate of absorbing energy, the temperature is constant✓ 1

Exam tip: In a fair-test question, name the variable that is controlled and say that it is kept the same for both tanks.

174IGPC3-2-08r36-07Conduction, convection and radiation Paper CMedium1 mark

A vaccine cool box is left outside in direct sunshine. Which outer surface will keep the contents coolest?

Show mark scheme
Marking pointMarkNotes
A shiny silver surface reflects most of the incident thermal radiation and is a poor absorber, so least energy enters the box. Black surfaces are the best absorbers whether matt or shiny. Matt dark grey also absorbs a large share of the radiation.✓ 1Answer D

Answer: D

Common error: A shiny surface only helps if it is also light-coloured; a shiny black surface still absorbs radiation well.

175IGPC3-2-08r36-08Conduction, convection and radiation Paper CMedium1 mark

An immersion heater at the bottom of a water tank heats the water around it. Which statement explains why the heated water rises?

waterimmersion heater
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Heated water expands, so the same mass occupies a larger volume, which lowers its density. The surrounding cooler, denser water pushes it upwards, which sets up a convection current. Heat is a form of energy transfer and not a substance. The particles do not get lighter; they stay the same mass. The water is well below boiling point and the rising is not caused by steam.✓ 1Answer A

Answer: A

Common error: Do not write that heat rises; say that the warmer fluid is less dense so it rises.

176IGPC3-2-08r36-10Conduction, convection and radiation Paper CMedium7 marks

A solar water heater has a flat metal panel on a roof. Water in copper pipes behind the panel is warmed and flows to a storage tank.

roofstorage tankmatt black panelcopper pipes
Diagram not to scale
(a)

The front of the panel is painted matt black. Explain why this makes the heater work better.

(2)
(b)

State the name of the process by which energy from the Sun reaches the panel.

(1)
(c)

Suggest why the pipes are made of copper rather than plastic.

(1)
(d)

Explain how a convection current moves warm water from the panel to the tank.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
matt black is a good absorber of thermal (infrared) radiation✓ 1
so more energy is absorbed and the water heats up faster✓ 1
Part (b)
radiation✓ 1
Part (c)
copper is a good conductor of thermal energy, so energy passes quickly into the water✓ 1
Part (d)
water in the panel is heated and expands✓ 1
it becomes less dense than the cooler water so it rises✓ 1
cooler, denser water sinks and moves in to take its place, which sets up a circulation✓ 1

Common error: In convection explanations, say the water becomes less dense; do not say that the heat rises.

177IGPC3-2-08r36-11Thermal expansion, specific heat capacity and changes of state Paper CMedium7 marks

A nurse wipes a patient's skin with an alcohol swab before an injection. The skin feels cool as the alcohol dries.

(a)

Explain, in terms of particles, why evaporation of the alcohol makes the skin feel cool.

(3)
(b)

State two differences between evaporation and boiling.

(2)
(c)

Give two changes that would make the alcohol evaporate faster.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the fastest or most energetic particles escape from the surface of the liquid✓ 1
the particles left behind have a lower average kinetic energy✓ 1
so the temperature of the liquid falls and energy flows from the skin into it✓ 1
Part (b)
first valid difference, for example evaporation happens at any temperature but boiling happens only at the boiling point✓ 1
second valid difference, for example evaporation happens only at the surface but boiling happens throughout the liquid, or evaporation is slow but boiling is rapid✓ 1
Part (c)
first valid change, for example a higher temperature or a larger surface area✓ 1
second valid change, for example air blowing over the surface or drier air✓ 1

Exam tip: Evaporation cools because the particles that escape are the ones with the most energy.

178IGPC3-2-08r36-13Particle model and gas pressure Paper CMedium6 marks

A steel pressure tank in a water-supply system contains air above the water. The air is trapped in the top of the tank.

(a)

Explain, in terms of particles, how the trapped air exerts a pressure on the walls of the tank.

(2)
(b)

More water is pumped into the tank so the volume of the air decreases. The temperature of the air stays constant. Explain why the pressure of the air increases.

(2)
(c)

The sun then warms the tank without changing the volume of the air. State what happens to the pressure of the air and give a reason.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
air particles move randomly and collide with the walls✓ 1
each collision exerts a force on the wall, and pressure is the force on each unit of area✓ 1
Part (b)
the same number of particles is in a smaller volume✓ 1
so the particles collide with the walls more often, which increases the force on each unit of area✓ 1
Part (c)
pressure increases✓ 1
particles move faster so they hit the walls more often and with greater force✓ 1

Common error: A gas has a pressure because of particle collisions with the walls, not because the particles push each other.

179IGPC3-2-08r36-14Thermal expansion, specific heat capacity and changes of state Paper CMedium6 marks

A gaming console has a safety cut-out containing a bimetallic strip made from brass joined to iron. For the same rise in temperature, brass expands more than iron.

(a)

Explain why the bimetallic strip bends when it is heated.

(3)
(b)

Describe how the strip can be used to switch off the console when it overheats.

(2)
(c)

Explain, in terms of particles, why brass expands when it is heated.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
brass expands more than iron✓ 1
the two metals are fixed together so the strip cannot stay straight✓ 1
the strip bends with the brass on the outside of the curve✓ 1
Part (b)
when the temperature gets too high the strip bends away from the contact✓ 1
this opens the contacts and breaks the circuit✓ 1
Part (c)
particles vibrate with a larger amplitude and move slightly further apart✓ 1

Exam tip: State which metal expands more and which side of the curve it ends up on.

180IGPC3-2-09r41-03Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

A ship's engineer compares a 1.0 kg block of aluminium with a 1.0 kg block of lead. They start at the same temperature and the same quantity of thermal energy is supplied to each. Aluminium has a specific heat capacity of 900 J/(kg °C) and lead has 130 J/(kg °C). Which statement is correct?

Show mark scheme
Marking pointMarkNotes
The temperature rise is energy divided by (mass × specific heat capacity). Both have the same mass and energy, so the material with the smaller specific heat capacity, lead, rises more. A larger specific heat capacity means a smaller rise, so the first wrong statement is reversed. Equal mass and energy do not give equal rises because the specific heat capacities differ. Density does not determine the temperature rise.✓ 1Answer A

Answer: A

Common error: A large specific heat capacity means a substance heats up slowly, not quickly.

181IGPC3-2-09r41-07Conduction, convection and radiation Paper CMedium1 mark

Four identical metal cans each hold 0.50 kg of water at 90 °C and stand in the same cool theatre dressing room. The outer surfaces of the cans are dull black, shiny silver, dull white and shiny white. Which can cools fastest?

Show mark scheme
Marking pointMarkNotes
Dull black surfaces are the best emitters of infrared radiation, so that can loses energy fastest. Shiny silver is a poor emitter, so that can cools slowly. Dull white and shiny white surfaces are poorer emitters than dull black, so those cans cool more slowly. The shininess or the pale colour does not make a surface emit better.✓ 1Answer A

Answer: A

Exam tip: Good absorbers of radiation are also good emitters: dull black is best, shiny silver is worst.

182IGPC3-2-09r41-11Particle model and gas pressure Paper CMedium7 marks

Air is stored in a sealed rigid steel tank on a ship.

(a)

Describe the movement of the particles in a gas.

(2)
(b)

Explain how the air in the tank produces a pressure on the walls of the tank.

(2)
(c)

The tank is left in strong sunlight and the temperature of the air rises. Explain, in terms of particles, why the pressure increases.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
the particles move randomly in all directions✓ 1
the particles move quickly and are far apart, colliding with each other and the walls✓ 1
Part (b)
particles collide with the walls and exert a force on them✓ 1
pressure is force per unit area, so the many collisions over the wall area give a pressure✓ 1
Part (c)
the particles gain kinetic energy and move faster✓ 1
the particles hit the walls more often✓ 1
the particles hit the walls with greater force each time✓ 1

Common error: Do not say the particles expand when heated; the volume of a rigid tank stays the same.

183IGPC3-2-09r41-15Conduction, convection and radiation Paper CMedium5 marks

A theatre lamp is cooled by a metal heat sink that has thin fins painted dull black.

(a)

Name the method by which thermal energy can be transferred through a vacuum.

(1)
(b)

Explain why painting the fins dull black helps the heat sink to cool the lamp.

(2)
(c)

Suggest why the heat sink has many thin fins rather than being one solid block of the same mass.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
radiation✓ 1
Part (b)
dull black surfaces are good emitters of infrared radiation✓ 1
so more energy is radiated away from the fins each second✓ 1
Part (c)
the fins give a larger surface area✓ 1
so more energy is transferred to the air each second by radiation and convection✓ 1

Exam tip: Link every surface answer to a property: colour and texture affect radiation, surface area affects how much of the surface is exposed.

184IGPC3-2-07r42-03Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

Sections of a steel footbridge are laid with small gaps between them. Which statement gives the reason for the gaps?

Show mark scheme
Marking pointMarkNotes
When the temperature rises the steel particles vibrate more and the metal expands. The gaps give room for this expansion so the sections do not buckle. Steel does not contract on heating. Gaps of this size have no useful effect on convection, and they are not designed to reduce conduction along the bridge.✓ 1Answer A

Answer: A

Common error: State that the material expands on heating; never say the particles themselves get bigger.

185IGPC3-2-07r42-04Particle model and gas pressure Paper CMedium1 mark

A sealed steel cylinder of gas of fixed volume stands on a ship's deck and is heated by the Sun. Which statement explains why the gas pressure increases?

Show mark scheme
Marking pointMarkNotes
Heating raises the average kinetic energy of the particles, so they move faster. They strike the walls more frequently and each collision exerts a larger force, so the pressure rises. The particles do not expand, no extra particles enter a sealed cylinder, and there is no increased attraction to the walls.✓ 1Answer C

Answer: C

Exam tip: For pressure changes at constant volume, give two effects: collisions are more frequent and each is harder.

186IGPC3-2-07r42-08Conduction, convection and radiation Paper CMedium1 mark

Warm air collects under the roof of a large warehouse, and a vent is opened in the roof. Which statement explains why the warm air rises towards the vent?

Show mark scheme
Marking pointMarkNotes
Warming makes the air expand so its density falls. Air that is less dense than its surroundings is pushed upwards by the denser cooler air. The particles do not become lighter, they are not attracted to the roof, and air is a poor conductor so conduction is not the mechanism.✓ 1Answer C

Answer: C

Common error: Say the warm air is less dense; the individual particles do not change mass.

187IGPC3-2-07r42-10Conduction, convection and radiation Paper CMedium6 marks

An architect designs a glass-fronted atrium for a marine research centre. On sunny days the room is warmed and a radiator beneath the window warms the air.

(a)

Explain how thermal energy is conducted from the warm room side to the cold outside through a solid glass pane.

(2)
(b)

Describe how convection currents carry thermal energy around the room.

(3)
(c)

The glass is replaced by double glazing with a sealed layer of air between two panes. State why this reduces conduction.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Particles on the warm side vibrate with more energy✓ 1
energy is passed to neighbouring particles by vibrations / collisions through the glass✓ 1
Part (b)
Air above the radiator is warmed, expands and becomes less dense✓ 1
the warm air rises✓ 1
cooler, denser air sinks to take its place, setting up a circulation✓ 1
Part (c)
Air is a poor conductor (an insulator)✓ 1

Exam tip: Describe convection as a sequence: warms, expands, less dense, rises, then cooler air sinks.

188IGPC3-2-07r42-11Particle model and gas pressure Paper CMedium5 marks

Steel gas cylinders are loaded onto a delivery lorry. Each cylinder is sealed and contains gas.

(a)

Explain how the gas particles produce a pressure on the walls of a cylinder.

(2)
(b)

The lorry stands in strong sunshine and the temperature of the gas rises. The volume of the cylinder does not change. Explain why the pressure of the gas increases.

(2)
(c)

Suggest why the cylinders are loaded in a shaded area.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The particles move randomly and collide with the walls✓ 1
each collision exerts a force on the wall, and the force per unit area is the pressure✓ 1
Part (b)
The particles gain kinetic energy and move faster✓ 1
they hit the walls more often and with greater force✓ 1
Part (c)
To stop the pressure rising too high, which could damage or burst the cylinders✓ 1

Exam tip: In gas pressure explanations always mention the walls and collisions, not just 'particles moving faster'.

189IGPC3-2-07r42-16Thermal expansion, specific heat capacity and changes of state Paper CMedium5 marks

A thermostat in a seaweed-farm greenhouse contains a bimetallic strip made from brass and steel joined together. Brass expands more than steel for the same temperature rise.

brasssteelstraight at room temperature
Diagram not to scale
(a)

Explain why the strip bends when it is heated.

(2)
(b)

State which metal is on the outside of the curve when the strip is heated.

(1)
(c)

A long steel pipeline carrying hot water at a desalination plant is built with bends rather than being perfectly straight. Suggest why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Brass expands more than steel✓ 1
the two metals are fixed together, so the strip must bend✓ 1
Part (b)
Brass✓ 1
Part (c)
The pipe expands when it gets hot✓ 1
the bends take up the extra length without damaging the joints✓ 1

Common error: The metal that expands more always ends up on the outside of the curve.

190IGPC3-2-08r47-03Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

A steel cable hangs between two wind-farm masts. It sags more on a hot afternoon than on a cold morning. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
Heating increases the vibrations of the particles in a solid, so on average they sit slightly further apart and the whole cable lengthens, so it sags more. The particles themselves do not expand. No particles are added to the cable. Mass does not increase on heating, and the sag is due to the longer length, not extra weight.✓ 1Answer A

Answer: A

Exam tip: In expansion questions, say the particles move further apart on average; the particles themselves stay the same size.

191IGPC3-2-08r47-04Conduction, convection and radiation Paper CMedium1 mark

A swimming-pool company covers a pool with a floating plastic sheet so that sunlight warms the water. Which outer surface of the sheet would absorb the most infrared radiation from the Sun?

Show mark scheme
Marking pointMarkNotes
Dull black surfaces are the best absorbers of infrared radiation. Shiny silver is a good reflector and a poor absorber. Shiny white and dull white surfaces reflect much of the radiation, so they absorb less than dull black.✓ 1Answer C

Answer: C

Exam tip: Good absorbers are dull and black; good reflectors are shiny and light-coloured. Good absorbers are also good emitters.

192IGPC3-2-08r47-08Thermal expansion, specific heat capacity and changes of state Paper CMedium1 mark

Ice on a turbine blade is melted by a heater. The ice is at 0 °C and the heater supplies energy at a constant rate. What happens while the ice is melting?

Show mark scheme
Marking pointMarkNotes
During melting the energy supplied overcomes the forces holding the particles in the solid, so the temperature stays constant at the melting point. A steady temperature rise would only occur once all of the ice has melted. The heater continues to supply energy throughout. The particles keep vibrating or moving; their average kinetic energy is simply not changing.✓ 1Answer A

Answer: A

Exam tip: A flat section on a heating curve means energy is going into changing state, not into raising the temperature.

193IGPC3-2-08r47-12Thermal expansion, specific heat capacity and changes of state Paper CMedium6 marks

A swimming pool has a concrete deck. A pool attendant uses an alcohol-in-glass thermometer to measure the water temperature.

alcoholnarrow glass tube
Simplified diagram
(a)

Narrow gaps are left between the sections of the concrete deck. Explain why.

(2)
(b)

The thermometer is placed in warm water. Explain, in terms of particles, why the alcohol rises up the tube.

(3)
(c)

State which of a solid, a liquid or a gas expands the most for the same rise in temperature.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
concrete expands when it gets hot✓ 1
the gaps allow room for expansion so the deck does not crack or buckle✓ 1
Part (b)
the alcohol particles gain energy and move faster✓ 1
the particles move further apart on average, so the alcohol expands✓ 1
the alcohol expands more than the glass, so the level rises up the tube✓ 1
Part (c)
gas✓ 1

Common error: The particles do not get bigger when heated; they move or vibrate more and spread further apart.

194IGPC3-2-08r47-14Particle model and gas pressure Paper CMedium6 marks

A librarian shines a bright lamp through a glass cell of air containing smoke. Through a microscope she sees tiny bright specks of smoke moving about in a random, jerky way.

(a)

Describe the movement of the smoke specks.

(1)
(b)

Explain what causes this movement.

(2)
(c)

The air in the cell is warmed. State and explain the change in the movement of the specks.

(2)
(d)

Suggest why the air particles themselves cannot be seen in the microscope.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
random, jerky (zig-zag) motion in changing directions✓ 1
Part (b)
invisible air particles move randomly at high speed✓ 1
they collide with the smoke specks, and the random collisions give an unbalanced force that changes the speck's direction✓ 1
Part (c)
the movement becomes more vigorous✓ 1
the air particles move faster, so they hit the specks harder and more often✓ 1
Part (d)
they are far too small to be seen✓ 1

Exam tip: The specks are visible but the air particles are not; the speck's motion is evidence that the air particles are moving.

195IGPC3-2-08r47-15Conduction, convection and radiation Paper CMedium6 marks

A solar heating panel for a swimming pool consists of plastic-covered copper pipes laid on a roof. Water from the pool is pumped through the pipes. There is a layer of foam insulation between the pipes and the roof. The pipes are painted dull black.

(a)

Explain why the pipes are painted dull black.

(2)
(b)

Explain why a layer of foam is placed between the pipes and the roof.

(2)
(c)

The pipes are made of copper rather than plastic. Suggest why.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
dull black surfaces are good absorbers of infrared radiation✓ 1
so more energy from the Sun is absorbed and transferred to the water✓ 1
Part (b)
foam (which traps air) is a poor conductor / good insulator✓ 1
so less thermal energy is lost by conduction from the pipes to the roof✓ 1
Part (c)
copper is a good conductor of thermal energy✓ 1
so energy passes quickly from the surface of the pipe to the water inside✓ 1

Exam tip: Name the property (good absorber, poor conductor) and then link it to the energy transfer it changes.

196IGPC3-2-09r47-07Conduction, convection and radiation Paper CMedium1 mark

A mountaineer wraps an emergency blanket with a shiny silver surface on the inside, next to the body. How does this reduce energy loss from the body?

Show mark scheme
Marking pointMarkNotes
Shiny silver surfaces are poor emitters and poor absorbers of thermal radiation and reflect most of it, so radiation from the body is reflected back instead of being lost. Being a good conductor would help energy escape, a thin shiny layer does not trap warm air by convection, and a shiny surface absorbs very little radiation.✓ 1Answer C

Answer: C

Common error: Shiny surfaces are poor absorbers and poor emitters of radiation; dull black surfaces are the good absorbers and emitters.

197IGPC3-2-09r47-12Thermal expansion, specific heat capacity and changes of state Paper CMedium7 marks

A student heats 0.10 kg of a wax in a water bath and records its temperature every 2 minutes.

time / min02468101214
temperature / °C3040506060606979

024681012140102030405060708090time / mintemperature / °C
Graph plotted from the data in the question
(a)

State the melting point of the wax.

(1)
(b)

Explain, in terms of particles, why the temperature stays constant between 6 minutes and 10 minutes although energy is still being supplied.

(2)
(c)

Calculate the average rate of temperature rise during the first 6 minutes.

