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International GCSE Physics · Pearson Edexcel 4PH1 · Topic 5
315 free, high-quality past-paper-style questions on Solids, liquids and gases for Edexcel IGCSE Physics, each with a full mark scheme. Covers: (d) Ideal Gas Molecules; (b) Density and Pressure; (c) Change of State. For when you have finished the past papers, or want to practise one topic at a time.Notes 5(b) →Notes 5(c) →Notes 5(d) →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
An engineer lowers a small pressure sensor to a fixed depth of 12 m in a still reservoir. At that depth she turns the sensor so that its sensing face points upwards, then sideways, then downwards.
Which statement describes the readings?
| Marking point | Mark | Notes |
|---|---|---|
| In a liquid at rest, the pressure at a given point acts equally in all directions. The pressure depends only on the depth, the density of the liquid and g. Because the depth does not change, the reading is the same whichever way the face points. Options A and B wrongly assume that pressure only acts in one direction. Option D wrongly assumes that pressure only acts vertically. | ✓ 1 | Answer C |
Answer: C
Exam tip: Pressure in a fluid depends on depth, not on direction. If the depth stays the same in a still liquid, the pressure stays the same.
A small piece of solid gallium metal is held in a warm hand and melts. Its temperature stays at 30 °C while it melts, even though energy is still being transferred to it from the hand.
Which statement describes what the energy transferred during melting does?
| Marking point | Mark | Notes |
|---|---|---|
| During a change of state the temperature stays constant, so the average kinetic energy of the particles does not change. The energy supplied does work against the attractive forces between the particles. This increases their potential energy, so the particles are no longer held in fixed positions and can move past one another. The first option confuses heating with a temperature rise. The third option wrongly assumes that the particles themselves get bigger. The fourth option wrongly assumes that a constant temperature means no energy is being stored. | ✓ 1 | Answer B |
Answer: B
Common error: A constant temperature does not mean that no energy is being absorbed. During melting the energy increases the potential energy of the particles. Their kinetic energy does not increase.
Engineers test a wax for use in a heat-storage panel. They melt the wax and then record its temperature every minute as it cools.
Readings in °C: 85, 72, 63, 56, 56, 56, 56, 49, 43
What is the freezing point of the wax?
| Marking point | Mark | Notes |
|---|---|---|
| While a substance freezes, its temperature stays constant, because energy is released as the particles form bonds. The readings stay at 56 °C for four minutes, so the freezing point is 56 °C. 85 °C is only the starting temperature. 63 °C is a reading taken while the liquid is still cooling. 43 °C is the solid cooling after freezing has finished. | ✓ 1 | Answer C |
Answer: C
Exam tip: In cooling data, the freezing point is the temperature that repeats over several consecutive readings.
A student investigates the cooling of palmitic acid, a solid used in making cosmetics. She melts some palmitic acid in a boiling tube in a hot water bath. She then removes the tube and lets the palmitic acid cool until it is solid again.
State the two measuring instruments she needs.
Describe how she should collect the data needed to plot a temperature–time graph.
State what her readings would show while the palmitic acid is freezing.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Thermometer / temperature probe | ✓ 1 | |
| Stopwatch / stopclock / timer / data logger with internal clock | ✓ 1 | |
| Part (b) | ||
| Place the thermometer (or probe) in the liquid palmitic acid and start the stopwatch, or start the data logger | ✓ 1 | |
| Record the temperature at regular time intervals, e.g. every 30 s | ✓ 1 | |
| Continue until well after the palmitic acid has become solid / the temperature has fallen below the freezing point | ✓ 1 | |
| Part (c) | ||
| The temperature stays constant (at the freezing point) / the readings do not change for a period of time | ✓ 1 | |
Exam tip: Keep taking readings after the change of state. Without readings after it, the flat section cannot be identified.
A cyclist takes a cold can of drink from a cool box on a warm day. After a few minutes, small drops of water appear on the outside of the can.
State the name of the change of state that forms the drops.
Explain, in terms of particles, how the drops form.