(2)
(d)

Describe one difference between the arrangement of the particles in the solid wax and in the liquid wax.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 60 °C✓ 1
Part (b)
B1 the energy is used to overcome the forces between the particles (to change the state)✓ 1
B1 so the average kinetic energy of the particles does not increase, and the temperature stays constant✓ 1
Part (c)
M1 rate = change in temperature / time = (60 − 30) / 6✓ 1
A1 5.0 °C / min✓ 1
Part (d)
B1 in the solid the particles are in a regular pattern and in fixed positions (vibrating on the spot)✓ 1
B1 in the liquid the particles are in an irregular arrangement and can move past each other✓ 1

Exam tip: Between 6 and 10 minutes the energy changes the state of the wax, not its temperature; say that the energy is used to overcome forces between particles.

198IGPC3-2-09r47-14Conduction, convection and radiation Paper CMedium6 marks

The metal cabinet of a telecoms relay station on a mountain is heated by an electric heater fixed near the floor of the room next to it.

(a)

Name the process by which thermal energy passes through the solid metal wall of the cabinet.

(1)
(b)

Explain how convection currents carry energy from the heater to the rest of the room.

(3)
(c)

The outside of the cabinet is painted white and shiny. Outline why this reduces the energy the cabinet absorbs from sunshine.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 conduction✓ 1
Part (b)
B1 air near the heater is warmed and expands, so its density decreases✓ 1
B1 the warm, less dense air rises✓ 1
B1 cooler, denser air sinks to take its place, so a circulation is set up✓ 1
Part (c)
B1 white and shiny surfaces are poor absorbers of thermal radiation✓ 1
B1 they reflect most of the radiation that arrives✓ 1

Exam tip: For convection answers, use the sequence warm, expands, less dense, rises, then cooler air sinks.

199IGPC3-2-07r15-08Thermal expansion, specific heat capacity and changes of state Paper CHard1 mark

A strip made of brass bonded to iron is straight at room temperature. Brass expands more than iron for the same temperature rise. The strip is heated evenly. Which description is correct?

brassironstraight at room temperature
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
A is correct: the brass lengthens more, so it must lie on the outer side of the curve. B puts the metals the wrong way round. C is wrong because unequal expansion of bonded metals makes the strip bend. D is wrong because the strength of the metal is irrelevant to the direction of bending.✓ 1Answer A

Answer: A

Exam tip: Whichever metal expands more is always on the outside of the curve when the strip is heated.

200IGPC3-2-08r36-16Conduction, convection and radiation Paper CHard11 marks

A pharmacist tests materials for a medicine cool box. Boxes of identical size, with walls of the same thickness, are each filled with the same mass of ice at 0 °C and left in a warm room. The time taken for all the ice to melt is recorded.
expanded polystyrene foam: 310 min
plywood: 120 min
cardboard: 85 min
aluminium sheet: 30 min

PolystyrenePlywoodCardboardAluminium0100200300time for ice to melt / min3101208530
Bar chart drawn from the data in the question
(a)

State which material is the best insulator and give a reason using the data.

(2)
(b)

State two variables that must be kept the same to make this a fair test.

(2)
(c)

Expanded polystyrene foam contains many tiny pockets of trapped air. Explain why this makes it a good insulator.

(3)
(d)

Another foam box has walls twice as thick and the ice melts in 590 min. State what this suggests about the thickness of the walls and the rate of energy transfer.

(2)
(e)

The outside of the box is covered with shiny aluminium foil. Explain how this helps to keep the ice from melting.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
expanded polystyrene foam✓ 1
the ice took longest to melt, so energy passed through the walls most slowly✓ 1
Part (b)
first valid variable, for example mass of ice✓ 1
second valid variable, for example room temperature, starting temperature of the ice or thickness of the walls✓ 1
Part (c)
air is a poor conductor because its particles are far apart✓ 1
so energy passes slowly from particle to particle by collisions✓ 1
the pockets are so small that the air cannot circulate, so there is very little convection✓ 1
Part (d)
thicker walls give a longer melting time✓ 1
so the rate of energy transfer into the box is smaller✓ 1
Part (e)
shiny surfaces are poor absorbers of thermal radiation and good reflectors✓ 1
so less energy from the room reaches the ice✓ 1

Exam tip: Explain insulation through the particles: widely spaced air particles conduct poorly, and trapped air cannot form convection currents.

201IGPC3-2-09r47-08Thermal expansion, specific heat capacity and changes of state Paper CHard1 mark

0.20 kg of water at 80 °C is poured into 0.30 kg of water at 20 °C in an insulated container. No energy is lost to the surroundings. What is the final temperature of the mixture?

Show mark scheme
Marking pointMarkNotes
Energy lost by hot water = energy gained by cold water, so 0.20 × (80 − T) = 0.30 × (T − 20). This gives 16 − 0.20T = 0.30T − 6, so 0.50T = 22 and T = 44 °C. The value 50 °C is the simple average of the temperatures, 56 °C comes from swapping the two masses, and 100 °C comes from adding the temperatures, which would be hotter than either sample.✓ 1Answer D

Answer: D

Exam tip: The final temperature must lie between the two starting temperatures and closer to the temperature of the larger mass.

202IGPC3-2-09r47-16Thermal expansion, specific heat capacity and changes of state Paper CHard7 marks

A steel telecoms cable of mass 12 kg is hung between two towers on a hot summer day. The specific heat capacity of steel is 490 J / (kg °C).

(a)

Explain, in terms of particles, why the cable gets longer when it is heated.

(2)
(b)

The cable is fitted on the hot summer day and is left slightly slack. Explain why this is necessary.

(2)
(c)

Sunshine heats the cable at a rate of 98 W for 25 minutes. Assume all of this energy raises the temperature of the cable. Calculate the rise in temperature of the cable.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the particles vibrate with greater amplitude (more kinetic energy)✓ 1
B1 so the average separation of the particles increases, and the cable expands✓ 1
Part (b)
B1 in winter the cable contracts (gets shorter)✓ 1
B1 a taut cable would be put under great tension and could snap or pull the towers✓ 1
Part (c)
M1 energy = power × time = 98 × (25 × 60) = 147000 J✓ 1
M1 temperature rise = energy / (mass × specific heat capacity) = 147000 / (12 × 490)✓ 1
A1 25 °C✓ 1

Common error: Convert minutes to seconds before using energy = power × time, or the temperature rise will be 60 times too small.

203IGPC-036-01-02Thermal expansion, specific heat capacity and changes of state Paper EEasy1 mark

Equal volumes of a solid, a liquid and a gas are each heated through the same temperature rise. The gas is kept at constant pressure.
Which list puts them in order of expansion, from largest to smallest?

Show mark scheme
Marking pointMarkNotes
Gas particles are far apart and the forces between them are very weak, so a gas expands by far the most. Liquid particles are close together but not held in fixed positions, so a liquid expands more than a solid. Solid particles are held strongly in fixed positions, so a solid expands least. The order is gas, liquid, solid.✓ 1Answer C

Answer: C

Common error: Do not assume that a solid expands most because its particles are packed most tightly. Stronger forces between the particles mean less expansion.

204IGPC3-2-07r42-06Thermal expansion, specific heat capacity and changes of state Paper EEasy1 mark

A sample of seawater of mass 0.50 kg has a specific heat capacity of 4000 J/(kg °C). How much energy is needed to raise its temperature by 8.0 °C?

Show mark scheme
Marking pointMarkNotes
E = mcΔθ = 0.50 × 4000 × 8.0 = 1.6 × 104 J. The value 2.5 × 102 J comes from dividing by the temperature change. The value 6.4 × 104 J comes from dividing by the mass. The value 3.2 × 104 J comes from leaving out the mass.✓ 1Answer D

Answer: D

Exam tip: Write the equation E = mcΔθ and substitute all three quantities before calculating.

205IGPC-036-02-05Conduction, convection and radiation Paper EMedium1 mark

Why are metals much better thermal conductors than non-metal solids?

Show mark scheme
Marking pointMarkNotes
In all solids, energy is passed on as neighbouring atoms or ions vibrate against each other (lattice vibrations). Metals also contain free electrons. These gain kinetic energy at the hot end, travel quickly through the metal and collide with ions further along, transferring energy. This extra mechanism makes metals much better conductors. The atoms in a solid stay in fixed positions and do not travel along it.✓ 1Answer A

Answer: A

Common error: In a solid the atoms vibrate about fixed positions. It is the free electrons that move through a metal, not the atoms.

206IGPC-036-02-08Conduction, convection and radiation Paper EMedium1 mark

Over many years, the average temperature of the Earth stays approximately constant. Which statement explains this?

Show mark scheme
Marking pointMarkNotes
An object stays at a constant temperature when the rate at which it gains energy equals the rate at which it loses energy. The Earth absorbs radiation from the Sun and emits infrared radiation into space. When these two rates are equal, the average temperature does not change. If emission were greater than absorption, the Earth would cool.✓ 1Answer A

Answer: A

Exam tip: Constant temperature does not mean no radiation is absorbed or emitted. It means the two rates are equal.

207IGPC-036-00-05Particle model and gas pressure Paper EMedium1 mark

A fixed mass of gas has a volume of 120 cm3 at a pressure of 100 kPa. It is compressed slowly to a volume of 40 cm3. The temperature stays constant.
What is the new pressure of the gas?

Show mark scheme
Marking pointMarkNotes
At constant temperature, pV = constant, so p1V1 = p2V2.
100 × 120 = p2 × 40, so p2 = 300 kPa.
The volume falls to one third, so the pressure triples. Option A treats pressure as directly proportional to volume. Option B adds the change in volume to the pressure. Option D is p × V, which is not a pressure.
✓ 1Answer C

Answer: C

Exam tip: Check the direction of your answer. A smaller volume must give a larger pressure, and a larger volume must give a smaller pressure.

208IGPC-036-00-06Particle model and gas pressure Paper EMedium1 mark

Pollen grains suspended in water are seen to move randomly. Each pollen grain has a mass much larger than a water molecule.
Which statement explains why the pollen grains move?

Show mark scheme
Marking pointMarkNotes
Each pollen grain is a massive particle bombarded by huge numbers of light, fast-moving water molecules. The impacts on different sides do not balance at every instant, so there is a net force that changes randomly. This gives the grain a random, jerky motion. Water molecules are much lighter than pollen grains (B). The same effect is seen with non-living particles such as smoke (C). The motion is caused by collisions, not attraction (D).✓ 1Answer A

Answer: A

Exam tip: A full explanation of Brownian motion needs two ideas: the observed particle is massive, and the molecules hitting it are light and fast-moving.

209IGPC-036-00-10Particle model and gas pressure Paper EMedium1 mark

A fixed mass of gas in a cylinder is at a pressure of 90 kPa. The piston is pulled out slowly so that the volume of the gas increases by 50%. The temperature remains constant.
What is the new pressure?

Show mark scheme
Marking pointMarkNotes
The new volume is 1.5 times the original volume. With pV constant, the new pressure is 90 ÷ 1.5 = 60 kPa. Option A halves the pressure, treating 'increases by 50%' as 'doubles'. Option C assumes that constant temperature means constant pressure. Option D multiplies by 1.5 instead of dividing.✓ 1Answer B

Answer: B

Common error: An increase of 50% means multiplying the volume by 1.5, not by 2. So the pressure is divided by 1.5.

210IGPC-036-01-06Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

An aluminium block has a mass of 2.0 kg. It is heated from 20 °C to 35 °C.
The specific heat capacity of aluminium is 900 J/(kg °C).
How much thermal energy does the block gain?

Show mark scheme
Marking pointMarkNotes
Use ΔE = mcΔθ.
The temperature rise is Δθ = 35 − 20 = 15 °C.
ΔE = 2.0 × 900 × 15 = 27 000 J.
The other options come from common mistakes:
• 13 500 J leaves out the mass.
• 63 000 J uses the final temperature (35 °C) instead of the temperature rise.
• 120 J divides by Δθ instead of multiplying.
✓ 1Answer B

Answer: B

Common error: Δθ is the change in temperature. Always subtract the starting temperature from the final temperature first.

211IGPC-036-01-07Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

Two blocks, P and Q, have the same mass. Each block is given the same quantity of thermal energy. The temperature of P rises by 5 °C and the temperature of Q rises by 20 °C.
Which statement about their specific heat capacities is correct?

Show mark scheme
Marking pointMarkNotes
Specific heat capacity is c = ΔE/(mΔθ). Here ΔE and m are the same for both blocks, so c is inversely proportional to Δθ.
P has one quarter of the temperature rise of Q, so the specific heat capacity of P is four times that of Q.
Only the temperature change matters, not the starting temperature.
✓ 1Answer A

Answer: A

Exam tip: For the same energy and the same mass, a smaller temperature rise means a larger specific heat capacity.

212IGPC-036-01-08Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

Some water is left in an open dish in a room.
Which change would not increase the rate of evaporation of the water?

Show mark scheme
Marking pointMarkNotes
A lid traps the vapour above the water. Escaped particles cannot move away and many return to the liquid, so the rate of evaporation falls.
The other three changes all increase the rate:
• Warming gives more particles enough energy to escape.
• Moving air carries vapour particles away from the surface.
• A wider dish has a larger surface area, so more particles can escape.
✓ 1Answer D

Answer: D

Exam tip: Three factors increase the rate of evaporation: higher temperature, larger surface area and more air movement.

213IGPC-036-01-13Thermal expansion, specific heat capacity and changes of state Paper EMedium7 marks

A student measures the specific heat capacity of aluminium. An electric heater and a thermometer are placed in holes in a 1.0 kg aluminium block.
The heater is connected to a 12 V supply and the current is 4.0 A. The heater is switched on for 5.0 minutes. The temperature of the block rises from 22.0 °C to 38.0 °C.

A12 Vthermometerheateraluminium block, 1.0 kg
Diagram not to scale
(a)

Calculate the electrical energy supplied by the heater.

(2)
(b)

Calculate the specific heat capacity of aluminium from these results.

(2)
(c)

Some energy is transferred from the block to the surroundings. Explain why this makes the calculated value larger than the true value.

(2)
(d)

Suggest one improvement to the experiment that would reduce this effect.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = IVt = 4.0 × 12 × 300✓ 1
E = 14 400 J✓ 1
Part (b)
c = E/(mΔθ) = 14 400 / (1.0 × 16.0)✓ 1
c = 900 J/(kg °C)✓ 1
Part (c)
Not all of the energy supplied goes into the block, so the temperature rise is smaller than it would otherwise be.✓ 1
The calculation divides the full energy supplied by this smaller Δθ, giving a value of c that is too large.✓ 1
Part (d)
Any one, e.g. wrap the block in insulation (lagging) / place the block on an insulating mat / cover the top surface of the block with an insulating lid.✓ 1

Exam tip: Write the equation, substitute the values, then give the answer with its unit. Each of these steps can earn credit.

214IGPC-036-00-13Particle model and gas pressure Paper EMedium5 marks

A syringe contains 60 cm3 of air at a pressure of 1.0 ×105 Pa. The outlet is sealed and the plunger is pushed in slowly until the volume of air is 25 cm3. The temperature of the air does not change.

outlet sealedpushed in slowlyair, 60 cm³1.0 × 10⁵ Pafinal volume 25 cm³
Diagram not to scale
(a)

State the conditions under which the pressure multiplied by the volume of a gas remains constant.

(1)
(b)

Calculate the new pressure of the air.

(2)
(c)

Explain, in terms of the air molecules, why the pressure has increased.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The gas has a fixed mass and its temperature stays constant. Both conditions are needed for the mark .✓ 1
Part (b)
p1V1 = p2V2, so 1.0 ×105 × 60 = p2 × 25✓ 1
p2 = 2.4 ×105 Pa✓ 1
Part (c)
The same number of molecules now occupy a smaller volume, so they hit each unit area of the wall more often .✓ 1
The average speed of the molecules is unchanged because the temperature is constant, so each collision exerts the same average force (or causes the same change in momentum). The greater collision rate therefore means a greater force per unit area .✓ 1

Exam tip: The volumes can be left in cm³ as long as both are in the same unit. The answer then keeps the unit of the pressure you started with.

215IGPC-036-00-16Particle model and gas pressure Paper EMedium7 marks

A sealed box contains a gas at a pressure of 1.2 ×105 Pa. One flat wall of the box has an area of 0.020 m2.

(a)

State what is meant by pressure.

(1)
(b)

Calculate the force exerted by the gas on this wall.

(2)
(c)

Explain, in terms of the momentum of the gas molecules, how this force arises.

(3)
(d)

State the total change in momentum of the molecules striking this wall each second.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Pressure is the force acting per unit area, at right angles to the surface .✓ 1
Part (b)
F = p × A = 1.2 ×105 × 0.020✓ 1
F = 2400 N (2.4 ×103 N)✓ 1
Part (c)
Molecules moving randomly hit the wall and rebound, so their momentum changes . The wall exerts a force on each molecule to change its momentum, so each molecule exerts an equal and opposite force on the wall . Very many collisions every second give a steady average total force equal to the rate of change of momentum .✓ 3
Part (d)
2400 kg m/s (each second), because force equals the rate of change of momentum .✓ 1

Exam tip: Force equals change in momentum ÷ time. So a force of 2400 N means the molecules' momentum changes by 2400 kg m/s every second.

216IGPC-036-01-14Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

A cyclist wraps a wet cloth around a metal water bottle on a hot day. The water in the cloth evaporates and the drink inside stays cool.

(a)

Describe evaporation in terms of particles.

(2)
(b)

Explain why evaporation from the cloth cools the bottle and the drink inside.

(2)
(c)

The cyclist rides faster, so more air moves over the cloth. State and explain the effect on the rate of cooling.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The particles in a liquid have a range of (kinetic) energies.✓ 1
The more-energetic particles at the surface overcome the attractive forces between particles and escape from the liquid into the air (accept: escape from the surface, at temperatures below the boiling point).✓ 1
Part (b)
The more-energetic particles leave, so the particles left behind in the cloth have a lower average kinetic energy and the water in the cloth is at a lower temperature.✓ 1
Thermal energy is then transferred from the warmer bottle and drink to the cooler wet cloth.✓ 1
Part (c)
The rate of cooling increases.✓ 1
The moving air carries vapour particles away from the cloth, so fewer return to the liquid and evaporation is faster (accept: moist air near the cloth is replaced by drier air).✓ 1

Common error: Do not say the particles "lose heat". Say that the more-energetic particles escape, so the average kinetic energy of the particles left behind falls.

217IGPC3-2-02c-06Particle model and gas pressure Paper EMedium1 mark

A fixed mass of gas at constant temperature has a pressure of 150 kPa and a volume of 2.0 litres. The gas is compressed to 0.50 litres. What is its new pressure?

Show mark scheme
Marking pointMarkNotes
Correct: D. At constant temperature pV is constant: 150 × 2.0 = p × 0.50, so p = 600 kPa. A is wrong: this divides the pressure by 4 because the volume is divided by 4; pressure and volume are inversely proportional, so it should rise. B is wrong: this subtracts the volumes, treating the change as a simple addition to the pressure. C is wrong: this multiplies by 2, using the original 2.0 litres only instead of the ratio 2.0 ÷ 0.50.✓ 1Answer D

Answer: D

Exam tip: When the volume falls by a factor of 4 at constant temperature, the pressure rises by a factor of 4.

218IGPC3-2-02c-11Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

A block of aluminium of mass 1.5 kg is heated by an electric heater rated at 450 W. The temperature of the block rises from 18 °C to 54 °C. The specific heat capacity of aluminium is 900 J/(kg °C).

(a)

Calculate the thermal energy gained by the block.

(3)
(b)

Calculate the time for which the heater must be switched on, assuming no energy is lost.

(2)
(c)

Suggest why the actual time needed would be longer than your answer to (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
temperature change = 54 − 18 = 36 °C✓ 1
energy = mass × specific heat capacity × temperature change✓ 1
1.5 × 900 × 36 = 48600 J✓ 1
Part (b)
time = energy ÷ power✓ 1
48600 ÷ 450 = 108 s (ECF from (a))✓ 1
Part (c)
some energy is transferred to the surroundings / not all the heater's energy reaches the block✓ 1

Exam tip: Always find the temperature change first. If energy is lost to the surroundings, the real heating time is longer.

219IGPC3-2-07r7-04Particle model and gas pressure Paper EMedium1 mark

A space probe's sampler traps 600 cm3 of gas at a pressure of 1.0×105 Pa. The gas is compressed slowly to 150 cm3 with its temperature kept constant. What is the new pressure?

Show mark scheme
Marking pointMarkNotes
At constant temperature pV is constant: p = 1.0×105 × 600 ÷ 150 = 400 000 Pa = 4.0×105 Pa. Option A is wrong: this comes from inverting the ratio, so pressure falls instead of rising. Option B is wrong: pressure does change when volume changes at constant temperature. Option C is wrong: this squares the volume ratio (42); pressure is inversely proportional to volume, not its square.✓ 1Answer D

Answer: D

Exam tip: For pV = constant, check the direction: a smaller volume must give a larger pressure.

220IGPC3-2-07r7-10Thermal expansion, specific heat capacity and changes of state Paper EMedium5 marks

A sports scientist heats 1.5 kg of water for an ice-bath experiment. The water is heated from 18 °C to 78 °C in an electric kettle of power 2.1 kW. The specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the thermal energy gained by the water.

(2)
(b)

Assume that all the energy from the kettle goes into the water. Calculate the time taken.

(2)
(c)

Suggest why the real time taken is longer than the time calculated in (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = m × c × Δθ = 1.5 × 4200 × 60✓ 1
= 378 000 J (3.78×105 J)✓ 1
Part (b)
time = energy ÷ power = 378 000 ÷ 2100 (ECF from (a))✓ 1
= 180 s✓ 1
Part (c)
some energy is lost to the surroundings / heats the kettle itself✓ 1

Common error: Convert kW to W before dividing: 2.1 kW = 2100 W.

221IGPC3-2-07r7-15Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

In winter, rail workers use an electric heater to melt ice that has formed on a set of points. The mass of ice is 0.80 kg at 0 °C. The specific latent heat of fusion of ice is 3.3×105 J/kg. The heater has power 1.1 kW and only 60% of its output energy reaches the ice.

(a)

Calculate the energy needed to melt the ice completely.

(2)
(b)

Calculate the time taken for the ice to melt completely.

(2)
(c)

Explain, in terms of particles, why the temperature of the ice stays at 0 °C while it melts.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = m × l = 0.80 × 3.3×105✓ 1
= 2.64×105 J (264 000 J)✓ 1
Part (b)
useful power = 0.60 × 1100 = 660 W✓ 1
time = 264 000 ÷ 660 = 400 s (ECF from (a))✓ 1
Part (c)
the energy supplied is used to overcome the forces between particles / increase their potential energy✓ 1
the average kinetic energy of the particles does not change, so the temperature stays constant✓ 1

Exam tip: During a change of state the temperature stays constant because energy changes potential, not kinetic, energy.

222IGPC3-2-07r9-05Particle model and gas pressure Paper EMedium1 mark

Air is trapped in a syringe at a pressure of 1.0 × 105 Pa and a volume of 120 cm3. The plunger is pushed in slowly so that the temperature stays constant and the volume becomes 40 cm3. What is the new pressure of the trapped air?

outlet sealedpushed in slowlyair, 120 cm³1.0 × 10⁵ Pafinal volume 40 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
C is correct: at constant temperature pV is constant, so p = 1.0 × 105 × 120 / 40 = 3.0 × 105 Pa. A is wrong because it uses the inverted ratio of volumes (pressure would fall). B is wrong because it adds the original pressure to the correct final pressure. D is wrong because it is the product p × V for the starting state, not the final pressure.✓ 1Answer C

Answer: C

Exam tip: For constant temperature, write p₁V₁ = p₂V₂ first; a smaller volume must give a larger pressure, which checks the ratio.

223IGPC3-2-07r9-13Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

A solar-heated storage tank holds 150 kg of water. The water is heated from 18 °C to 54 °C. The specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the thermal energy gained by the water. Give your answer in megajoules (MJ) to 3 significant figures.

(2)
(b)

The panels supply energy to the water at a steady rate of 2.0 kW, with no energy losses. Use the unrounded energy from (a) to calculate the time needed. Give your answer in minutes.

(2)
(c)

In practice the heating takes 252 minutes. Calculate the percentage efficiency of the heating.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Temperature change = 54 − 18 = 36 °C✓ 1
E = 150 × 4200 × 36 = 22.7 MJ✓ 1
Part (b)
t = E / P = 2.268 × 107 / 2000 = 11 340 s✓ 1
= 189 minutes (ECF from (a))✓ 1
Part (c)
efficiency = useful (ideal) energy or time / actual energy or time × 100✓ 1
= 189 / 252 × 100 = 75 % (ECF from (b))✓ 1

Common error: Convert kilowatts to watts and seconds to minutes carefully; 2.0 kW is 2000 W.

224IGPC3-2-07r12-04Particle model and gas pressure Paper EMedium1 mark

A sealed syringe contains 60 cm3 of air at a pressure of 100 kPa. The plunger is pushed in slowly so that the volume becomes 24 cm3. The temperature of the air does not change. What is the new pressure of the air?

outlet sealedpushed in slowlyair, 60 cm³100 kPafinal volume 24 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At constant temperature pV is constant, so p = 100 × 60 ÷ 24 = 250 kPa. 40 kPa comes from inverting the ratio (100 × 24 ÷ 60). 136 kPa comes from adding the volume decrease (36) to 100 kPa, which is not how pressure and volume are related. 100 kPa wrongly assumes the pressure does not change.✓ 1Answer B

Answer: B

Exam tip: For a fixed mass of gas at constant temperature, a smaller volume means a larger pressure: use p₁V₁ = p₂V₂.

225IGPC3-2-07r12-05Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A farmer warms 150 kg of water in a trough from 8 °C to 14 °C. The specific heat capacity of water is 4200 J / (kg °C). How much thermal energy is transferred to the water?

Show mark scheme
Marking pointMarkNotes
Energy = mcΔθ = 150 × 4200 × 6 = 3 780 000 J = 3780 kJ. 8820 kJ uses 14 °C instead of the temperature change. 630 kJ leaves out the temperature change (150 × 4200). 25.2 kJ leaves out the mass (4200 × 6).✓ 1Answer A

Answer: A

Common error: Use the change in temperature, not the final temperature, in the specific heat capacity equation.

226IGPC3-2-07r12-12Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

After landing, a steel brake disc of mass 12 kg is at 40 °C and then absorbs 2.16 × 106 J of energy from friction. The specific heat capacity of steel is 450 J / (kg °C). Assume that all of the energy becomes thermal energy in the disc.

(a)

Define specific heat capacity.

(2)
(b)

Calculate the rise in temperature of the disc.

(2)
(c)

Calculate the final temperature of the disc. Use the answer to (b); ECF from (b).

(1)
(d)

Suggest why the actual final temperature is lower than the value in (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The thermal energy needed to raise the temperature✓ 1
of 1 kg of the substance by 1 °C✓ 1
Part (b)
Δθ = E ÷ (mc) = 2.16 × 106 ÷ (12 × 450)✓ 1
Δθ = 400 °C✓ 1
Part (c)
40 + 400 = 440 °C✓ 1
Part (d)
Some energy is transferred to the surroundings (air, wheel, etc.) rather than staying in the disc✓ 1

Exam tip: Calculate the temperature change first, then add it to the starting temperature to find the final temperature.

227IGPC3-2-05r13-05Particle model and gas pressure Paper EMedium1 mark

A cyclist's pump contains 240 cm3 of air at a pressure of 1.0×105 Pa. The air is compressed slowly to 60 cm3 with the temperature constant. What is the final pressure of the air?

outlet blockedpushed in slowlyair, 240 cm³1.0 × 10⁵ Pafinal volume 60 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
Correct: pV is constant, so p = 1.0×105 × 240 / 60 = 400000 Pa = 4.0×105 Pa. 2.5×104 Pa comes from inverting the ratio; 3.0×105 Pa uses the fractional change in volume instead of the ratio; 1.0×105 Pa assumes pressure does not change.✓ 1Answer C

Answer: C

Exam tip: Check your answer is sensible: if the volume falls to a quarter at constant temperature, the pressure must rise by four times.

228IGPC3-2-05r13-13Thermal expansion, specific heat capacity and changes of state Paper EMedium5 marks

A data centre uses a water-cooling loop. The specific heat capacity of water is 4200 J / (kg °C).

(a)

Calculate the thermal energy absorbed when 15 kg of water is warmed from 18 °C to 26 °C.

(2)
(b)

Another server row releases 1.26×106 J. Calculate the mass of water that can absorb this energy with a temperature rise of 6.0 °C.

(2)
(c)

State why water is a suitable coolant.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = mass × specific heat capacity × temperature change = 15 × 4200 × 8✓ 1
= 504000 J✓ 1
Part (b)
mass = energy / (c × ΔT) = 1260000 / (4200 × 6.0)✓ 1
= 50 kg✓ 1
Part (c)
high specific heat capacity so it absorbs a large amount of energy per kg for a small rise in temperature✓ 1

Exam tip: Write the formula and substitute values before giving the answer; each step can earn a mark.

229IGPC3-2-05r13-15Particle model and gas pressure Paper EMedium4 marks

A hand pump contains 120 cm3 of air at a pressure of 1.0×105 Pa. The outlet is blocked and the air is compressed slowly to a volume of 30 cm3. The temperature of the air stays constant.

outlet blockedpushed in slowlyair, 120 cm³1.0 × 10⁵ Pafinal volume 30 cm³
Diagram not to scale
(a)

Explain, in terms of particles, why the pressure increases as the volume decreases.

(2)
(b)

Calculate the final pressure of the air.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
particles are in a smaller space so hit the walls more often✓ 1
same speed so same force per collision but more collisions per second gives larger pressure✓ 1
Part (b)
pV constant: p = 1.0×105 × 120 / 30✓ 1
= 4.0×105 Pa (400000 Pa)✓ 1

Exam tip: State that the temperature is constant before using pV = constant.

230IGPC3-2-07r13-03Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A sports scientist heats a 0.40 kg sample of an alloy. A transfer of 2.4 kJ of thermal energy raises its temperature by 10 °C. What is the specific heat capacity of the alloy?

Show mark scheme
Marking pointMarkNotes
A is correct: c = Q / (m ΔT) = 2400 / (0.40 × 10) = 600 J / (kg °C). B leaves out the mass (2400 / 10). C multiplies Q by m instead of dividing (2400 × 0.40 / 10). D multiplies by the temperature change and divides by the mass (2400 × 10 / 0.40).✓ 1Answer A

Answer: A

Exam tip: Convert kJ to J before substituting into Q = mcΔT, and check that your answer has the units J / (kg °C).

231IGPC3-2-07r13-04Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A probe on a comet supplies thermal energy to a block of ice at 0 °C. The ice melts to water at 0 °C. What happens to the thermal energy supplied during melting?

Show mark scheme
Marking pointMarkNotes
D is correct: during a change of state the energy goes into overcoming the forces between molecules, which raises their potential energy, while the average kinetic energy and temperature stay constant. A and B would require the temperature to rise. C is wrong because the water formed is still at 0 °C.✓ 1Answer D

Answer: D

Common error: Temperature measures the average kinetic energy of the particles. During melting the temperature is constant, so the energy supplied is not increasing kinetic energy.

232IGPC3-2-07r13-07Particle model and gas pressure Paper EMedium1 mark

Gas is trapped in a syringe at a pressure of 1.0 × 105 Pa. Its volume is 60 cm3. The plunger is pushed in slowly, so that the temperature stays constant, until the volume is 24 cm3. What is the new pressure of the gas?

outlet sealedpushed in slowlyair, 60 cm³1.0 × 10⁵ Pafinal volume 24 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
B is correct: pV is constant, so p = 1.0 × 105 × 60 / 24 = 2.5 × 105 Pa. A uses the volume ratio the wrong way round (24/60). C is the increase in pressure (1.0 × 105 × 36/24), not the new pressure. D assumes that pressure is unchanged because the temperature is constant.✓ 1Answer B

Answer: B

Exam tip: For a fixed mass of gas at constant temperature, when the volume decreases the pressure must increase, so check that your answer is bigger than the starting pressure.

233IGPC3-2-07r13-11Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

A sports centre warms the water in a hydrotherapy tub. The tub holds 250 kg of water at 18 °C. The water is heated to 38 °C by a 5.0 kW heater. The specific heat capacity of water is 4200 J / (kg °C).

(a)

Calculate the thermal energy needed to raise the temperature of the water from 18 °C to 38 °C.

(2)
(b)

Calculate the time taken if all the energy from the heater is transferred to the water.

(2)
(c)

In practice the warming takes 80 minutes. Calculate the thermal energy lost to the surroundings during this time. Use the unrounded answer to (a).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = mcΔT✓ 1
250 × 4200 × 20 = 2.1 × 107 J✓ 1
Part (b)
t = E / P✓ 1
2.1 × 107 / 5000 = 4200 s (70 minutes) (ECF from (a))✓ 1
Part (c)
Energy supplied = 5000 × 80 × 60 = 2.4 × 107 J✓ 1
2.4 × 107 − 2.1 × 107 = 3.0 × 106 J✓ 1

Exam tip: Convert minutes to seconds before using power × time, and keep the same units for every energy value.

234IGPC3-2-07r13-12Thermal expansion, specific heat capacity and changes of state Paper EMedium8 marks

A probe on a comet heats a 0.60 kg sample of ice, starting at −20 °C, using a heater that transfers 120 W to the sample. All the energy from the heater goes into the sample. For ice, the specific heat capacity is 2100 J / (kg °C). The specific latent heat of fusion of ice is 3.3 × 105 J / kg.

(a)

Calculate the thermal energy needed to raise the temperature of the ice from −20 °C to 0 °C.

(2)
(b)

Calculate the thermal energy needed to melt all the ice at 0 °C.

(2)
(c)

Calculate the total time for which the heater must be switched on to warm and then completely melt the sample.

(2)
(d)

Explain, in terms of molecules, why the temperature stays at 0 °C while the ice is melting.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Q = mcΔT✓ 1
0.60 × 2100 × 20 = 25 200 J✓ 1
Part (b)
Q = ml✓ 1
0.60 × 3.3 × 105 = 1.98 × 105 J✓ 1
Part (c)
Total energy = 25 200 + 198 000 = 223 200 J✓ 1
t = 223 200 / 120 = 1860 s (ECF from (a) and (b))✓ 1
Part (d)
The energy supplied is used to overcome the forces between molecules / break the structure of the solid✓ 1
so it increases potential energy, not the average kinetic energy, and temperature depends on kinetic energy✓ 1

Common error: Do not forget the warming stage. The energy to melt the ice is only part of the total when the ice starts below 0 °C.

235IGPC3-2-05r14-04Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A data-centre cooling loop holds 2.0 kg of water of specific heat capacity 4200 J/(kg °C). The water is heated from 20 °C to 35 °C by the servers. How much thermal energy does the water gain?

Show mark scheme
Marking pointMarkNotes
Energy = mcΔT = 2.0 × 4200 × (35 − 20) = 126 000 J = 126 kJ, so B. A (63 kJ) omits the mass. C (168 kJ) uses 20 °C instead of the 15 °C change. D (294 kJ) uses the final temperature 35 °C instead of the change.✓ 1Answer B

Answer: B

Exam tip: Use the temperature change, final minus initial, not either temperature on its own.

236IGPC3-2-05r14-05Particle model and gas pressure Paper EMedium1 mark

A cyclist's pump holds 400 cm3 of air at a pressure of 100 kPa. The outlet is blocked and the piston is pushed in slowly until the air occupies 160 cm3. The temperature stays constant. What is the new pressure of the air?

outlet blockedpushed in slowlyair, 400 cm³100 kPafinal volume 160 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
For a fixed mass of gas at constant temperature pV is constant: p2 = 100 × 400 ÷ 160 = 250 kPa, so C. A (40 kPa) multiplies by the volume ratio the wrong way round. B assumes the pressure does not change. D (340 kPa) adds the volume decrease of 240 to the pressure.✓ 1Answer C

Answer: C

Exam tip: When volume decreases at constant temperature, the pressure must increase; check that your answer is larger.

237IGPC3-2-05r14-11Particle model and gas pressure Paper EMedium5 marks

A sealed syringe contains 60 cm3 of air at a pressure of 100 kPa. The piston is pushed in slowly until the volume of the air is 24 cm3. The air is a fixed mass and its temperature stays constant.

outlet sealedpushed in slowlyair, 60 cm³100 kPafinal volume 24 cm³
Diagram not to scale
(a)

Calculate the final pressure of the air.

(2)
(b)

Explain, in terms of particles, why the pressure rises as the volume decreases.

(2)
(c)

Suggest why the piston must be pushed slowly.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
pV is constant, so p1V1 = p2V2 with correct substitution, e.g. 100 × 60 = p2 × 24✓ 1
p2 = 250 kPa✓ 1
Part (b)
the particles are in a smaller space so they hit the walls more often✓ 1
so the total force per unit area on the walls is greater✓ 1
Part (c)
so that the temperature of the air stays constant (energy has time to escape to the surroundings)✓ 1

Exam tip: State that the temperature and the mass of gas are constant whenever you use pV = constant.

238IGPC3-2-05r14-12Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

A cyclist brakes hard on a long descent. The steel brake disc has a mass of 0.40 kg and a specific heat capacity of 450 J/(kg °C). The braking transfers 18 kJ of thermal energy to the disc. Assume none of this energy is lost to the surroundings.

(a)

Calculate the increase in temperature of the disc.

(3)
(b)

The disc was at 15 °C before braking. Determine its final temperature.

(1)
(c)

In practice the final temperature is lower than your answer to (b). Suggest two reasons.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔT = E ÷ (mc)✓ 1
ΔT = 18 000 ÷ (0.40 × 450)✓ 1
ΔT = 100 °C✓ 1
Part (b)
115 °C (ECF from (a))✓ 1
Part (c)
any two from: some energy is lost to the air by convection; some energy is lost by radiation; some energy is conducted to the wheel, brake pads or other parts✓ 2

Common error: Give the temperature rise from (a) a unit of °C, then add it to the starting temperature in (b); do not forget to add the starting value.

239IGPC3-2-07r14-05Particle model and gas pressure Paper EMedium1 mark

A wind turbine's pitch system uses a gas-filled accumulator. A fixed mass of gas at 2.0 × 105 Pa occupies 40 litres. It is compressed slowly to 16 litres with the temperature kept constant. What is the final pressure?

Show mark scheme
Marking pointMarkNotes
B is correct: at constant temperature pV is constant, so p2 = 2.0 × 105 × 40 / 16 = 5.0 × 105 Pa. A is wrong because it uses p2 = p1V2/V1 (pressure proportional to volume). C is wrong because it assumes pressure is unchanged at constant temperature. D is wrong because it multiplies the pressure by the 24-litre decrease in volume.✓ 1Answer B

Answer: B

Common error: When the volume decreases at constant temperature, the pressure must increase; check that your answer goes the right way.