State what happens to the temperature of the drink as more drops form.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Condensation / condensing | ✓ 1 | |
| Part (b) | ||
| Water vapour particles in the air hit the cold surface of the can and lose energy / slow down | ✓ 1 | |
| The particles move closer together and are attracted to each other, forming liquid water | ✓ 1 | |
| Part (c) | ||
| It increases / the drink warms up | ✓ 1 | |
Common error: The water does not leak through the can. It comes from water vapour already in the air.
A planetary rover has a weight of 3800 N on the surface of Mars. It rests evenly on six wheels. Each wheel has a contact area of 0.020 m2 with the ground.
What is the pressure exerted on the ground by the rover?
| Marking point | Mark | Notes |
|---|---|---|
| The total contact area is 6 × 0.020 = 0.12 m2. Pressure = force ÷ area = 3800 ÷ 0.12 = 31 700 Pa ≈ 3.2 ×104 Pa. Option A divides area by force, B multiplies force by area, and D uses the area of only one wheel. | ✓ 1 | Answer C |
Answer: C
Common error: When a load is shared by several supports, add their contact areas before dividing. Using only one wheel's area makes the pressure six times too large.
The concrete wall of a dam is built much thicker at the bottom than at the top.
Which statement explains why?
| Marking point | Mark | Notes |
|---|---|---|
| The pressure in a liquid increases with depth (p = ρgh), so the force on each square metre of wall is greatest near the bottom. The wall must be strongest there. Water is almost incompressible, so A is wrong. The width of the reservoir does not affect the pressure, so C is wrong. Pressure in a liquid acts in all directions, not only downwards, so D is wrong. | ✓ 1 | Answer B |
Answer: B
Common error: Liquid pressure depends on depth and density, not on how much liquid there is or how wide the container is.
A surgeon fits a titanium component into a patient's hip joint. The component has a mass of 0.090 kg. The density of titanium is 4500 kg/m3.
State the equation linking density, mass and volume.
Calculate the volume of the titanium component. Give your answer in m3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| density = mass ÷ volume (ρ = m/V) | ✓ 1 | |
| Part (b) | ||
| rearrange to V = m ÷ ρ | ✓ 1 | |
| V = 0.090 ÷ 4500 | ✓ 1 | |
| V = 2.0 ×10−5 m3 | ✓ 1 | |
Exam tip: Write the rearranged equation as a separate line before substituting. This can earn a mark even if the arithmetic later goes wrong.
An engineer wants to find the density of an irregularly shaped plastic part made on a 3D printer. The mass of the part is 15.0 g. The engineer lowers the part into a measuring cylinder of water until it is completely under the surface. The water level rises from 46.0 cm3 to 58.5 cm3.
Calculate the volume of the part.
Calculate the density of the part in g/cm3.
Convert your calculated density of the part into kg/m3.
An air bubble stays stuck to the surface of the part while it is under the water. Explain the effect of this on the calculated density.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 58.5 − 46.0 = 12.5 cm3 | ✓ 1 | |
| Part (b) | ||
| ρ = m ÷ V = 15.0 ÷ 12.5 | ✓ 1 | |
| ρ = 1.20 g/cm3 (accept 1.2) | ✓ 1 | |
| Part (c) | ||
| 1200 kg/m3 (allow follow-through from the density calculated) | ✓ 1 | |
| Part (d) | ||
| the measured volume is too large / the water rises more than it should | ✓ 1 | |
| so the calculated density is too small | ✓ 1 | |
Exam tip: To convert from g/cm³ to kg/m³, multiply by 1000. Water has a density of 1.0 g/cm³, which is 1000 kg/m³. Use this as a quick check on your answer.
A titanium part from a racing bicycle frame has a mass of 0.90 kg and a volume of 200 cm3.
What is the density of titanium?
| Marking point | Mark | Notes |
|---|---|---|
| Volume = 200 cm3 = 200 × 10−6 m3 = 2.0 × 10−4 m3. Density = mass ÷ volume = 0.90 ÷ 2.0 × 10−4 = 4500 kg/m3. Option A uses cm3 without converting, giving 0.0045 kg/cm3 and then a wrong power of ten. Option B multiplies mass by volume. Option C divides volume by mass. | ✓ 1 | Answer D |
Answer: D
Common error: 1 m³ = 1 000 000 cm³, not 100 cm³. Convert cm³ to m³ by multiplying by 10⁻⁶ before dividing.