240IGPC3-2-07r14-11Thermal expansion, specific heat capacity and changes of state Paper EMedium5 marks

A small paddling pool contains 600 kg of water at 18.0 °C. A heater of power 6.0 kW warms the water to 24.0 °C. The specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the thermal energy gained by the water. Give the exact value in J.

(2)
(b)

Assume all the energy from the heater is transferred to the water. Use your unrounded answer to (a) to calculate the time taken.

(2)
(c)

Suggest why the time taken in practice would be longer.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔE = mcΔθ = 600 × 4200 × 6.0✓ 1
= 1.512 × 107 J (15 120 000 J)✓ 1
Part (b)
t = E / P = 1.512 × 107 / 6000 (ECF from (a))✓ 1
= 2520 s (42 minutes)✓ 1
Part (c)
Some energy is lost to the surroundings (by evaporation, conduction or radiation)✓ 1

Common error: Use the temperature change, not the final temperature, in mcΔθ.

241IGPC3-2-07r14-12Particle model and gas pressure Paper EMedium5 marks

A wind turbine's pitch system uses an accumulator containing a fixed mass of gas at constant temperature. The gas is at a pressure of 1.8 × 106 Pa and occupies 25 litres. It is then compressed to 10 litres.

Before: p = 1.8 × 10⁶ PaV = 25 litresgasAfter: p = ?V = 10 litresgas
Diagram not to scale
(a)

Explain, in terms of gas molecules, why the pressure increases when the gas is compressed.

(2)
(b)

Calculate the final pressure.

(2)
(c)

The gas is initially at a temperature of 17 °C. Convert this temperature to kelvin.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The same number of molecules in a smaller volume collide with each unit area of wall more often✓ 1
so the average force per unit area on the walls is greater✓ 1
Part (b)
p1V1 = p2V2, so p2 = 1.8 × 106 × 25 / 10✓ 1
= 4.5 × 106 Pa✓ 1
Part (c)
17 + 273 = 290 K✓ 1

Exam tip: You do not need to convert litres to m³ in pV = constant, provided both volumes use the same unit.

242IGPC3-2-07r15-06Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A robot arm has two parts of equal mass: one made of aluminium (specific heat capacity 900 J/(kg °C)) and one made of steel (specific heat capacity 450 J/(kg °C)). They start at the same temperature and each absorbs the same amount of energy. Which statement is correct?

Show mark scheme
Marking pointMarkNotes
Since ΔT = E/(mc), with E and m equal, ΔT is inversely proportional to c, so the steel (half the c) rises twice as much: A is correct. B ignores the different values of c. C reverses the relationship. D is wrong because the ratio of temperature rises does not depend on the actual energy.✓ 1Answer A

Answer: A

Common error: A larger specific heat capacity means a smaller temperature rise for the same energy, not a larger one.

243IGPC3-2-07r15-12Thermal expansion, specific heat capacity and changes of state Paper EMedium5 marks

A solid aluminium block of mass 0.85 kg is heated by an electric heater of power 150 W for 4.0 minutes. The block starts at 18 °C. The specific heat capacity of aluminium is 900 J/(kg °C). Assume all the energy from the heater is transferred to the block.

to power supplyheater (150 W)thermometeraluminium blockmass 0.85 kg
Diagram not to scale
(a)

Calculate the energy supplied by the heater.

(2)
(b)

Calculate the rise in temperature of the block. Give your answer to 3 significant figures. Use 36000 J for the energy.

(2)
(c)

In a real experiment the measured temperature rise is smaller than your answer. Suggest one reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Converts 4.0 minutes to 240 s✓ 1
Energy = 150 × 240 = 36000 J✓ 1
Part (b)
ΔT = E / (m c) = 36000 / (0.85 × 900)✓ 1
ΔT = 47.1 °C (47.06 before rounding)✓ 1
Part (c)
Some energy is transferred to the surroundings (or heats the heater itself) and does not heat the block✓ 1

Exam tip: Convert minutes to seconds before using power × time.

244IGPC3-2-07r15-14Particle model and gas pressure Paper EMedium5 marks

A pneumatic gripper on a fish-sorting robot contains 240 cm3 of air at a pressure of 1.00 × 105 Pa. The air is compressed to a volume of 64 cm3 without any change in temperature.

Before: p = 1.00 × 10⁵ PaV = 240 cm³gasAfter: p = ?V = 64 cm³
Diagram not to scale
(a)

Explain, in terms of particles, why the pressure of the air increases when the volume decreases.

(2)
(b)

Calculate the new pressure of the air.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The average speed of the particles is unchanged because the temperature is constant✓ 1
The particles hit the walls more often per unit area, so the pressure increases✓ 1
Part (b)
Uses p1V1 = p2V2✓ 1
p2 = (1.00 × 105 × 240) / 64✓ 1
p2 = 3.75 × 105 Pa (375000 Pa)✓ 1

Exam tip: Volumes in the same unit on both sides of pV = constant do not need converting to m³.

245IGPC3-2-07r15-15Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

A fishing boat packs its catch with 2.5 kg of ice at 0 °C. The ice melts completely and the melt water then warms from 0 °C to 6.0 °C. The specific latent heat of fusion of ice is 3.3 × 105 J/kg and the specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the energy absorbed while the ice melts.

(2)
(b)

Calculate the energy absorbed by the melt water as it warms from 0 °C to 6.0 °C.

(2)
(c)

Calculate the total energy absorbed, giving your answer to 3 significant figures.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Energy = m L = 2.5 × 3.3 × 105✓ 1
= 8.25 × 105 J (825000 J)✓ 1
Part (b)
Energy = m c ΔT = 2.5 × 4200 × 6.0✓ 1
= 63000 J✓ 1
Part (c)
Adds 825000 J and 63000 J✓ 1
Total = 888000 J (8.88 × 105 J)✓ 1

Common error: Melting needs m L at constant temperature; do not use m c ΔT for the melting stage.

246IGPC3-2-06r16-05Particle model and gas pressure Paper EMedium1 mark

A cyclist's hand pump contains air of volume 80 cm³ at a pressure of 1.0 × 105 Pa. The air is compressed to 20 cm³ without a change in temperature. What is the new pressure?

Before: p = 1.0 × 10⁵ PaV = 80 cm³gasAfter: p = ?V = 20 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
B is correct: pV is constant, so p = 1.0 × 10^5 × 80 ÷ 20 = 4.0 × 10^5 Pa. A inverts the ratio. C gives the product pV, forgetting to divide by the new volume. D wrongly assumes pressure is unchanged because the temperature is constant.✓ 1Answer B

Answer: B

Exam tip: At constant temperature, p₁V₁ = p₂V₂; a smaller volume must give a larger pressure.

247IGPC3-2-06r16-12Thermal expansion, specific heat capacity and changes of state Paper EMedium7 marks

A cyclist and bicycle have a total mass of 82 kg and travel at 11 m/s. The cyclist brakes to a stop. All the kinetic energy is transferred equally to two steel brake discs, each of mass 0.15 kg and specific heat capacity 450 J/(kg °C).

(a)

Define specific heat capacity.

(1)
(b)

Calculate the kinetic energy of the cyclist and bicycle before braking.

(2)
(c)

Calculate the temperature rise of each disc. Use the unrounded kinetic energy and give your answer to 3 significant figures.

(3)
(d)

Suggest why the actual temperature rise is less than your answer.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy needed to raise the temperature of 1 kg of a substance by 1 °C✓ 1
Part (b)
M1 kinetic energy = ½ × 82 × 112✓ 1
A1 4961 J✓ 1
Part (c)
M1 energy per disc = 4961 ÷ 2 = 2480.5 J✓ 1
M1 Δθ = energy ÷ (mc) = 2480.5 ÷ (0.15 × 450)✓ 1
A1 36.75 so 36.7 °C✓ 1
Part (d)
some energy is lost to the surroundings (air) by convection or radiation✓ 1

Exam tip: Write the equation ΔQ = mcΔθ first, then substitute; this earns the method mark even if the arithmetic slips.

248IGPC3-2-06r16-13Thermal expansion, specific heat capacity and changes of state Paper EMedium7 marks

In a data centre, a server dissipates 4.5 kW. Water of specific heat capacity 4200 J/(kg °C) flows through a cooling plate, entering at 18 °C and leaving at 24 °C.

(a)

Calculate the mass of water that must flow through the plate each second to remove all the energy. Give your answer to 3 significant figures.

(3)
(b)

The flow stops and the 40 kg of water in a storage tank, initially at 18 °C, absorbs all the energy from the server. Calculate the time for its temperature to reach 30 °C.

(3)
(c)

Suggest why water is a good liquid for cooling servers.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 temperature rise = 6 °C✓ 1
M1 4500 = m × 4200 × 6✓ 1
A1 0.179 kg✓ 1
Part (b)
M1 energy = 40 × 4200 × 12 = 2.016 × 106 J✓ 1
M1 time = energy ÷ power = 2.016 × 106 ÷ 4500✓ 1
A1 448 s✓ 1
Part (c)
high specific heat capacity, so it absorbs a lot of energy for a small temperature rise✓ 1

Common error: Use the temperature change (a difference), not the final temperature, in mcΔθ.

249IGPC3-2-07r17-07Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

On a railway, a point heater melts ice that is at 0 °C on a set of points. The heater has a power of 850 W and all of its energy goes into melting the ice. There is 0.50 kg of ice. The specific latent heat of fusion of ice is 3.4 × 105 J/kg.
The heater is switched on for 2.0 minutes. What mass of ice melts?

Show mark scheme
Marking pointMarkNotes
Energy supplied = 850 × 120 = 1.02 × 105 J. Mass melted = 1.02 × 105 / 3.4 × 105 = 0.30 kg, so B is correct. A is the mass of ice that is left, not the mass that melts. C assumes all the ice melts, ignoring the energy supplied. D comes from using 2.0 s instead of 2.0 minutes (850 × 2.0 / 3.4 × 105).✓ 1Answer B

Answer: B

Common error: Convert time to seconds before using energy = power × time.

250IGPC3-2-07r17-11Thermal expansion, specific heat capacity and changes of state Paper EMedium7 marks

A sports science centre cools 50 kg of water in an ice bath from 22.0 °C to 12.0 °C using a chiller. The specific heat capacity of water is 4200 J/(kg °C). The chiller removes energy from the water at a constant rate of 1.4 kW.

(a)

Calculate the thermal energy removed from the water.

(3)
(b)

Calculate the time the chiller takes to cool the water. Give your answer in minutes.

(3)
(c)

Suggest one reason why the cooling takes longer than the time calculated in (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Temperature change = 10.0 °C✓ 1
E = mcΔT = 50 × 4200 × 10.0✓ 1
E = 2.1 × 106 J✓ 1
Part (b)
t = E / P (ECF from (a))✓ 1
t = 2.1 × 106 / 1400 = 1500 s✓ 1
t = 25 minutes✓ 1
Part (c)
Energy flows into the water from the warmer surroundings (by conduction, convection or radiation) / the tub itself also has to be cooled✓ 1

Common error: Power is in watts (joules per second): change kW to W and seconds to minutes carefully.

251IGPC3-2-07r17-13Particle model and gas pressure Paper EMedium6 marks

The air inside a bicycle pump has a volume of 600 cm3 and a pressure of 1.0 × 105 Pa. The outlet of the pump is blocked, and the handle is pushed in slowly so that the volume of the air becomes 250 cm3. Assume that the temperature of the air stays constant.

Before: p = 1.0 × 10⁵ PaV = 600 cm³gasAfter: p = ?V = 250 cm³gas
Diagram not to scale
(a)

Calculate the new pressure of the air. Give your answer to 2 significant figures.

(3)
(b)

Explain, in terms of particles, why the pressure increases as the volume decreases.

(2)
(c)

In practice the pump becomes warm. Suggest why the temperature of the air does not stay constant.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
p1V1 = p2V2✓ 1
p2 = (1.0 × 105 × 600) / 250✓ 1
p2 = 2.4 × 105 Pa (240 000 Pa)✓ 1
Part (b)
The same number of particles are in a smaller volume, so they collide with the walls more often per second (per unit area)✓ 1
The temperature is constant, so the average speed and the force of each collision stay the same✓ 1
Part (c)
Work is done on the air as it is compressed, which raises its temperature✓ 1

Common error: Volumes may be left in cm³ in p₁V₁ = p₂V₂ because the units cancel, but both volumes must use the same unit.

252IGPC3-2-06r25-05Particle model and gas pressure Paper EMedium1 mark

The cylinder of a theatre fog machine contains 120 cm3 of gas at a pressure of 100 kPa. A piston compresses the gas to 40 cm3 while the temperature stays constant. What is the new pressure?

Before: p = 100 kPaV = 120 cm³gasAfter: p = ?V = 40 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At constant temperature pV is constant, so p = 100 × 120 ÷ 40 = 300 kPa. Dividing the other way round gives 33 kPa, and subtracting or adding the volumes (180 kPa, 140 kPa) ignores the inverse proportionality between pressure and volume.✓ 1Answer A

Answer: A

Exam tip: At constant temperature, reducing the volume to one third multiplies the pressure by three; check that your answer goes the right way.

253IGPC3-2-06r25-11Particle model and gas pressure Paper EMedium6 marks

The hand pump of an inflatable life raft contains 450 cm3 of air at atmospheric pressure, 101 kPa. The air is at 17 °C. The outlet is blocked and the piston is pushed in slowly until the air occupies 150 cm3. Assume the temperature stays constant.

Before: p = 101 kPaV = 450 cm³gasAfter: p = ?V = 150 cm³
Diagram not to scale
(a)

Calculate the temperature of the air in kelvin.

(1)
(b)

Calculate the pressure of the air in the pump when its volume is 150 cm3.

(2)
(c)

Explain, in terms of particles, why the pressure increases when the air is compressed.

(2)
(d)

In practice the pump is pushed in quickly and the air warms slightly. State how the pressure compared with your answer to (b) would change.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
17 + 273 = 290 K✓ 1
Part (b)
pV = constant (p1V1 = p2V2), so p = 101 × 450 ÷ 150✓ 1
303 kPa✓ 1
Part (c)
the same number of particles are in a smaller volume, so they hit the walls more often✓ 1
the particles' average speed is unchanged, so each collision exerts the same force; more collisions per second give a larger pressure✓ 1
Part (d)
the pressure would be higher (greater) than 303 kPa✓ 1

Exam tip: Convert temperatures to kelvin by adding 273, and only use pV = constant when the temperature and the amount of gas are fixed.

254IGPC3-2-06r25-12Thermal expansion, specific heat capacity and changes of state Paper EMedium6 marks

An electric immersion heater of power 3.0 kW is used to warm 20 kg of water in a tank in a ship's galley from 12 °C to 57 °C. The specific heat capacity of water is 4200 J/(kg °C). Assume all the energy from the heater warms the water.

to power supplyimmersion heater, 3.0 kWthermometerwater, 20 kg
Diagram not to scale
(a)

Calculate the energy gained by the water.

(3)
(b)

Calculate the time taken, in seconds.

(2)
(c)

Suggest why the actual time taken is longer than your answer to (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
temperature rise = 57 − 12 = 45 °C✓ 1
energy = mcΔθ = 20 × 4200 × 45✓ 1
3.78 × 106 J (3780000 J)✓ 1
Part (b)
time = energy ÷ power = 3780000 ÷ 3000✓ 1
1260 s✓ 1
Part (c)
some energy from the heater is lost to the surroundings (tank, air), so more time is needed✓ 1

Exam tip: Convert kilowatts to watts before using energy = power × time.

255IGPC3-2-09r33-05Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A cool box used in refrigerated freight contains 2.0 kg of ice at 0 °C. The specific latent heat of fusion of ice is 340 kJ/kg. How much energy is needed to melt all of the ice, with no change in temperature?

Show mark scheme
Marking pointMarkNotes
Energy = mass × specific latent heat = 2.0 × 340 = 680 kJ. 340 kJ forgets to multiply by the mass, 170 kJ divides by the mass instead of multiplying, and 342 kJ adds the mass to the latent heat value.✓ 1Answer B

Answer: B

Common error: Latent heat calculations use E = m × L; the temperature change is not part of the calculation.

256IGPC3-2-09r33-10Thermal expansion, specific heat capacity and changes of state Paper EMedium5 marks

A heater is used to warm the water in a marine aquarium. The tank contains 40 kg of water. The specific heat capacity of water is 4200 J/(kg °C). The temperature of the water rises from 18.0 °C to 24.0 °C.

(a)

Calculate the energy transferred to the water.

(2)
(b)

The heater has a power of 240 W and all of its energy is transferred to the water. Calculate the time taken. Use the exact value from (a).

(2)
(c)

Suggest why the time taken in practice would be longer than in (b).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of E = mcΔθ with a temperature rise of 6.0 °C✓ 1
A1 E = 1 008 000 J (1.008 × 106 J)✓ 1
Part (b)
M1 use of t = E / P✓ 1
A1 t = 4200 s (accept 70 minutes); ECF from (a)✓ 1
Part (c)
Some energy is transferred to the surroundings / the tank / the air instead of the water✓ 1

Exam tip: Write the temperature change as a difference (final minus initial) before substituting into E = mcΔθ.

257IGPC3-2-09r33-11Thermal expansion, specific heat capacity and changes of state Paper EMedium7 marks

A logistics company ships vaccines in an insulated box. The box contains 1.5 kg of ice at 0 °C. The specific latent heat of fusion of ice is 3.4 × 105 J/kg. Energy enters the box through its walls at a constant rate of 30 W, and all of this energy is used to melt the ice.

(a)

State what is meant by specific latent heat of fusion.

(2)
(b)

Calculate the energy needed to melt all of the ice.

(2)
(c)

Calculate the time taken for all of the ice to melt.

(2)
(d)

Suggest why ice at 0 °C is better for keeping the vaccines cool than the same mass of water at 0 °C.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The energy needed to change 1 kg of a substance from solid to liquid✓ 1
with no change in temperature✓ 1
Part (b)
M1 use of E = mL✓ 1
A1 E = 510 000 J (5.1 × 105 J)✓ 1
Part (c)
M1 use of t = E / P✓ 1
A1 t = 17 000 s (1.7 × 104 s); ECF from (b)✓ 1
Part (d)
Melting ice absorbs a large amount of energy without its temperature rising, whereas the water would warm up as it absorbs energy✓ 1

Common error: Do not use the specific heat capacity formula for a change of state; the temperature does not change while the ice melts.

258IGPC3-2-08r36-03Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A copper heat sink in a gaming PC has a mass of 0.40 kg. The specific heat capacity of copper is 390 J/(kg °C). The heat sink absorbs 7800 J of thermal energy from the processor and no energy is lost to the surroundings. What is the rise in temperature of the heat sink?

Show mark scheme
Marking pointMarkNotes
Temperature rise = energy ÷ (mass × specific heat capacity) = 7800 ÷ (0.40 × 390) = 50 °C. Multiplying the energy by the mass and dividing by c gives 8.0 °C, which is the wrong rearrangement. Leaving out the mass gives 20 °C. Leaving out the specific heat capacity gives 19 500 °C.✓ 1Answer C

Answer: C

Exam tip: Write the equation ΔE = mcΔθ first and rearrange it before substituting numbers.

259IGPC3-2-08r36-05Particle model and gas pressure Paper EMedium1 mark

A syringe is sealed at its nozzle and holds 60 cm³ of air at a pressure of 100 kPa. The plunger is pushed in slowly until the volume is 24 cm³. The temperature stays constant. What is the new pressure of the air?