A small electronic pressure sensor is held at a fixed depth of 1.5 m in a swimming pool. The sensor is turned so that it faces upwards, then sideways, then downwards.
Which statement about the sensor readings is correct?
| Marking point | Mark | Notes |
|---|---|---|
| At a given point in a liquid, the pressure acts equally in all directions. The pressure depends only on the depth and the density of the liquid, not on which way a surface faces. The other options wrongly treat pressure in a liquid as acting in only one direction. | ✓ 1 | Answer B |
Answer: B
Common error: Pressure in a fluid is not just 'weight pushing down'. At a given depth it pushes equally on surfaces facing in any direction.
A nurse pushes on the plunger of a syringe to inject a vaccine.
State the equation linking pressure, force and area.
The nurse pushes on the plunger with a force of 12 N. The area of the plunger in contact with the liquid is 1.5 cm2.
Calculate the pressure exerted on the liquid by the plunger. Give your answer in pascals.
The tip of the syringe needle has a very small area.
Explain why this helps the needle to pass through skin easily.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| pressure = force ÷ area or p = F/A | ✓ 1 | |
| Part (b) | ||
| area = 1.5 × 10−4 m2 | ✓ 1 | |
| p = 12 ÷ 1.5 × 10−4 | ✓ 1 | |
| p = 80 000 Pa or 8.0 × 104 Pa | ✓ 1 | |
| Part (c) | ||
| for the same force, a smaller area gives a larger pressure | ✓ 1 | |
| the large pressure is enough to break / pierce the skin | ✓ 1 | |
Common error: 1 cm² = 1 × 10⁻⁴ m², not 1 × 10⁻² m². Square the conversion factor as well as the unit.
A student is given a small, irregularly shaped fragment of a meteorite. She wants to find its density.
Describe how the student can use a measuring cylinder and water to find the volume of the fragment.
The fragment has a mass of 154 g. The water level rises from 45.0 cm3 to 67.0 cm3.
Calculate the density of the fragment in g/cm3.
Suggest one reason why the student's value for the density may be inaccurate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| part-fill the measuring cylinder with water and record the volume | ✓ 1 | |
| lower the fragment in fully so it is submerged, record the new volume and subtract the first reading from it | ✓ 1 | |
| Part (b) | ||
| volume = 67.0 − 45.0 = 22.0 cm3 | ✓ 1 | |
| density = 154 ÷ 22.0 | ✓ 1 | |
| density = 7.0 g/cm3 | ✓ 1 | |
| Part (c) | ||
| any one from: air bubbles trapped on the fragment / water splashed out of the cylinder when it was added / the volume of the thread was included / the meniscus was not read at eye level | ✓ 1 | |
Exam tip: Mass in g divided by volume in cm³ gives density directly in g/cm³. No unit conversion is needed if the question asks for g/cm³.
Liquid nitrogen is used in hospitals to freeze and remove warts. It boils at −196 °C.
What is this temperature on the Kelvin scale?
| Marking point | Mark | Notes |
|---|---|---|
| To convert from °C to K, add 273: −196 + 273 = 77 K. A Kelvin temperature cannot be negative, and 469 K comes from subtracting −196 instead of adding it. 196 K simply drops the minus sign. | ✓ 1 | Answer A |
Answer: A
Common error: Adding 273 to a negative Celsius value gives a smaller number. Check that your Kelvin answer is positive and lower than 273 K for any temperature below 0 °C.
A student looks through a microscope at smoke particles trapped in a small glass cell full of air. The particles move in small, jerky, random paths.
What causes this motion?
| Marking point | Mark | Notes |
|---|---|---|
| Each smoke particle is hit by fast-moving, randomly moving air molecules, which are too small to see. At any instant more molecules hit one side than the other, so there is a resultant force in a random direction and the particle jerks about. Convection would produce a smooth, large-scale drift rather than random jerks. Repulsion or attraction between particles does not explain random changes of direction. | ✓ 1 | Answer B |
Answer: B
Exam tip: Write 'air molecules' (the invisible particles) and 'smoke particles' (the visible ones) so that it is clear which particles are doing the hitting.