Before: p = 100 kPaV = 60 cm³gasAfter: p = ?V = 24 cm³gas
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At constant temperature pressure × volume is constant, so p = 100 × 60 ÷ 24 = 250 kPa. Subtracting the volume change from the pressure gives 64 kPa, which is not how the quantities are related. Inverting the ratio gives 40 kPa, which would mean the pressure fell when the gas was compressed. Multiplying the pressure by the new volume gives 2400, which ignores the original volume.✓ 1Answer B

Answer: B

Exam tip: If you squash a gas into a smaller volume at constant temperature, the pressure must go up. Use that to check your answer.

260IGPC3-2-09r41-04Particle model and gas pressure Paper EMedium1 mark

A theatre's pneumatic stage lift traps 0.50 m3 of air at a pressure of 2.0 × 105 Pa in a cylinder. The air is compressed slowly to 0.20 m3 with its temperature kept constant. What is the new pressure of the air?

Before: p = 2.0 × 10⁵ PaV = 0.50 m³gasAfter: p = ?V = 0.20 m³gas
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At constant temperature pV is constant, so p = (2.0 × 105 × 0.50) / 0.20 = 5.0 × 105 Pa. Multiplying the pressure by 0.20/0.50 instead gives 0.80 × 105 Pa because the ratio is inverted. Multiplying the pressure by 0.50 only gives 1.0 × 105 Pa. Using the volume ratio 0.50/0.20 = 2.5 alone, without multiplying by the starting pressure, gives 2.5 which is not a pressure.✓ 1Answer D

Answer: D

Exam tip: Compressing a gas at constant temperature must increase its pressure, so check your answer is larger than the starting value.

261IGPC3-2-09r41-10Thermal expansion, specific heat capacity and changes of state Paper EMedium7 marks

A theatre technician tests an aluminium block of mass 0.80 kg that will be fitted inside a stage lamp. An electric heater of power 60 W is fitted in the block and switched on for 5.0 minutes. All the energy from the heater is transferred to the block. The temperature of the block rises from 20 °C to 45 °C.

to power supplyheater (60 W)thermometeraluminium blockmass 0.80 kg
Diagram not to scale
(a)

Calculate the energy transferred by the heater.

(2)
(b)

Calculate the specific heat capacity of the aluminium, in J/(kg °C). Use your answer to (a) (ECF from (a)).

(3)
(c)

In a real experiment some energy escapes to the surroundings. Suggest one way of reducing this energy loss.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = power × time (M1)✓ 1
60 × (5.0 × 60) = 18 000 J (A1)✓ 1
Part (b)
c = E / (m Δθ) (M1)✓ 1
temperature change Δθ = 25 °C (A1)✓ 1
c = 18 000 / (0.80 × 25) = 900 J/(kg °C) (A1)✓ 1
Part (c)
wrap the block in insulation (lagging) (B1)✓ 1
so less energy is transferred to the surroundings by conduction, convection or radiation (B1)✓ 1

Exam tip: Convert minutes to seconds before using energy = power × time.

262IGPC3-2-09r41-12Particle model and gas pressure Paper EMedium6 marks

A volcanologist collects a sample of volcanic gas in a rigid flask of volume 0.25 m3 at a pressure of 6.0 × 105 Pa. The gas is released slowly into a large evacuated chamber and expands to fill it, and its pressure falls to 1.5 × 105 Pa. The temperature of the gas stays constant.

V = 0.25 m³p = 6.0 × 10⁵ Parigid flaskvalvelarge evacuated chamber
Diagram not to scale
(a)

State the quantity that stays constant for a fixed mass of gas when its pressure changes at constant temperature.

(1)
(b)

Calculate the final volume occupied by the gas.

(3)
(c)

Explain, in terms of particles, why the pressure falls when the gas occupies a larger volume at constant temperature.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
pressure × volume (pV)✓ 1
Part (b)
p1V1 = p2V2 (M1)✓ 1
6.0 × 105 × 0.25 = 1.5 × 105 × V (A1)✓ 1
V = 1.0 m3 (A1)✓ 1
Part (c)
the particles are further apart so travel further between collisions with the walls✓ 1
so there are fewer collisions per second with each unit area of wall✓ 1

Exam tip: Keep units consistent on both sides of p₁V₁ = p₂V₂, then the answer takes the unit of the volume you started with.

263IGPC3-2-07r42-05Particle model and gas pressure Paper EMedium1 mark

A research diver releases an air bubble of volume 12 cm3 at a depth where the pressure is 3.0 × 105 Pa. The bubble rises to the surface, where the pressure is 1.0 × 105 Pa. The temperature of the air in the bubble stays constant. What is the volume of the bubble at the surface?

Show mark scheme
Marking pointMarkNotes
At constant temperature pV is constant, so V = 12 × 3.0 × 105 / 1.0 × 105 = 36 cm3. A volume of 4.0 cm3 comes from inverting the ratio. Staying at 12 cm3 ignores the pressure change. A volume of 24 cm3 comes from adding the pressure difference instead of using a ratio.✓ 1Answer A

Answer: A

Common error: When pressure falls at constant temperature the volume rises in proportion; do not add or subtract the values.

264IGPC3-2-07r42-12Thermal expansion, specific heat capacity and changes of state Paper EMedium5 marks

A marine biology laboratory keeps a tank containing 50 kg of seawater. The specific heat capacity of the seawater is 4200 J/(kg °C). A heater of power 1.5 kW warms the water from 14.0 °C to 20.0 °C.

to power supplyheater, 1.5 kWthermometerseawater, 50 kg
Diagram not to scale
(a)

Calculate the thermal energy needed to warm the water.

(2)
(b)

Calculate the time the heater takes if all its energy warms the water.

(2)
(c)

Suggest why, in practice, the time taken is longer.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 E = mcΔθ = 50 × 4200 × 6.0✓ 1
A1 1.26 × 106 J✓ 1
Part (b)
M1 t = E / P = 1.26 × 106 / 1500✓ 1
A1 840 s (ECF from (a))✓ 1
Part (c)
Some energy is transferred to the surroundings (the tank walls or air)✓ 1

Exam tip: Convert kilowatts to watts before using power = energy ÷ time.

265IGPC3-2-07r42-15Particle model and gas pressure Paper EMedium6 marks

A compressed-air tank in a port workshop contains air at a pressure of 2.4 × 105 Pa and a volume of 0.050 m3. The air is compressed slowly to a volume of 0.020 m3. The temperature of the air stays constant.

Before: p = 2.4 × 10⁵ PaV = 0.050 m³gasAfter: p = ?V = 0.020 m³
Diagram not to scale
(a)

State which quantity, for the air particles, stays constant when the temperature is constant.

(1)
(b)

Calculate the new pressure of the air.

(3)
(c)

Explain, in terms of particles, why the pressure increases when the volume decreases at constant temperature.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The average kinetic energy of the particles✓ 1
Part (b)
M1 pV = constant, so p1V1 = p2V2✓ 1
M1 2.4 × 105 × 0.050 = p2 × 0.020✓ 1
A1 6.0 × 105 Pa✓ 1
Part (c)
The same number of particles are in a smaller volume, so they are closer together✓ 1
they hit the walls more often per unit area, giving a greater force per unit area✓ 1

Exam tip: Use any consistent units for volume in pV = constant, but keep pressure units the same on both sides.

266IGPC3-2-08r47-05Particle model and gas pressure Paper EMedium1 mark

A sealed syringe in a library workshop contains 90 cm3 of air at 1.2 × 105 Pa. The plunger is pushed in slowly until the volume is 30 cm3. The temperature stays constant. What is the new pressure of the air?

Before: p = 1.2 × 10⁵ PaV = 90 cm³gasAfter: p = ?V = 30 cm³
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At constant temperature pV is constant, so p = (1.2 × 105 × 90) ÷ 30 = 3.6 × 105 Pa. Inverting the ratio of volumes gives 4.0 × 104 Pa, which would mean the pressure falls on compression. The pressure does not stay the same, because fewer cubic centimetres hold the same particles. Using the volume change (60 cm3) in place of the final volume gives 1.8 × 105 Pa, which is wrong.✓ 1Answer D

Answer: D

Common error: Use the initial and final volumes themselves in p₁V₁ = p₂V₂, not the change in volume.

267IGPC3-2-08r47-10Thermal expansion, specific heat capacity and changes of state Paper EMedium9 marks

A hot tub contains 800 kg of water at 18 °C. A 6.0 kW electric heater raises the water temperature to 36 °C. Assume that all of the heater's energy is transferred to the water. The specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the thermal energy needed to heat the water.

(3)
(b)

Calculate the time the heater takes to supply this energy. Give your answer in hours, using the unrounded energy from (a).

(3)
(c)

In practice the water takes 3.5 hours to reach 36 °C. Calculate the efficiency of the heating process as a percentage.

(2)
(d)

Suggest one reason why the heating takes longer than the calculated time.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
temperature change = 18 °C✓ 1
energy = mass × specific heat capacity × temperature change = 800 × 4200 × 18✓ 1
6.05 × 107 J (60 480 000 J)✓ 1
Part (b)
time = energy ÷ power (with power = 6000 W)✓ 1
60 480 000 ÷ 6000 = 10 080 s✓ 1
2.80 hours✓ 1
Part (c)
efficiency = useful time (or energy) ÷ actual time (or energy)✓ 1
2.8 ÷ 3.5 × 100 = 80%✓ 1
Part (d)
energy is lost from the water to the surroundings, e.g. by evaporation from the surface, conduction through the tub walls or convection to the air✓ 1

Common error: Keep units consistent: use watts and seconds together, then convert to hours at the end.

268IGPC3-2-09r47-02Particle model and gas pressure Paper EMedium1 mark

A sealed syringe contains 60 cm3 of air at a pressure of 100 kPa. The plunger is pushed in slowly until the volume is 24 cm3. The temperature of the air stays constant. What is the new pressure of the air?

Before: p = 100 kPaV = 60 cm³gasAfter: p = ?V = 24 cm³gas
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
At constant temperature pV is constant, so p = 100 × 60 / 24 = 250 kPa. The value 40 kPa comes from inverting the ratio of volumes, 136 kPa comes from adding the 36 cm3 decrease in volume to the pressure, and 2400 kPa comes from multiplying the pressure by the final volume.✓ 1Answer B

Answer: B

Exam tip: When the volume decreases at constant temperature the pressure must increase; check that your answer is larger than the starting pressure.

269IGPC3-2-09r47-05Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A puddle of water on a mountain path becomes cooler as it evaporates. Which statement explains this cooling?

Show mark scheme
Marking pointMarkNotes
Only the fastest, most energetic particles can overcome the attraction of their neighbours and leave the liquid. The remaining particles therefore have a lower average kinetic energy, which means a lower temperature. Slower particles do not preferentially escape, the loss is not shared equally, and the forces between particles are not changed by evaporation.✓ 1Answer A

Answer: A

Common error: Evaporation takes place at any temperature, and it is the escape of the most energetic particles that causes the cooling.

270IGPC3-2-09r47-06Thermal expansion, specific heat capacity and changes of state Paper EMedium1 mark

A 800 W heater supplies energy to 0.30 kg of a solid that is already at its melting point. The solid takes 150 s to melt completely. No energy is lost to the surroundings. What is the specific latent heat of fusion of the substance?

Show mark scheme
Marking pointMarkNotes
Energy supplied = 800 × 150 = 120000 J. Specific latent heat = energy / mass = 120000 / 0.30 = 400000 J/kg = 4.0 × 105 J/kg. The value 1.2 × 105 leaves out the mass, 3.6 × 104 multiplies by the mass instead of dividing, and 18 J/kg divides the power by the time and the mass.✓ 1Answer B

Answer: B

Exam tip: Check the units: latent heat is energy per kilogram, so the energy (power × time) must be divided by the mass.

271IGPC3-2-09r47-10Particle model and gas pressure Paper EMedium6 marks

A weather balloon carries a radio transmitter. At ground level the balloon holds 2.4 m3 of helium at a pressure of 1.0 × 105 Pa. As the balloon rises, the temperature of the gas stays constant.

(a)

The pressure of the gas falls to 4.0 × 104 Pa. Calculate the new volume of the gas.

(2)
(b)

Explain, in terms of particles, why the pressure of the gas falls as the balloon expands.

(2)
(c)

The balloon bursts when its volume reaches 8.0 m3. Calculate the gas pressure at which the balloon bursts. Use the data from the start of the question.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p1V1 = p2V2, so V = 1.0 × 105 × 2.4 / 4.0 × 104✓ 1
A1 6.0 m3✓ 1
Part (b)
B1 the same particles are now spread through a larger volume, so they hit the walls less often✓ 1
B1 the average speed of the particles is unchanged, so fewer collisions per second means a lower pressure✓ 1
Part (c)
M1 p = 1.0 × 105 × 2.4 / 8.0✓ 1
A1 30000 Pa (3.0 × 104 Pa)✓ 1

Exam tip: Write the equation p₁V₁ = p₂V₂ first, then substitute; this keeps the pressures and volumes in the correct places.

272IGPC3-2-09r47-13Thermal expansion, specific heat capacity and changes of state Paper EMedium7 marks

A mountaineer melts 0.80 kg of snow at 0 °C on a stove and then heats the water to 75 °C. The specific latent heat of fusion of ice is 3.3 × 105 J/kg and the specific heat capacity of water is 4200 J / (kg °C). Assume that all the energy from the stove goes to the snow and water.

(a)

Calculate the energy needed to melt the snow.

(2)
(b)

Calculate the energy needed to heat the melted water from 0 °C to 75 °C.

(2)
(c)

The stove supplies energy at a rate of 1200 W. Calculate the total time taken for the whole process. Use the unrounded values from (a) and (b).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 energy = mass × specific latent heat = 0.80 × 3.3 × 105✓ 1
A1 264000 J (2.64 × 105 J)✓ 1
Part (b)
M1 energy = 0.80 × 4200 × 75✓ 1
A1 252000 J (2.52 × 105 J)✓ 1
Part (c)
M1 total energy = 264000 + 252000 = 516000 J (ECF from (a) and (b))✓ 1
M1 time = energy / power = 516000 / 1200✓ 1
A1 430 s✓ 1

Common error: While the snow melts its temperature stays at 0 °C, so the latent heat and the heating of the water must be calculated as two separate stages.

273IGPC-036-02-09Conduction, convection and radiation Paper EHard1 mark

Two solid cubes, X and Y, are made of the same material and have the same dull black surface. Both are at a temperature of 60 °C. Cube X has sides of 2.0 cm and cube Y has sides of 4.0 cm. What is the ratio \(\frac{\text{rate of infrared emission from Y}}{\text{rate of infrared emission from X}}\)?

cube Xside 2.0 cmcube Yside 4.0 cmboth cubes: same material, dull black surface, 60 °C
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
The surfaces have the same temperature and the same finish. The rate of emission therefore depends only on the surface area. Doubling the side length multiplies the surface area by 22 = 4, so Y emits 4 times as fast. Option 8 comes from comparing volumes and option 2 from comparing side lengths.✓ 1Answer B

Answer: B

Common error: Radiation is emitted from a surface, so compare areas, not volumes. The larger cube does cool more slowly, but only because its mass increases by 8 times while its area increases by 4 times.

274IGPC-036-02-15Conduction, convection and radiation Paper EHard8 marks

A solar water-heating panel on a roof contains copper pipes painted matt black, inside a box with a glass cover. Water flows slowly through the pipes.
The panel has an area of 2.5 m2. On a sunny day, 800 W of solar radiation falls on each 1 m2 of panel. Only 40% of the energy that falls on the panel is transferred to the water.
The specific heat capacity of water is 4200 J/(kg °C).

(a)

Explain why the pipes are painted matt black.

(1)
(b)

Calculate the power transferred to the water.

(2)
(c)

In 1 hour, 30 kg of water flows through the panel. Calculate the rise in temperature of this water.

(3)
(d)

The water flow is switched off but the Sun still shines on the panel. The temperature of the panel rises and then becomes constant. Explain why the temperature stops rising.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A matt black surface is a good absorber of (infrared) radiation .✓ 1
Part (b)
Power falling on the panel = 800 × 2.5 = 2000 W✓ 1
Power to the water = 0.40 × 2000 = 800 W✓ 1
Part (c)
Energy = P × t = 800 × 3600 = 2.88 ×106 J✓ 1
ΔT = E ÷ (mc) = 2.88 ×106 ÷ (30 × 4200)✓ 1
ΔT = 23 °C✓ 1
Part (d)
As the panel gets hotter, it loses energy (by radiation, conduction and convection) at a greater rate . The temperature becomes constant when the rate of energy loss equals the rate at which radiation is absorbed .✓ 2

Exam tip: Convert the time to seconds (1 hour = 3600 s) before using E = Pt. Apply the 40% before working out the energy.

275IGPC-036-00-08Particle model and gas pressure Paper EHard1 mark

A gas molecule of mass 4.8 ×10−26 kg travels at 500 m/s at right angles towards the wall of a container. It rebounds from the wall at the same speed.
What is the magnitude of the change in momentum of the molecule?

wall500 m/s500 m/smolecule: mass 4.8 × 10⁻²⁶ kg
Molecule before and after the collision (path drawn apart for clarity)
Show mark scheme
Marking pointMarkNotes
Momentum is a vector. Before the collision the momentum is +mv and after it is −mv, so the change has magnitude 2mv.
2 × 4.8 ×10−26 × 500 = 4.8 ×10−23 kg m/s.
Option A ignores the change of direction because the speed is unchanged. Option B gives only mv. Option D doubles the correct answer again.
✓ 1Answer C

Answer: C

Common error: The speed is unchanged, but the velocity reverses. The momentum change in a head-on rebound is 2mv, not zero.

276IGPC-036-01-09Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A 50 W immersion heater is placed in 0.50 kg of a liquid and switched on for 4.0 minutes. The temperature of the liquid rises from 20 °C to 32 °C.
Assume all the energy from the heater goes into the liquid.
What is the specific heat capacity of the liquid?

50 W immersion heaterthermometerliquid, 0.50 kg
Diagram not to scale
Show mark scheme
Marking pointMarkNotes
First find the energy supplied:
E = Pt = 50 × 240 = 12 000 J.
The temperature rise is Δθ = 32 − 20 = 12 °C.
c = E/(mΔθ) = 12 000 / (0.50 × 12) = 2000 J/(kg °C).
The other options come from common mistakes:
• 33 J/(kg °C) uses the time in minutes instead of seconds.
• 750 J/(kg °C) uses the final temperature (32 °C) instead of the rise.
• 1000 J/(kg °C) leaves out the mass.
✓ 1Answer D

Answer: D

Common error: Change the time into seconds before using E = Pt.

277IGPC-036-00-14Particle model and gas pressure Paper EHard6 marks

A diver's air tank has a volume of 12 litres. It is filled with air at a pressure of 2.0 ×107 Pa. During the dive, the diver breathes air at the same pressure as the surrounding water, which is 3.0 ×105 Pa.

(a)

State one assumption you must make to use the relationship pV = constant for the air from the tank.

(1)
(b)

Calculate the volume that all the air from the full tank would occupy at a pressure of 3.0 ×105 Pa.

(2)
(c)

Air stops flowing out once the pressure in the tank falls to 3.0 ×105 Pa, so 12 litres of air at this pressure stay in the tank. Calculate the volume of air available for the diver to breathe.

(2)
(d)

The diver breathes 20 litres of air per minute at this depth. Calculate how long the air lasts.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Any one: the temperature of the air stays constant / the mass (amount) of air is fixed, with no air leaking or added .✓ 1
Part (b)
V2 = p1V1 ÷ p2 = (2.0 ×107 × 12) ÷ (3.0 ×105)✓ 1
V2 = 800 litres✓ 1
Part (c)
Available volume = 800 − 12, i.e. subtracting the air that stays in the tank✓ 1
= 788 litres✓ 1
Part (d)
Time = 788 ÷ 20 = 39 minutes (accept 39.4 min)✓ 1

Common error: The tank cannot be emptied completely. It always keeps its own volume of air at the outside pressure, so subtract this before working out the time.

278IGPC-036-01-16Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

An electric kettle has a power rating of 2.2 kW. It contains 1.5 kg of water at 18 °C. Only 90% of the electrical energy supplied is transferred to the water.
Specific heat capacity of water = 4200 J/(kg °C)

(a)

Calculate the thermal energy needed to raise the temperature of the water to 100 °C.

(2)
(b)

Calculate the time taken for the kettle to bring the water to 100 °C.

(3)
(c)

The kettle is left switched on after the water reaches 100 °C. Explain why the temperature of the water does not rise any further.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔE = mcΔθ = 1.5 × 4200 × (100 − 18)✓ 1
ΔE = 516 600 J (5.2 × 105 J).✓ 1
Part (b)
Useful power = 0.90 × 2200 = 1980 W✓ 1
t = E/P = 516 600 / 1980✓ 1
t = 261 s (accept 2.6 × 102 s).✓ 1
Part (c)
The water is boiling, so the energy supplied is used to change liquid water into steam.✓ 1
This energy overcomes the forces between the particles, separating them (accept: increases their potential energy), so their average kinetic energy does not increase.✓ 1

Common error: Use only the useful power (90% of 2.2 kW) when calculating the time. Also change kW into W first.

279IGPC-036-01-15Thermal expansion, specific heat capacity and changes of state Paper EHard6 marks

A copper block of mass 0.40 kg is heated in a hot-water bath until its temperature is 95 °C. It is then quickly placed in an insulated cup containing 0.60 kg of water at 20 °C. The water is stirred until the block and the water reach the same temperature.
Specific heat capacity of copper = 390 J/(kg °C)
Specific heat capacity of water = 4200 J/(kg °C)

(a)

Describe what happens to the water particles as the temperature of the water rises.

(2)
(b)

Assume that all the energy lost by the copper is gained by the water. Calculate the final temperature.

(3)
(c)

Suggest one reason why the measured final temperature is lower than the calculated value.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The water particles gain energy (the internal energy of the water increases).✓ 1
The average kinetic energy of the particles increases, so on average they move faster.✓ 1
Part (b)
Energy lost by copper = energy gained by water: 0.40 × 390 × (95 − T) = 0.60 × 4200 × (T − 20)✓ 1
156(95 − T) = 2520(T − 20), so 14 820 + 50 400 = 2676T✓ 1
T = 24.4 °C (accept 24 °C).✓ 1
Part (c)
Any one of: the copper cooled in the air while it was being moved; some energy was transferred to the cup, stirrer or thermometer; some energy was lost to the surroundings because the insulation is not perfect.✓ 1

Exam tip: Write the temperature changes as (95 − T) and (T − 20). Then set energy lost equal to energy gained and solve for T.

280IGPC3-2-02c-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

Equal masses of substance X (specific heat capacity 400 J/(kg °C)) and substance Y (specific heat capacity 1200 J/(kg °C)) are given the same amount of thermal energy. Neither changes state. How does the temperature rise of X compare with that of Y?

Show mark scheme
Marking pointMarkNotes
Correct: B. Δθ = Q ÷ (mc). With Q and m equal, Δθ is inversely proportional to c, so X rises by 1200 ÷ 400 = 3 times as much. A is wrong: this takes the ratio the wrong way round; the lower specific heat capacity gives the larger rise. C is wrong: equal energy and equal mass give equal rises only if the specific heat capacities are equal. D is wrong: this subtracts specific heat capacities, which is not meaningful and gives a temperature value from the wrong quantity.✓ 1Answer B

Answer: B

Exam tip: A small specific heat capacity means a large temperature rise for a given energy input.

281IGPC3-2-02c-12Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

0.10 kg of ice at 0 °C is added to 0.30 kg of water at 40 °C in a well-insulated container and stirred until all the ice has melted. The specific latent heat of fusion of ice is 3.36 × 105 J/kg and the specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the energy needed to melt the ice without changing its temperature.

(2)
(b)

Calculate the energy released as the 0.30 kg of water cools from 40 °C to 0 °C. Use your answers to explain why all the ice melts.

(2)
(c)

Calculate the final temperature of the mixture.

(4)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = mass × specific latent heat✓ 1
0.10 × 3.36 × 105 = 33600 J✓ 1
Part (b)
0.30 × 4200 × 40 = 50400 J✓ 1
this is greater than 33600 J, so there is enough energy to melt all the ice✓ 1
Part (c)
energy lost by warm water = energy to melt ice + energy to warm melted ice water✓ 1
0.30 × 4200 × (40 − T) = 33600 + 0.10 × 4200 × T✓ 1
50400 − 1260T = 33600 + 420T, so 16800 = 1680T✓ 1
T = 10 °C✓ 1

Exam tip: Set up an energy balance: energy lost by the warm water equals energy gained by the ice, including melting and then warming.

282IGPC3-2-02c-15Particle model and gas pressure Paper EHard8 marks

A cylinder with a gas-tight piston traps 360 cm3 of air at a pressure of 1.2 × 105 Pa. The piston is pushed in slowly so that the temperature of the gas stays constant. The volume is reduced to 150 cm3. The area of the piston is 4.0 cm2.

pushed in slowlyair, 360 cm³1.2 × 10⁵ Pafinal volume 150 cm³piston area 4.0 cm²
Diagram not to scale
(a)

Explain, in terms of the air molecules, why the pressure of the gas increases.

(3)
(b)

Calculate the final pressure of the gas.

(3)
(c)

Calculate the force exerted by the gas on the piston at the final volume.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the same number of molecules are in a smaller volume✓ 1
so they collide with the walls more often (more collisions per second per unit area)✓ 1
each collision is unchanged because the temperature (average speed) is constant, so the force per unit area increases✓ 1
Part (b)
pV is constant at constant temperature✓ 1
1.2 × 105 × 360 = p × 150✓ 1
p = 2.88 × 105 Pa✓ 1
Part (c)
area = 4.0 cm2 = 4.0 × 10−4 m2✓ 1
force = pressure × area = 2.88 × 105 × 4.0 × 10−4 = 115.2 N (ECF from (b))✓ 1

Exam tip: At constant temperature use pV = constant. Convert cm² to m² by dividing by 10 000 before using force = pressure × area.

283IGPC3-2-07r7-06Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

An electric heater of power 800 W is used to melt 0.50 kg of ice at 0 °C. Only 80% of the energy from the heater goes into the ice, and the ice is completely melted after 210 s. What is the specific latent heat of fusion of ice from these results?

Show mark scheme
Marking pointMarkNotes
Useful energy = 0.80 × 800 × 210 = 134 400 J; l = E ÷ m = 134 400 ÷ 0.50 = 268 800 J/kg ≈ 2.7×105 J/kg. Option A is wrong: this forgets to divide by the mass (0.80 × 800 × 210 = 134 400 J) and treats energy as J/kg. Option C is wrong: this ignores the 80% efficiency. Option D is wrong: this divides by 0.80 instead of multiplying, giving too much useful energy.✓ 1Answer B

Answer: B

Common error: Multiply power by time for the energy supplied, then take only the useful fraction before dividing by the mass.

284IGPC3-2-07r7-12Particle model and gas pressure Paper EHard5 marks

A gym stores compressed gas in a tank of volume 0.050 m3 at a pressure of 8.0×105 Pa. The gas is used to inflate exercise balls. Each ball is inflated to a volume of 0.025 m3 at a pressure of 1.0×105 Pa. Gas stops flowing from the tank when the pressure in the tank has fallen to 1.0×105 Pa. Assume the temperature of the gas stays constant.

(a)

Calculate the volume that the gas in the full tank would occupy at a pressure of 1.0×105 Pa.

(2)
(b)

Determine the number of balls that can be completely inflated from the full tank.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
p1V1 = p2V2: 8.0×105 × 0.050 = 1.0×105 × V✓ 1
V = 0.40 m3✓ 1
Part (b)
gas still in tank at the end occupies 0.050 m3 at 1.0×105 Pa, so 0.40 − 0.050 = 0.35 m3 available (ECF from (a))✓ 1
number = 0.35 ÷ 0.025✓ 1
14 balls✓ 1

Common error: Do not forget that the tank is not empty at the end: it still holds gas at 1.0×10⁵ Pa.

285IGPC3-2-07r7-16Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

A telescope technician drops a 0.60 kg metal mount at 105 °C into 0.90 kg of water at 15 °C in an insulated container. The specific heat capacity of the metal is 450 J/(kg °C) and that of water is 4200 J/(kg °C).

(a)

State one assumption made when calculating the final temperature of the water and metal.

(1)
(b)

Show that the final temperature of the water and metal is 21 °C.

(3)
(c)

In a second experiment the same metal mount at 105 °C is placed in a different mass m of water at 15 °C and the final temperature is 18 °C. Calculate m. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
no thermal energy is lost to the surroundings / container, all energy lost by the metal is gained by the water✓ 1
Part (b)
energy lost by metal = 0.60 × 450 × (105 − T) and energy gained by water = 0.90 × 4200 × (T − 15)✓ 1
270(105 − T) = 3780(T − 15), so 28 350 − 270T = 3780T − 56 700✓ 1
4050T = 85 050, so T = 21 °C✓ 1
Part (c)
energy lost by metal = 0.60 × 450 × (105 − 18) = 23 490 J✓ 1
energy gained by water = m × 4200 × (18 − 15) = 12 600 m✓ 1
m = 23 490 ÷ 12 600 = 1.864 = 1.86 kg✓ 1

Common error: Use the temperature change of each substance, not the final temperature, when finding energy transferred.

286IGPC3-2-07r9-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A heater of constant power melts a block of ice at 0 °C completely in 5.5 minutes. The same heater then warms the melted water from 0 °C to 40 °C. No energy is lost to the surroundings. The specific latent heat of fusion of ice is 3.3 × 105 J/kg and the specific heat capacity of water is 4200 J/(kg °C). How long does the warming take?

Show mark scheme
Marking pointMarkNotes
D is correct: the heater power P is constant, so Pt1 = mL and Pt2 = mcΔθ. Dividing gives t2 = t1cΔθ/L = 5.5 × 4200 × 40 / 330 000 = 2.8 minutes. A is wrong because it assumes the same time because the power is the same. B is wrong because it uses a temperature rise of 100 °C. C is wrong because it inverts the ratio and uses L / (cΔθ).✓ 1Answer D

Answer: D

Exam tip: When the same mass and same heater are used, the mass cancels, so compare the energy per kilogram for each stage.

287IGPC3-2-07r9-15Particle model and gas pressure Paper EHard7 marks

A renewable energy company stores energy by compressing air. A volume of 8.0 m3 of air at a pressure of 1.0 × 105 Pa is slowly compressed into a tank of volume 0.25 m3. The temperature of the air stays constant.

(a)

Calculate the pressure of the air in the tank.

(2)
(b)

Explain, in terms of molecules, why the pressure rises when the same amount of air is squeezed into a smaller volume at constant temperature.

(3)
(c)

In practice the air becomes hotter when it is compressed quickly. State whether the pressure in the tank is then greater than, equal to or less than 3.2 × 106 Pa, and explain your answer.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
p1V1 = p2V2✓ 1
p2 = 1.0 × 105 × 8.0 / 0.25 = 3.2 × 106 Pa✓ 1
Part (b)
The average speed (kinetic energy) of the molecules is unchanged✓ 1
The same number of molecules are in a smaller volume, so they hit the walls more often✓ 1
The total force on each unit area of wall is greater, so the pressure is greater✓ 1
Part (c)
Greater than 3.2 × 106 Pa✓ 1
The molecules have greater average kinetic energy, so they hit the walls harder and more often✓ 1

Exam tip: State clearly what stays constant (here temperature) before choosing the gas equation.

288IGPC3-2-07r9-16Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

A barista adds 0.050 kg of milk at 4.0 °C to 0.20 kg of coffee at 85 °C. Assume that milk and coffee both have a specific heat capacity of 4200 J/(kg °C), and that no energy is lost to the surroundings in part (a).

(a)

Calculate the final temperature of the mixture. Give your answer to the nearest degree.

(3)
(b)

In a second attempt the measured final temperature is 66.0 °C. Calculate the energy transferred to the surroundings.

(3)
(c)

Suggest one change that reduces the energy transferred to the surroundings.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Energy lost by coffee = 0.20 × 4200 × (85 − T)✓ 1
Energy gained by milk = 0.050 × 4200 × (T − 4.0), and the two are equated✓ 1
T = 68.8, so 69 °C✓ 1
Part (b)
Energy lost by coffee = 0.20 × 4200 × (85 − 66.0) = 15 960 J✓ 1
Energy gained by milk = 0.050 × 4200 × (66.0 − 4.0) = 13 020 J✓ 1
Energy transferred to surroundings = 15 960 − 13 020 = 2940 J✓ 1
Part (c)
Use a lid or an insulated cup✓ 1

Common error: In mixing problems the temperature change is different for each liquid: hotter liquid (85 − T), colder liquid (T − 4.0).

289IGPC3-2-07r12-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

1.0 kg of water at 100 °C is boiled completely to give 1.0 kg of steam at 100 °C. Which row compares the steam with the water?

Show mark scheme
Marking pointMarkNotes
Temperature is a measure of average kinetic energy, so it is unchanged at 100 °C. The energy supplied (latent heat) is used to separate the particles against the forces between them, so potential energy increases. The other rows assume that the temperature or the energy stored between particles must change in the wrong way, or that no energy is stored although energy was supplied.✓ 1Answer C

Answer: C

Exam tip: During a change of state the temperature stays constant: the energy supplied changes the potential energy, not the kinetic energy.

290IGPC3-2-07r12-13Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

A farmer heats water in a sealed sterilising tank using an electric heater of power 2.0 kW. The water is already at its boiling point of 100 °C. The specific latent heat of vaporisation of water is 2.26 × 106 J/kg. Assume that all of the energy from the heater is used to vaporise the water.

(a)

Calculate the energy needed to change 0.50 kg of water at 100 °C into steam at 100 °C.

(2)
(b)

Calculate the time taken. Use the exact answer to (a); ECF from (a).

(2)
(c)

Suggest why the real time would be longer than the answer to (b).

(1)
(d)

Explain why evaporation from the surface of water at 20 °C cools the water that remains.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = mL = 0.50 × 2.26 × 106✓ 1
E = 1.13 × 106 J✓ 1
Part (b)
t = E ÷ P = 1.13 × 106 ÷ 2000✓ 1
t = 565 s✓ 1
Part (c)
Some of the heater's energy is lost to the surroundings, so more time is needed✓ 1
Part (d)
The fastest (most energetic) particles escape from the surface✓ 1
so the average kinetic energy of the remaining particles falls, so its temperature decreases✓ 1

Exam tip: Latent heat is used for a change of state with no change in temperature; use E = mL, not E = mcΔθ.

291IGPC3-2-07r12-15Particle model and gas pressure Paper EHard7 marks

A weather balloon released from an airfield contains a fixed mass of helium. At ground level the helium has a pressure of 101 kPa and a volume of 2.4 m3. At a certain height the pressure of the helium is 38 kPa. Assume that the temperature of the helium does not change.

(a)

Explain, using the particle model, why the pressure of a fixed mass of gas falls when its volume increases at constant temperature.

(3)
(b)

Calculate the volume of the helium at this height. Give your answer to 2 significant figures.

(3)
(c)

Suggest why the balloon is likely to burst if it rises much higher.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The average speed of the particles is unchanged✓ 1
The particles are further apart, so they hit the walls less often✓ 1
so the force on each unit area of wall is smaller, so the pressure is lower✓ 1
Part (b)
p1V1 = p2V2✓ 1
V2 = 101 × 2.4 ÷ 38 = 6.378 m3✓ 1
V2 = 6.4 m3✓ 1
Part (c)
The outside pressure falls further, so the gas expands until the balloon skin is stretched beyond its limit✓ 1

Exam tip: State the condition (constant temperature, fixed mass) before using pV = constant.

292IGPC3-2-05r13-07Particle model and gas pressure Paper EHard1 mark

A sealed rigid cylinder in a cooling system contains gas at 27 °C and a pressure of 1.5×105 Pa. The gas is heated to 327 °C. What is the final pressure?

Show mark scheme
Marking pointMarkNotes
Correct: at constant volume p is proportional to the kelvin temperature. T rises from 300 K to 600 K, so p doubles to 300000 Pa = 3.0×105 Pa. 1.8×106 Pa uses the ratio of Celsius temperatures; 3.3×106 Pa uses 600 K over 27; 1.5×105 Pa assumes constant volume means constant pressure.✓ 1Answer A

Answer: A

Common error: Always convert to kelvin before using a pressure-temperature ratio.

293IGPC3-2-05r13-14Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

In an experiment to find the specific heat capacity of a metal fragment from an excavation, a fragment of mass 0.40 kg is heated to 100 °C and placed in 0.50 kg of water at 20 °C. The final steady temperature is 25.0 °C. The specific heat capacity of water is 4200 J / (kg °C).

(a)

Calculate the thermal energy gained by the water.

(2)
(b)

Assume no energy is lost to the surroundings. Use the answer to (a) to calculate the specific heat capacity of the fragment.

(3)
(c)

Some energy is in fact lost to the surroundings. Suggest and explain the effect on the value of the specific heat capacity calculated from the water's temperature rise.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = m c ΔT = 0.50 × 4200 × 5.0✓ 1
= 10500 J✓ 1
Part (b)
energy lost by fragment = 10500 J (ECF from (a))✓ 1
c = energy / (m ΔT) = 10500 / (0.40 × 75)✓ 1
= 350 J / (kg °C)✓ 1
Part (c)
the water gains less energy than the fragment loses✓ 1
calculated value is lower than the true value✓ 1

Common error: The fragment's temperature change is from 100 °C to 25.0 °C, which is 75 °C, not the 5.0 °C change of the water.

294IGPC3-2-05r13-16Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

A cool box for a field expedition contains a block of ice of mass 0.80 kg at 0 °C. Thermal energy flows into the ice at a constant rate of 112 W and the ice takes 40 minutes to melt completely. The specific heat capacity of water is 4200 J / (kg °C).

(a)

Calculate the thermal energy absorbed by the ice during the 40 minutes.

(2)
(b)

Calculate the specific latent heat of fusion of ice.

(2)
(c)

State why the temperature of the ice stays at 0 °C while it melts.

(1)
(d)

After melting, the water continues to absorb energy at 112 W. Calculate the time taken for the water to warm by 5.0 °C.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = power × time = 112 × 2400✓ 1
= 268800 J✓ 1
Part (b)
l = energy / mass = 268800 / 0.80✓ 1
= 336000 J/kg (3.36×105 J/kg)✓ 1
Part (c)
energy is used to separate particles / overcome forces between particles, not to raise their kinetic energy✓ 1
Part (d)
energy = 0.80 × 4200 × 5.0 = 16800 J✓ 1
time = 16800 / 112 = 150 s✓ 1

Common error: Convert minutes to seconds before using energy = power × time.