A basketball is filled with air. The ball is sealed and kept at a constant temperature.
Describe the motion of the air molecules inside the ball.
Explain how the air molecules exert a pressure on the inside surface of the ball.
More air is pumped into the ball. The volume of the ball and the temperature of the air stay the same.
Explain why the pressure inside the ball increases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Molecules move randomly / continuously in random directions with a range of speeds | ✓ 1 | |
| Part (b) | ||
| Molecules collide with / bounce off the inside surface of the ball, exerting a force on it (accept: their momentum changes, so there is a force on the surface) | ✓ 1 | |
| Pressure is this force acting on each unit area of the surface / pressure = force ÷ area | ✓ 1 | |
| Part (c) | ||
| There are more molecules in the same volume (accept: the molecules are closer together / higher density) | ✓ 1 | |
| So there are more collisions with the inside surface each second, giving a greater force on each unit area (accept: molecules hit the surface more often because their average speed is unchanged) | ✓ 1 | |
Common error: Do not say the molecules move faster when more air is added. The temperature is constant, so the average speed of the molecules is unchanged. Only the number of collisions per second increases.
A portable oxygen cylinder used by paramedics has an internal volume of 10 dm3. When it is full, the oxygen is at a pressure of 15 000 kPa. Atmospheric pressure is 100 kPa.
State the equation linking the pressure and volume of a fixed mass of gas at constant temperature.
Calculate the volume the oxygen would occupy at atmospheric pressure.
State one assumption you made in your calculation.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p1V1 = p2V2 | ✓ 1 | |
| Part (b) | ||
| V2 = (15 000 × 10) ÷ 100 | ✓ 1 | |
| V2 = 1500 dm3 | ✓ 1 | |
| Part (c) | ||
| The temperature of the oxygen stays constant (OR no oxygen leaks out / the mass of gas is fixed) | ✓ 1 | |
Exam tip: Pressures and volumes do not have to be in SI units for p₁V₁ = p₂V₂. Use the same unit on both sides and the answer comes out in that unit.
The New Horizons space probe measured the surface temperature of Pluto as about −230 °C.
State the value of absolute zero in °C.
Explain, in terms of molecules, why there is a lowest possible temperature.
Convert the surface temperature of Pluto to kelvin.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| −273 °C (accept −273.15 °C) | ✓ 1 | |
| Part (b) | ||
| Temperature depends on the (average) kinetic energy of the molecules, so as temperature falls the molecules lose kinetic energy and move more slowly | ✓ 1 | |
| At absolute zero the molecules have minimum/zero kinetic energy (accept 'molecules stop moving'), so no more energy can be removed and the temperature cannot fall any further | ✓ 1 | |
| Part (c) | ||
| −230 + 273 = 43 K (accept 43.15 K using 273.15) | ✓ 1 | |
Common error: The unit is 'K', not '°K'. The Kelvin scale does not use a degree sign.
A small bracket for a drone is made on a 3D printer from nylon. The bracket has a mass of 54 g and a volume of 48 cm3.
Which of these is the density of the nylon?
| Marking point | Mark | Notes |
|---|---|---|
| Density = mass ÷ volume = 54 g ÷ 48 cm3 = 1.125 g/cm3, which is about 1.1 g/cm3. Option A comes from dividing volume by mass. Option C uses the value in kg/m3 (1100 kg/m3) but keeps the unit g/cm3. Option D comes from multiplying mass by volume. | ✓ 1 | Answer B |
Answer: B
Common error: Density is mass divided by volume, not the other way round. Check that your answer is sensible: most plastics have a density close to that of water, which is 1 g/cm³.
A mobile crane is set up on a soft grass verge. Before it lifts a load, the crane lowers steel legs called outriggers. Each outrigger stands on a flat pad. When the crane is lifting, one pad pushes down on the ground with a force of 48 000 N. The area of this pad is 0.60 m2.