295IGPC3-2-07r13-14Particle model and gas pressure Paper EHard6 marks

A cyclist uses a pump to inflate a tyre. The temperature of the air stays constant.

(a)

One stroke of the pump squeezes 0.15 litres of air at a pressure of 1.0 × 105 Pa into a chamber of volume 0.050 litres. Calculate the pressure of the air in the chamber.

(2)
(b)

The tyre has a fixed volume of 2.0 litres and starts with air at 1.0 × 105 Pa. Each stroke adds 0.15 litres of air measured at 1.0 × 105 Pa. Calculate the number of strokes needed to raise the pressure in the tyre to 3.4 × 105 Pa.

(3)
(c)

Suggest one reason why more strokes than this are needed in practice.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
p1V1 = p2V2✓ 1
p2 = 1.0 × 105 × 0.15 / 0.050 = 3.0 × 105 Pa✓ 1
Part (b)
Volume of air at 1.0 × 105 Pa needed = 3.4 × 105 × 2.0 / 1.0 × 105 = 6.8 litres✓ 1
Extra air needed = 6.8 − 2.0 = 4.8 litres✓ 1
4.8 / 0.15 = 32 strokes✓ 1
Part (c)
Some air leaks past the valve / seals of the pump or tyre✓ 1

Exam tip: Treat the air already in the tyre as part of the total amount of gas: compare all of the air at one common pressure.

296IGPC3-2-07r13-15Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

A student investigates the specific heat capacity of a metal block of mass 2.0 kg. An electric heater rated 60 W is placed in a hole in the block. The heater is on for 420 s and the temperature of the block rises by 12.0 °C. Assume that the heater transfers all its electrical energy as thermal energy.

thermometermetal block, 2.0 kgsupplyheater (60 W)
Diagram not to scale
(a)

Calculate the electrical energy supplied by the heater.

(2)
(b)

Calculate the specific heat capacity of the metal using these results.

(2)
(c)

The accepted value for this metal is 900 J / (kg °C). Explain why the student's value is higher.

(2)
(d)

Suggest one change that would give a more accurate result.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = Pt✓ 1
60 × 420 = 25 200 J✓ 1
Part (b)
c = E / (mΔT)✓ 1
25 200 / (2.0 × 12.0) = 1050 J / (kg °C) (ECF from (a))✓ 1
Part (c)
Some thermal energy is lost to the surroundings✓ 1
so the temperature rise is smaller than if all the energy stayed in the block, giving a larger calculated c✓ 1
Part (d)
Wrap the block in insulation / lagging✓ 1

Common error: If energy is lost, the temperature rise is smaller than expected, so the calculated specific heat capacity is too high, not too low.

297IGPC3-2-07r13-16Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

A tram of mass 18 000 kg is travelling at 20 m / s and brakes to rest. During braking, 80% of its kinetic energy is transferred equally to four steel brake discs. Each disc has a mass of 36 kg and a specific heat capacity of 500 J / (kg °C). Assume no thermal energy is lost from the discs during braking.

(a)

Calculate the kinetic energy of the tram before braking.

(2)
(b)

Calculate the thermal energy transferred to each disc.

(2)
(c)

Calculate the rise in temperature of one disc.

(2)
(d)

After the tram has stopped the discs cool down. Explain why the temperature falls more slowly as time passes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
KE = ½mv2✓ 1
0.5 × 18 000 × 202 = 3.6 × 106 J✓ 1
Part (b)
0.80 × 3.6 × 106 = 2.88 × 106 J✓ 1
2.88 × 106 / 4 = 7.2 × 105 J (ECF from (a))✓ 1
Part (c)
ΔT = E / (mc)✓ 1
7.2 × 105 / (36 × 500) = 40 °C (ECF from (b))✓ 1
Part (d)
The rate of loss of thermal energy depends on the temperature difference between the disc and the surroundings✓ 1
As the disc cools this difference decreases, so the rate of energy loss decreases✓ 1

Exam tip: In multi-step calculations, write the energy value you get at each stage so that later marks can still be earned if an earlier slip occurs.

298IGPC3-2-05r14-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

An archaeologist's lab melts a sample of ice at 0 °C using an electric heater of power 167 W that is switched on for 400 s. All the thermal energy from the heater goes into melting the ice. The specific latent heat of fusion of ice is 334 J/g. What mass of ice melts?

Show mark scheme
Marking pointMarkNotes
Energy supplied = 167 × 400 = 66 800 J. Mass = E ÷ L = 66 800 ÷ 334 = 200 g, so B. A (16 g) divides by 4.2 J/(g °C), the specific heat capacity, instead of the latent heat. C (400 g) divides by the power instead of the latent heat. D (0.20 g) comes from mixing up kilojoules and joules.✓ 1Answer B

Answer: B

Common error: Latent heat is used for a change of state at constant temperature; the specific heat capacity is for a temperature change.

299IGPC3-2-05r14-14Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

A conservator melts solid beeswax to seal an ancient pot. The wax has a mass of 0.25 kg and is at 25 °C. Its melting point is 65 °C. For the wax, the specific heat capacity in the solid state is 2500 J/(kg °C) and the specific latent heat of fusion is 2.0 × 105 J/kg. The wax is heated by a 250 W heater. Assume all the energy from the heater goes into the wax.

(a)

Calculate the energy needed to raise the temperature of the solid wax to its melting point.

(2)
(b)

Calculate the energy needed to melt all of the wax once it is at its melting point.

(2)
(c)

Calculate the total time for which the heater must be on.

(2)
(d)

Explain, in terms of particles, why the temperature of the wax stays constant while it melts although energy is still being supplied.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = mcΔT = 0.25 × 2500 × (65 − 25)✓ 1
E = 25 000 J✓ 1
Part (b)
E = mL = 0.25 × 2.0 × 105✓ 1
E = 50 000 J✓ 1
Part (c)
t = total energy ÷ power = (25 000 + 50 000) ÷ 250✓ 1
t = 300 s (ECF from (a) and (b))✓ 1
Part (d)
the energy supplied is used to break the bonds (overcome the forces) between particles✓ 1
so the average kinetic energy of the particles does not increase✓ 1

Common error: Do not use the specific heat capacity for the melting stage: during a change of state the energy is calculated with the latent heat.

300IGPC3-2-05r14-16Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

During a hill climb, sweat evaporates from a cyclist's skin.

(a)

Explain, in terms of particles, why evaporation cools the liquid that remains.

(3)
(b)

State two factors that increase the rate of evaporation of sweat.

(2)
(c)

The specific latent heat of vaporisation of sweat is 2.3 × 106 J/kg. Calculate the energy taken from the cyclist when 0.040 kg of sweat evaporates.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
the particles with the most kinetic energy (fastest) escape from the surface✓ 1
the particles that remain have a lower average kinetic energy✓ 1
so the temperature of the remaining liquid falls✓ 1
Part (b)
any two from: higher temperature; larger surface area; moving air or wind across the skin; lower humidity (drier air)✓ 2
Part (c)
E = mL = 0.040 × 2.3 × 106✓ 1
E = 9.2 × 104 J (92 000 J)✓ 1

Common error: Evaporation happens at any temperature and only at the surface; do not confuse it with boiling.

301IGPC3-2-07r14-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A humidifier in a rare-book store contains water already at 100 °C. Its heater has power 2.0 kW and 80% of the electrical energy goes into the water. The specific latent heat of vaporisation of water is 2.3 × 106 J/kg. How long does it take to turn 0.40 kg of this water into steam?

Show mark scheme
Marking pointMarkNotes
A is correct: energy needed = 0.40 × 2.3 × 106 = 9.2 × 105 J; useful power = 0.80 × 2000 = 1600 W; time = 920 000 / 1600 = 575 s. B is wrong because it ignores the 80% efficiency. C is wrong because it multiplies the time by 0.80 instead of dividing. D is wrong because it omits both the mass and the efficiency.✓ 1Answer A

Answer: A

Common error: If only part of the supplied energy is useful, divide the energy needed by the useful power (efficiency × power).

302IGPC3-2-07r14-14Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

A steam humidifier in a library archive contains 0.25 kg of water at 20 °C. Its heater has power 2.5 kW and all the energy goes to the water. The specific heat capacity of water is 4200 J/(kg °C) and the specific latent heat of vaporisation of water is 2.3 × 106 J/kg.

(a)

Calculate the energy needed to raise the temperature of the water to its boiling point of 100 °C.

(2)
(b)

After reaching 100 °C, 0.050 kg of the water is turned into steam. Calculate the energy needed for this.

(2)
(c)

Calculate the total time for the heater to supply the energy in (a) and (b).

(2)
(d)

Explain why the temperature of the water stays at 100 °C while it is boiling.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
ΔE = 0.25 × 4200 × 80✓ 1
= 84 000 J✓ 1
Part (b)
ΔE = ml = 0.050 × 2.3 × 106✓ 1
= 115 000 J✓ 1
Part (c)
Total energy = 84 000 + 115 000 = 199 000 J (ECF from (a) and (b))✓ 1
t = E / P = 199 000 / 2500 = 79.6 s✓ 1
Part (d)
The energy supplied is used to overcome the forces between particles (increase potential energy), not to increase their average kinetic energy✓ 1

Common error: During a change of state the energy supplied changes the arrangement of particles, not the temperature.

303IGPC3-2-07r14-15Thermal expansion, specific heat capacity and changes of state Paper EHard6 marks

A student heats a 1.2 kg block of metal using an electric heater of power 60 W for 5.0 minutes. The temperature of the block rises from 20.0 °C to 35.0 °C. Assume that all the electrical energy goes into the block.

to power supplyheater (60 W)thermometermetal blockmass 1.2 kg
Diagram not to scale
(a)

Calculate the electrical energy supplied to the heater.

(2)
(b)

Calculate the specific heat capacity of the metal.

(2)
(c)

A data book gives 900 J/(kg °C) for this metal. Suggest why the student's value is higher.

(1)
(d)

Suggest one improvement to the experiment.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = Pt = 60 × 300✓ 1
= 18 000 J✓ 1
Part (b)
c = E / (mΔθ) = 18 000 / (1.2 × 15.0) (ECF from (a))✓ 1
= 1000 J/(kg °C)✓ 1
Part (c)
Some energy was lost to the surroundings, so the block gained less than 18 000 J and the temperature rise was smaller than assumed✓ 1
Part (d)
Wrap the block in thermal insulation / lag it, or add a lid, to reduce energy loss✓ 1

Exam tip: Convert minutes to seconds before using E = Pt.

304IGPC3-2-07r15-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

Water evaporates from a wet cloth at a temperature below the boiling point of water. Which statement explains why the water left in the cloth becomes cooler?

Show mark scheme
Marking pointMarkNotes
Option D is correct: only the fastest surface particles have enough energy to escape, so the average kinetic energy of those left falls and the water cools. Option A is wrong because the slowest particles do not escape first, and removing slow particles would raise the average. Option B is wrong because air does not slow the particles in this way. Option C is wrong because the forces between particles do not strengthen.✓ 1Answer D

Answer: D

Exam tip: Link evaporation to the escape of the most energetic particles from the surface at any temperature.

305IGPC3-2-07r15-16Thermal expansion, specific heat capacity and changes of state Paper EHard7 marks

An immersion heater of power 2.0 kW is used to heat 18 kg of water in a fish-processing plant from 12 °C to 62 °C. This takes 38 minutes. The specific heat capacity of water is 4200 J/(kg °C).

to power supplyimmersion heater, 2.0 kWthermometerwater, 18 kg
Diagram not to scale
(a)

Calculate the energy gained by the water.

(2)
(b)

Calculate the electrical energy supplied by the heater in 38 minutes.

(2)
(c)

Calculate the efficiency of the heating process as a percentage, to 3 significant figures.

(2)
(d)

Suggest one change that would increase the efficiency.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
Energy = m c ΔT = 18 × 4200 × 50✓ 1
= 3.78 × 106 J (3780000 J)✓ 1
Part (b)
Converts 38 minutes to 2280 s and uses E = P t = 2000 × 2280✓ 1
= 4.56 × 106 J (4560000 J)✓ 1
Part (c)
Efficiency = useful energy / total energy × 100 = 3.78 × 106 / 4.56 × 106 × 100✓ 1
= 82.9 %✓ 1
Part (d)
Insulate (lag) the container or fit a lid so less energy is lost to the surroundings✓ 1

Exam tip: Check that power is in watts and time in seconds before multiplying; 2.0 kW is 2000 W.

306IGPC3-2-06r16-08Particle model and gas pressure Paper EHard1 mark

A sealed rigid cylinder holds gas at a temperature of 27 °C and a pressure of p. The gas is heated to 87 °C. What is the new pressure?

Show mark scheme
Marking pointMarkNotes
C is correct: at constant volume p ∝ T in kelvin, so the ratio is 360 ÷ 300 = 1.2. A and B use Celsius values (27 ÷ 87 and 87 ÷ 27). D uses the Celsius rise divided by the starting Celsius value, 60 ÷ 27.✓ 1Answer C

Answer: C

Common error: Gas-law calculations need kelvin; Celsius ratios are meaningless.

307IGPC3-2-06r16-14Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

A conservator melts 0.30 kg of solid wax, initially at 20 °C, using a heater of power 120 W. The wax has specific heat capacity 2400 J/(kg °C) as a solid, a melting point of 62 °C and a specific latent heat of fusion of 1.8 × 105 J/kg. Assume all the energy from the heater goes into the wax.

(a)

Define specific latent heat of fusion.

(1)
(b)

Calculate the energy needed to raise the temperature of the solid wax to 62 °C.

(2)
(c)

Calculate the energy needed to melt all the wax at 62 °C.

(2)
(d)

Calculate the time the heater takes to melt all the wax. ECF from (b) and (c).

(2)
(e)

Suggest why the real time is longer.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy needed to change 1 kg of a substance from solid to liquid at constant temperature (at its melting point)✓ 1
Part (b)
M1 0.30 × 2400 × 42✓ 1
A1 30 240 J✓ 1
Part (c)
M1 0.30 × 1.8 × 105✓ 1
A1 54 000 J✓ 1
Part (d)
M1 total energy = 30 240 + 54 000 = 84 240 J✓ 1
A1 time = 84 240 ÷ 120 = 702 s✓ 1
Part (e)
some energy from the heater is lost to the surroundings✓ 1

Exam tip: For a heating-then-melting problem, split the energy into two separate calculations and add them.

308IGPC3-2-06r16-16Particle model and gas pressure Paper EHard7 marks

A cyclist's pump contains 120 cm³ of air at atmospheric pressure, 1.00 × 105 Pa. The air is compressed to 30 cm³ at constant temperature before being pushed into a tyre.

Before: p = 1.00 × 10⁵ PaV = 120 cm³gasAfter: p = ?V = 30 cm³
Diagram not to scale
(a)

Calculate the pressure of the compressed air.

(2)
(b)

Explain, in terms of particles, why the pressure increases when the volume decreases.

(2)
(c)

The air in the tyre is at a pressure of 3.00 × 105 Pa and a temperature of 15 °C. The tyre warms to 45 °C and its volume stays constant. Calculate the new pressure. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p1V1 = p2V2✓ 1
A1 4.00 × 105 Pa✓ 1
Part (b)
particles are closer together in the smaller volume✓ 1
so they hit each unit area of the walls more often (greater force per unit area)✓ 1
Part (c)
M1 convert to kelvin: 288 K and 318 K✓ 1
M1 p1/T1 = p2/T2 so p2 = 3.00 × 105 × 318 ÷ 288✓ 1
A1 331250 so 3.31 × 105 Pa✓ 1

Common error: Using °C in the gas laws gives wrong answers; add 273 to each temperature.

309IGPC3-2-07r17-08Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A physiotherapy clinic warms 0.50 kg of water from 20 °C to 60 °C using an electric heater of power 1.2 kW running for 2.0 minutes. The specific heat capacity of water is 4200 J/(kg °C).
How much energy is transferred to the surroundings and not to the water?

Show mark scheme
Marking pointMarkNotes
Electrical energy supplied = 1200 × 120 = 144 kJ. Energy gained by the water = 0.50 × 4200 × 40 = 84 kJ. Energy lost = 144 − 84 = 60 kJ, so A is correct. B is only the energy gained by the water. C is the total energy supplied, ignoring what the water gained. D adds the two energies instead of subtracting them.✓ 1Answer A

Answer: A

Exam tip: Work out the energy supplied and the energy gained separately, then subtract to find the loss.

310IGPC3-2-07r17-14Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

In an astronomy laboratory, a sample of ice of mass 0.25 kg at −20 °C is used to model the nucleus of a comet. It is heated by an electric heater of power 500 W. Assume all the energy from the heater goes into the sample.
Specific heat capacity of ice = 2100 J/(kg °C)
Specific latent heat of fusion of ice = 3.4 × 105 J/kg

(a)

Calculate the energy needed to raise the temperature of the ice from −20 °C to 0 °C.

(2)
(b)

Calculate the energy needed to melt all the ice at 0 °C.

(2)
(c)

Calculate the total time for which the heater must be on to melt all the ice, starting from −20 °C.

(2)
(d)

Explain, in terms of particles, why the temperature stays at 0 °C while the ice is melting.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = mcΔT = 0.25 × 2100 × 20✓ 1
E = 10 500 J✓ 1
Part (b)
E = mL = 0.25 × 3.4 × 105✓ 1
E = 85 000 J✓ 1
Part (c)
Total energy = 10 500 + 85 000 = 95 500 J (ECF from (a) and (b))✓ 1
t = 95 500 / 500 = 191 s✓ 1
Part (d)
The energy supplied is used to overcome the forces between the particles (break the bonds in the solid)✓ 1
rather than to increase the average kinetic energy of the particles✓ 1

Exam tip: For a change of state there is no temperature change, so use mL; use mcΔT only when the temperature is changing.

311IGPC3-2-07r17-16Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

A runner produces excess thermal energy at a rate of 600 W for 30 minutes. 80% of this energy is removed by the evaporation of sweat. The rest stays in the runner's body.
Specific latent heat of vaporisation of sweat = 2.4 × 106 J/kg
Mass of the runner = 60 kg
Specific heat capacity of the runner's body = 3600 J/(kg °C)

(a)

Show that the excess thermal energy produced in 30 minutes is 1.08 × 106 J.

(2)
(b)

Calculate the mass of sweat that evaporates.

(3)
(c)

Calculate the rise in temperature of the runner's body. Give your answer to 2 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
30 minutes = 1800 s✓ 1
E = Pt = 600 × 1800 = 1.08 × 106 J✓ 1
Part (b)
Energy removed by evaporation = 0.80 × 1.08 × 106 = 8.64 × 105 J✓ 1
m = E / L = 8.64 × 105 / 2.4 × 106✓ 1
m = 0.36 kg✓ 1
Part (c)
Energy stored = 0.20 × 1.08 × 106 = 2.16 × 105 J✓ 1
ΔT = E / mc = 2.16 × 105 / (60 × 3600)✓ 1
ΔT = 1.0 °C✓ 1

Common error: Use the correct share of the energy in each calculation: 80% for evaporation (mL) and 20% for the temperature rise (mcΔT).

312IGPC3-2-06r25-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A ship's cold store uses 2.0 kg of ice at 0 °C to chill fish. The ice melts and the water warms to 10 °C. The specific latent heat of fusion of ice is 340000 J/kg and the specific heat capacity of water is 4200 J/(kg °C). How much energy does the ice and meltwater absorb in total?