State the equation linking pressure, force and area.
Calculate the pressure that the pad exerts on the ground.
The crane operator places a wider wooden mat under the pad. Explain why this makes the outrigger less likely to sink into the grass.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| pressure = force ÷ area (p = F/A) | ✓ 1 | |
| Part (b) | ||
| p = 48 000 ÷ 0.60 | ✓ 1 | |
| p = 80 000 Pa (80 kPa or 80 000 N/m2) | ✓ 1 | |
| Part (c) | ||
| The force is spread over a larger area (and the force stays the same) | ✓ 1 | |
| so the pressure on the ground is smaller and the ground is less likely to give way | ✓ 1 | |
Exam tip: When you explain a change in pressure, say what happens to both the force and the area. Then state the effect on the pressure.
A beekeeper melts a block of beeswax of mass 360 g and volume 400 cm3. What is the density of the beeswax in kg/m3?
| Marking point | Mark | Notes |
|---|---|---|
| Density = mass / volume = 0.360 kg / (4.00 × 10−4 m3) = 900 kg/m3, so C is correct. A is the value in g/cm3, with no conversion to kg/m3. B divides volume by mass, the wrong way up. D converts cm3 to m3 using a factor of 103 instead of 106. | ✓ 1 | Answer C |
Answer: C
Common error: Convert every quantity to SI units before substituting: 1 m³ = 10⁶ cm³, not 10³ cm³.
A hiker of mass 80 kg walks across deep snow. The total area of her two boots in contact with the snow is 0.040 m2. Use g = 10 N/kg.
Calculate the weight of the hiker.
Calculate the pressure that her boots exert on the snow.
The hiker now wears snowshoes. The total area of the two snowshoes in contact with the snow is 0.50 m2. Calculate the new pressure on the snow.
Explain why wearing snowshoes stops the hiker sinking so far into the snow.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = mg | ✓ 1 | |
| = 80 × 10 = 800 N | ✓ 1 | |
| Part (b) | ||
| p = F / A | ✓ 1 | |
| 800 / 0.040 = 20000 Pa | ✓ 1 | |
| Part (c) | ||
| p = 800 / 0.50 (ECF from (a)) | ✓ 1 | |
| 1600 Pa | ✓ 1 | |
| Part (d) | ||
| The same weight is spread over a larger area | ✓ 1 | |
| so the pressure on the snow is smaller (so she sinks less) | ✓ 1 | |
Exam tip: For an explain question on pressure, name the quantity that stays the same (the weight) and the one that changes (the area).
A beekeeper pours honey into an empty jar of capacity 2.5 litres until it is full. The empty jar has mass 0.40 kg. The full jar has mass 4.15 kg. 1 litre = 1.0 × 10−3 m3.
Calculate the mass of the honey.
Write the volume of the honey in m3.
Calculate the density of the honey.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.15 − 0.40 | ✓ 1 | |
| = 3.75 kg | ✓ 1 | |
| Part (b) | ||
| 2.5 × 10−3 m3 (0.0025 m3) | ✓ 1 | |
| Part (c) | ||
| ρ = m / V | ✓ 1 | |
| 3.75 / 0.0025 (ECF from (a) and (b)) | ✓ 1 | |
| 1500 kg/m3 | ✓ 1 | |
Common error: Subtract the mass of the container first. Using the mass of the full jar gives a density that is far too large.
The air in a tyre on an F1 car is at 25 °C when the car is parked. After a run the tyre is hot and the pressure of the air inside it has increased. The volume of the tyre does not change noticeably.
Convert 25 °C to a temperature in kelvin.
Explain, in terms of molecules, why the pressure of the air increases when the temperature rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 25 + 273 = 298 K | ✓ 1 | |
| Part (b) | ||
| The molecules gain kinetic energy and move faster | ✓ 1 | |
| They hit the tyre wall more often | ✓ 1 | |
| and with a greater force (change of momentum) in each collision | ✓ 1 | |
| so the force per unit area on the wall, the pressure, is larger | ✓ 1 | |
Exam tip: Link temperature to molecular speed first, then speed to collision frequency and force, and finish by naming pressure as force per unit area.
The casing of a robot's motor reaches a temperature of 47 °C. What is this temperature in kelvin?
| Marking point | Mark | Notes |
|---|---|---|
| Temperature in K = temperature in °C + 273, so 47 + 273 = 320 K, option C. A (226 K) subtracts 273 instead of adding. B (47 K) leaves the value unconverted. D (420 K) adds 373 instead of 273. | ✓ 1 | Answer C |
Answer: C
Common error: Add 273 to go from °C to K; never subtract it.
A fish farm heats 500 kg of water in a tank. The specific heat capacity of the water is 3900 J/kg °C. The temperature of the water rises by 4 °C. How much energy is transferred to the water?
| Marking point | Mark | Notes |
|---|---|---|
| Energy = m c Δθ = 500 × 3900 × 4 = 7 800 000 J = 7.8 × 106 J, option C. A divides by the temperature change instead of multiplying. B leaves out the specific heat capacity (m × Δθ only). D uses 277 (a kelvin temperature) as the temperature change. | ✓ 1 | Answer C |
Answer: C
Common error: The temperature change is the same in °C and in K, so no 273 is needed in this equation.
A hot-water bottle contains 1.2 kg of water at 65 °C. The water cools to 35 °C. The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy transferred from the water as it cools. Give your answer in kJ.
The bottle transfers energy to the room at an average rate of 42 W. Calculate how many minutes the water takes to cool from 65 °C to 35 °C.
Explain why water is a good choice of liquid for a hot-water bottle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Change in temperature = 30 °C | ✓ 1 | |
| Energy = m c Δθ = 1.2 × 4200 × 30 | ✓ 1 | |
| 151.2 kJ (accept 151 kJ) | ✓ 1 | |
| Part (b) | ||
| Time = energy ÷ power = 151 200 ÷ 42 (ECF from (a)) | ✓ 1 | |
| 3600 s = 60 minutes | ✓ 1 | |
| Part (c) | ||
| Water has a high specific heat capacity | ✓ 1 | |
| So a given mass stores (and releases) a large amount of energy for each °C fall in temperature | ✓ 1 | |
Exam tip: Convert kJ to J (or J to kJ) before substituting into an equation that uses J.
A fish-processing plant keeps its fish in a cold store at −18 °C.
Convert −18 °C to kelvin.
State the temperature of absolute zero in kelvin and in °C.
A box of fish is taken out of the cold store and warms from 4 °C to 22 °C. State the temperature rise of the fish in kelvin.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Temperature in K = temperature in °C + 273 | ✓ 1 | |
| 255 K | ✓ 1 | |
| Part (b) | ||
| 0 K | ✓ 1 | |
| −273 °C | ✓ 1 | |
| Part (c) | ||
| A change of 1 °C equals a change of 1 K, so the rise is 18 K | ✓ 1 | |
Common error: Only add 273 to temperatures; a temperature difference is the same number in °C and in K.
A smoke cell is a small glass cell filled with smoke and lit from the side. When viewed through a microscope, bright specks of smoke are seen moving about in random, jerky paths. Which statement gives the best explanation for this movement?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: air molecules are in constant random motion and hit a smoke speck from all sides; at any instant the hits are unequal, so the speck is pushed in a random direction. A is wrong because light does not exert a noticeable force that makes the motion random. C is wrong because the specks are far more massive than air molecules and are the ones being knocked about, not the cause. D is wrong because convection would give a smooth, steady drift, not random jerky paths. | ✓ 1 | Answer B |
Answer: B
Common error: Do not say that the smoke particles move by themselves. The visible motion is evidence for the motion of the invisible air molecules that hit them.
A delivery van with its load has a mass of 3200 kg. The van stands on level road on four tyres. The total area of the tyres in contact with the road is 0.080 m2. Use g = 10 N/kg.
Calculate the weight of the loaded van.
Calculate the pressure that the tyres exert on the road.
The road surface is damaged if the pressure on it is greater than 250 000 Pa. Calculate the smallest total contact area that the loaded van could have without damaging the road.