Show mark scheme
Marking pointMarkNotes
Melting: 2.0 × 340000 = 680000 J. Warming the water: 2.0 × 4200 × 10 = 84000 J. Total = 764000 J. Using only the melting energy or only the warming energy omits one stage, and omitting the mass in both stages gives 340000 + 42000 = 382000 J.✓ 1Answer C

Answer: C

Exam tip: When a substance changes state and then changes temperature, calculate each stage separately and add the energies.

313IGPC3-2-06r25-14Thermal expansion, specific heat capacity and changes of state Paper EHard9 marks

A volcanologist heats 0.25 kg of water at 20 °C in a pan using a heater of power 800 W, until the water boils at 100 °C, and then continues heating until 0.040 kg of the water has boiled away. Specific heat capacity of water = 4200 J/(kg °C). Specific latent heat of vaporisation of water = 2.26 × 106 J/kg. Assume all the energy from the heater is transferred to the water.

(a)

Calculate the energy needed to raise the temperature of the water from 20 °C to 100 °C.

(2)
(b)

Calculate the energy needed to boil away 0.040 kg of water at 100 °C.

(2)
(c)

Calculate the total energy supplied. Give the exact value.

(1)
(d)

Calculate the total time for which the heater is switched on. Use the exact value from (c).

(2)
(e)

Explain, in terms of particles, why the temperature stays at 100 °C while the water boils.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
E = mcΔθ = 0.25 × 4200 × 80✓ 1
84000 J✓ 1
Part (b)
E = mL = 0.040 × 2.26 × 106✓ 1
90400 J✓ 1
Part (c)
84000 + 90400 = 174400 J (ECF from (a) and (b))✓ 1
Part (d)
time = energy ÷ power = 174400 ÷ 800✓ 1
218 s✓ 1
Part (e)
the energy supplied is used to overcome the forces between particles so they can separate✓ 1
the average kinetic energy of the particles does not change, so the temperature stays constant✓ 1

Exam tip: Split a heating problem at each temperature change or change of state, and use mcΔθ for temperature changes but mL for changes of state.

314IGPC3-2-09r33-06Particle model and gas pressure Paper EHard1 mark

A fixed mass of gas in the cylinder of a ship's refrigeration unit is compressed to half its volume while its temperature is kept constant. The pressure doubles. Which statement explains the increase in pressure in terms of particles?

Show mark scheme
Marking pointMarkNotes
At constant temperature the average speed is unchanged. In half the volume the particles reach the walls more often, so the collision rate per unit area doubles and the pressure doubles. Particle speed depends on temperature, the number of particles is fixed, and particles are not squashed.✓ 1Answer D

Answer: D

Exam tip: For gas pressure answers, mention both how often particles hit the walls and how hard they hit.

315IGPC3-2-09r33-13Particle model and gas pressure Paper EHard7 marks

Divers collect gas from deep water in a flexible sampler. At the depth of collection the sampler holds 45 cm3 of gas at a pressure of 3.0 × 105 Pa. The sampler is brought to the surface, where the pressure is 1.0 × 105 Pa. Assume the temperature of the gas and the mass of gas stay constant.

(a)

Calculate the volume of the gas at the surface.

(2)
(b)

At the surface the sampler is compressed, still at constant temperature, until its volume is 54 cm3. Calculate the new pressure. Use 135 cm3 as the starting volume.

(2)
(c)

Explain, in terms of particles, why the pressure of a gas falls when its volume increases at constant temperature.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of p1V1 = p2V2✓ 1
A1 V = 135 cm3✓ 1
Part (b)
M1 use of pV = constant with 1.0 × 105 Pa and 135 cm3✓ 1
A1 p = 2.5 × 105 Pa (250 000 Pa); ECF from (a)✓ 1
Part (c)
The average speed of the particles is unchanged✓ 1
The particles travel further between collisions with the walls, so the collisions with the walls are less frequent✓ 1
The force per unit area on the walls is smaller✓ 1

Exam tip: Use any consistent volume unit with pV = constant, but keep the same unit on both sides.

316IGPC3-2-09r33-15Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

A marine biology laboratory heats a 0.50 kg block of ice, used to chill specimens, from −10 °C until it has become water at 20 °C. A 500 W heater is used and all of its energy is transferred to the ice and water. Use these data: specific heat capacity of ice = 2100 J/(kg °C); specific latent heat of fusion of ice = 3.4 × 105 J/kg; specific heat capacity of water = 4200 J/(kg °C).

(a)

Calculate the energy needed to warm the ice from −10 °C to 0 °C.

(2)
(b)

Calculate the energy needed to melt the ice at 0 °C.

(2)
(c)

Calculate the energy needed to warm the melted ice from 0 °C to 20 °C.

(2)
(d)

Calculate the total time for which the heater is switched on. Use the exact values from (a), (b) and (c).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of E = mcΔθ with a temperature rise of 10 °C✓ 1
A1 E = 10 500 J✓ 1
Part (b)
M1 use of E = mL✓ 1
A1 E = 170 000 J✓ 1
Part (c)
M1 use of E = mcΔθ with a temperature rise of 20 °C✓ 1
A1 E = 42 000 J✓ 1
Part (d)
M1 total energy = sum of the three energies (222 500 J) divided by power✓ 1
A1 t = 445 s; ECF from (a), (b) and (c)✓ 1

Exam tip: Split a heating problem into stages: warm the solid, melt it, then warm the liquid; add the energies at the end.

317IGPC3-2-08r36-12Thermal expansion, specific heat capacity and changes of state Paper EHard9 marks

An electric shower heats water as it flows through the heater. The heater has a power of 7.56 kW. Water enters at 12 °C and flows through at a rate of 0.060 kg every second. The specific heat capacity of water is 4200 J/(kg °C).

(a)

Define specific heat capacity.

(2)
(b)

Assume all of the energy from the heater is transferred to the water. Calculate the rise in temperature of the water.

(3)
(c)

State the temperature of the water leaving the shower in part (b).

(1)
(d)

In practice the water leaves the heater at 38 °C. Calculate the efficiency of the heater, to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
thermal energy needed to raise the temperature✓ 1
of 1 kg of a substance by 1 °C✓ 1
Part (b)
energy transferred to the water every second = 7560 J✓ 1
use of ΔE = mcΔθ with m = 0.060 kg✓ 1
Δθ = 7560 ÷ (0.060 × 4200) = 30 °C✓ 1
Part (c)
42 °C (ECF from (b))✓ 1
Part (d)
useful energy per second = 0.060 × 4200 × (38 − 12)✓ 1
= 6552 J✓ 1
efficiency = 6552 ÷ 7560 × 100 = 86.7 %✓ 1

Common error: Use the temperature rise (final minus initial) in mcΔθ, not the final temperature.

318IGPC3-2-08r36-15Thermal expansion, specific heat capacity and changes of state Paper EHard9 marks

A pharmacist heats 0.25 kg of a solid compound using a 60 W electric heater. Assume that all the energy from the heater is transferred to the compound. The compound begins to melt at 5.0 min and is completely liquid at 10.0 min. The temperature is recorded every minute.
time / min: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
temperature / °C: 20, 28, 36, 44, 52, 60, 60, 60, 60, 60, 60, 66, 72

01234567891011121301020304050607080time / mintemperature / °C
Graph plotted from the data in the question
(a)

State the melting point of the compound.

(1)
(b)

Explain why the temperature stays constant while the compound is melting although the heater is still supplying energy.

(2)
(c)

Calculate the specific latent heat of fusion of the compound.

(3)
(d)

Calculate the specific heat capacity of the liquid, using the readings from 10 min to 12 min.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
60 °C✓ 1
Part (b)
the energy is used to overcome the forces between the particles (change the state)✓ 1
not to increase the kinetic energy of the particles, so the temperature does not change✓ 1
Part (c)
energy supplied during melting = 60 × 300 = 18 000 J✓ 1
use of l = E ÷ m✓ 1
l = 18 000 ÷ 0.25 = 72 000 J/kg (7.2 × 104 J/kg)✓ 1
Part (d)
energy supplied = 60 × 120 = 7200 J✓ 1
temperature rise = 12 °C and use of c = E ÷ (mΔθ)✓ 1
c = 7200 ÷ (0.25 × 12) = 2400 J/(kg °C)✓ 1

Common error: Convert minutes to seconds before using E = Pt: 5.0 min to 10.0 min is 300 s, not 5 s.

319IGPC3-2-09r41-05Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A theatre steam machine condenses 0.020 kg of steam at 100 °C to water and then the water cools to 40 °C. The specific latent heat of vaporisation of water is 2.26 × 106 J/kg and the specific heat capacity of water is 4200 J/(kg °C). How much thermal energy is released in total?

Show mark scheme
Marking pointMarkNotes
Condensing releases 0.020 × 2.26 × 106 = 45 200 J. Cooling the water through 60 °C releases 0.020 × 4200 × 60 = 5040 J. The total is 50 240 J, which is 50.2 kJ. Including only the condensation gives 45.2 kJ. Using a 100 °C fall in temperature for the cooling gives 53.6 kJ. Including only the cooling gives 5.04 kJ.✓ 1Answer B

Answer: B

Common error: When steam condenses and then cools, two separate energy terms are needed: latent heat for the change of state and mcΔθ for the cooling.

320IGPC3-2-09r41-08Particle model and gas pressure Paper EHard1 mark

A volcanologist seals a fixed mass of gas in a rigid bottle at 27 °C and a pressure of 1.00 × 105 Pa. The bottle is warmed to 87 °C. Take absolute zero as −273 °C. What is the new gas pressure?

Show mark scheme
Marking pointMarkNotes
The volume is fixed, so pressure is proportional to the kelvin temperature. The temperatures are 300 K and 360 K, so p = 1.00 × 105 × 360 / 300 = 1.20 × 105 Pa. Using the Celsius ratio 87/27 gives 3.22 × 105 Pa. Inverting the ratio gives 0.83 × 105 Pa. Adding the 60 °C rise on as a 60% increase gives 1.60 × 105 Pa.✓ 1Answer C

Answer: C

Common error: Gas law calculations need temperatures in kelvin; ratios of Celsius temperatures are meaningless.

321IGPC3-2-09r41-14Thermal expansion, specific heat capacity and changes of state Paper EHard8 marks

A ship's desalination unit contains 5.0 kg of seawater at 20 °C. An electric heater of power 6.0 kW heats the water to its boiling point of 100 °C and then boils off 1.5 kg of it as steam. For this question, treat the seawater as having the same specific heat capacity, 4200 J/(kg °C), and the same specific latent heat of vaporisation, 2.26 × 106 J/kg, as pure water. Assume all the energy from the heater is transferred to the water.

(a)

Calculate the energy needed to raise the temperature of the water from 20 °C to 100 °C.

(2)
(b)

Calculate the energy needed to boil off 1.5 kg of the water at 100 °C.

(2)
(c)

Calculate the time for which the heater must be switched on. Use your answers to (a) and (b) (ECF from (a) and (b)).

(3)
(d)

In practice the time needed is longer than your answer to (c). Suggest one reason.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = m c Δθ (M1)✓ 1
5.0 × 4200 × 80 = 1.68 × 106 J (A1)✓ 1
Part (b)
energy = m L (M1)✓ 1
1.5 × 2.26 × 106 = 3.39 × 106 J (A1)✓ 1
Part (c)
total energy = 1.68 × 106 + 3.39 × 106 = 5.07 × 106 J (A1)✓ 1
t = E / P = 5.07 × 106 / 6000 (M1)✓ 1
t = 845 s (A1)✓ 1
Part (d)
some energy is lost to the surroundings or used to heat the container✓ 1

Exam tip: Split a heating-then-boiling process into two energy steps and add them; the temperature does not change during boiling.

322IGPC3-2-07r42-07Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A technician melts 0.20 kg of ice at 0 °C in a fish-tank cooler using a 400 W heater for 170 s. All of the energy from the heater melts the ice and none is lost. What is the specific latent heat of fusion of the ice?

Show mark scheme
Marking pointMarkNotes
Energy supplied = 400 × 170 = 6.8 × 104 J. Specific latent heat = E / m = 6.8 × 104 / 0.20 = 3.4 × 105 J/kg. The value 1.4 × 104 J/kg comes from multiplying by the mass. The value 6.8 × 104 is the energy in joules, with no division by mass. The value 2.0 × 103 J/kg comes from dividing the power by the mass and ignoring the time.✓ 1Answer B

Answer: B

Common error: Latent heat is energy per kilogram: divide the energy by the mass, and find the energy from power × time first.

323IGPC3-2-07r42-13Thermal expansion, specific heat capacity and changes of state Paper EHard9 marks

A logistics company ships vaccines in an insulated container that holds a block of ice of mass 2.0 kg at 0 °C. Energy leaks into the container at a constant rate of 170 W and all of it goes into the ice or the water formed. The specific latent heat of fusion of ice is 3.4 × 105 J/kg. The specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the energy needed to melt all of the ice.

(2)
(b)

Calculate the time taken to melt all of the ice.

(2)
(c)

Explain, in terms of particles, why the temperature stays at 0 °C while the ice melts.

(2)
(d)

After all the ice has melted, the water absorbs energy at the same rate. Calculate the further time taken for the water to warm from 0 °C to 5.0 °C. Give your answer to 3 significant figures.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 E = mL = 2.0 × 3.4 × 105✓ 1
A1 6.8 × 105 J✓ 1
Part (b)
M1 t = E / P = 6.8 × 105 / 170✓ 1
A1 4000 s (ECF from (a))✓ 1
Part (c)
The energy supplied is used to overcome the forces between the particles✓ 1
so the potential energy of the particles increases while their average kinetic energy, and so the temperature, stays constant✓ 1
Part (d)
M1 E = mcΔθ = 2.0 × 4200 × 5.0 (= 42 000 J)✓ 1
M1 t = E / P = 42 000 / 170✓ 1
A1 247 s✓ 1

Common error: During a change of state the supplied energy changes the potential energy of the particles, not their temperature.

324IGPC3-2-08r47-06Thermal expansion, specific heat capacity and changes of state Paper EHard1 mark

A 5.0 kW heater supplies all its energy to 250 kg of water in a small pool. The specific heat capacity of water is 4200 J/(kg °C). How long does the heater take to raise the temperature of the water by 2.0 °C?

Show mark scheme
Marking pointMarkNotes
Energy needed = mcΔθ = 250 × 4200 × 2.0 = 2.1 × 106 J. Time = energy ÷ power = 2.1 × 106 ÷ 5000 = 420 s. Leaving out the temperature change of 2.0 °C gives 210 s. Leaving out the mass gives 1.7 s. Using 5.0 for the power instead of 5000 W gives 4.2 × 105 s.✓ 1Answer B

Answer: B

Common error: Convert kilowatts to watts before dividing energy by power.

325IGPC3-2-08r47-13Thermal expansion, specific heat capacity and changes of state Paper EHard10 marks

Ice of mass 12 kg at 0 °C has formed on a wind-turbine blade. A 4.0 kW heater supplies energy to the ice. Assume that all of the heater's energy goes into the ice and then into the water formed. The specific latent heat of fusion of ice is 3.3 × 105 J/kg. The specific heat capacity of water is 4200 J/(kg °C).

(a)

Calculate the energy needed to melt all of the ice.

(2)
(b)

Calculate the time taken to melt all of the ice.

(2)
(c)

The water formed is then heated from 0 °C to 5.0 °C. Calculate the energy needed to do this.

(2)
(d)

Calculate the total time from switching on the heater until the water is at 5.0 °C. Give your answer to 3 significant figures.

(2)
(e)

Explain, in terms of particles, why the temperature stays at 0 °C while the ice is melting.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
energy = mass × specific latent heat = 12 × 3.3 × 105✓ 1
3.96 × 106 J✓ 1
Part (b)
time = energy ÷ power = 3.96 × 106 ÷ 4000✓ 1
990 s✓ 1
Part (c)
energy = 12 × 4200 × 5.0✓ 1
2.52 × 105 J✓ 1
Part (d)
total energy = 3.96 × 106 + 2.52 × 105 = 4.212 × 106 J✓ 1
4.212 × 106 ÷ 4000 = 1053 s, so 1050 s to 3 significant figures✓ 1
Part (e)
the energy supplied is used to overcome the forces between the particles✓ 1
so the average kinetic energy of the particles does not increase✓ 1

Exam tip: Treat melting and warming as separate stages: use E = mL for the change of state and E = mcΔθ for the temperature rise.

326IGPC3-2-08r47-16Particle model and gas pressure Paper EHard10 marks

A hydraulic accumulator on a wind turbine contains 0.050 m3 of nitrogen gas at a pressure of 4.0 × 105 Pa. The gas is compressed slowly to a volume of 0.020 m3 with its temperature constant.

Before: p = 4.0 × 10⁵ PaV = 0.050 m³gasAfter: p = ?V = 0.020 m³
Diagram not to scale
(a)

Calculate the new pressure of the gas.

(3)
(b)

Explain, in terms of particles, why the pressure increases as the gas is compressed at constant temperature.

(3)
(c)

The compressed gas is then released into the open air, where the pressure is 1.0 × 105 Pa. Calculate the volume the gas occupies, assuming that the temperature is the same as at the start.

(2)
(d)

In practice the gas gets warmer when it is compressed quickly. Explain this in terms of the particles.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
pressure × volume is constant (p1V1 = p2V2)✓ 1
4.0 × 105 × 0.050 = p2 × 0.020✓ 1
1.0 × 106 Pa✓ 1
Part (b)
the same number of particles is in a smaller volume✓ 1
the particles hit the walls more often✓ 1
so the force per unit area on the walls is greater✓ 1
Part (c)
use the original values: 4.0 × 105 × 0.050 = 1.0 × 105 × V✓ 1
V = 0.20 m3✓ 1
Part (d)
work is done on the gas, which gives the particles more kinetic energy✓ 1
the average speed of the particles increases, so the temperature rises✓ 1

Common error: The relationship pV = constant only applies when the temperature and the amount of gas stay the same.

327IGPC3-2-09r47-15Particle model and gas pressure Paper EHard8 marks

A mountaineer uses a cylinder of oxygen of internal volume 3.0 dm3. The cylinder contains oxygen at a pressure of 1.5 × 107 Pa. The oxygen is released through a regulator at atmospheric pressure, 1.0 × 105 Pa. The temperature is constant throughout.

(a)

Calculate the volume that all of the oxygen in the cylinder would occupy at atmospheric pressure.

(2)
(b)

The cylinder cannot be emptied completely: when the regulator stops the flow, the cylinder still holds 3.0 dm3 of oxygen at atmospheric pressure. Calculate the volume of oxygen, measured at atmospheric pressure, that the climber can use.

(2)
(c)

The climber uses 3.0 dm3 of oxygen each minute, measured at atmospheric pressure. Calculate how long the oxygen lasts. Use your answer to (b).

(2)
(d)

Explain, in terms of particles, why the pressure in the cylinder falls as oxygen is released.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 p1V1 = p2V2, so V = 1.5 × 107 × 3.0 / 1.0 × 105✓ 1
A1 450 dm3✓ 1
Part (b)
M1 volume used = 450 − 3.0 (ECF from (a))✓ 1
A1 447 dm3✓ 1
Part (c)
M1 time = 447 / 3.0✓ 1
A1 149 minutes (ECF from (b))✓ 1
Part (d)
B1 there are fewer particles in the same volume✓ 1
B1 so there are fewer collisions per second with the walls, giving a lower pressure✓ 1

Common error: The gas left in the cylinder at atmospheric pressure cannot be used, so subtract it before working out how long the supply lasts.