Explain why the pressure on the road decreases when the van is unloaded.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = mg | ✓ 1 | |
| W = 3200 × 10 = 32000 N | ✓ 1 | |
| Part (b) | ||
| p = F ÷ A | ✓ 1 | |
| p = 32000 ÷ 0.080 = 400000 Pa (4.0 × 105 Pa) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| A = F ÷ p (rearrangement) | ✓ 1 | |
| A = 32000 ÷ 250000 = 0.128 m2 (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| The weight (force) of the van decreases | ✓ 1 | |
| The contact area is unchanged, so pressure = force ÷ area is smaller | ✓ 1 | |
Common error: A larger pressure does not mean a larger contact area. For a given weight, pressure goes up when the area gets smaller.
In a smoke cell experiment, smoke from a burning straw is trapped in a small glass cell. The cell is lit from the side and viewed through a microscope. Small bright specks of smoke are seen.
Describe the movement of the bright specks.
Explain why the specks move in this way.
The cell is warmed slightly. State and explain how the movement of the specks changes.
The air molecules cannot be seen with the microscope, but the smoke specks can. Suggest why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The specks move in random directions / zig-zag paths | ✓ 1 | |
| The speed and direction change irregularly / unpredictably | ✓ 1 | |
| Part (b) | ||
| Air molecules are in constant random motion | ✓ 1 | |
| The molecules collide with the smoke specks | ✓ 1 | |
| The collisions on different sides are unequal at any instant, so there is a resultant force in a changing direction | ✓ 1 | |
| Part (c) | ||
| The specks move faster / more vigorously | ✓ 1 | |
| The air molecules have a higher average speed (kinetic energy) so hit harder or more often | ✓ 1 | |
| Part (d) | ||
| The smoke specks are much larger than air molecules, so they scatter enough light to be seen | ✓ 1 | |
Common error: The specks seen are not air molecules. They are larger particles, and their random movement is evidence for the motion of the molecules that hit them.
A camera on a cable is lowered from a bridge deck into a river to inspect a pier. A gauge on the camera reads a total pressure of 1.85 × 105 Pa. Atmospheric pressure is 1.00 × 105 Pa and the density of the river water is 1000 kg/m3. Use g = 10 N/kg.
Calculate the pressure due to the water alone.
Calculate the depth of the camera.
Explain why the pressure in the water increases with depth.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1.85 × 105 − 1.00 × 105 = 85000 Pa | ✓ 1 | |
| Part (b) | ||
| p = hρg, so h = p ÷ (ρg) | ✓ 1 | |
| h = 85000 ÷ (1000 × 10) | ✓ 1 | |
| h = 8.5 m (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| There is more water above the camera at greater depth | ✓ 1 | |
| The weight of this water acts on each unit area, so the pressure is greater | ✓ 1 | |
Exam tip: The gauge reading includes atmospheric pressure. Subtract it before using p = hρg.
Equal masses of aluminium (specific heat capacity 900 J/kg °C) and copper (specific heat capacity 390 J/kg °C) are supplied with the same amount of thermal energy. Which statement is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: ΔT = E/(mc), so for equal E and m the temperature rise is inversely proportional to c; 900/390 ≈ 2.3, so copper rises about 2.3 times as much. A is wrong: a large c means more energy is needed per degree, so the rise is smaller. C is wrong: the same energy gives the same rise only if c is also the same. D is wrong: density does not appear in the equation and the masses are equal. | ✓ 1 | Answer B |
Answer: B
Common error: A high specific heat capacity means a material is slow to change temperature, not quick.
A hospital bed pushes down on a patient's skin with a total force of 800 N. The patient is in contact with the bed over an area of 0.40 m2. What is the average pressure on the patient?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: p = F/A = 800 ÷ 0.40 = 2000 Pa. B multiplies force by area (800 × 0.40). C divides area by force (0.40 ÷ 800). D treats 0.40 m² as 40 cm² (it is 4000 cm²) and then divides 800 by 40. | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the equation p = F/A first and check that the force is in newtons and the area in square metres.
A nurse rubs alcohol gel onto her hands. The gel feels cold as it changes from a liquid to a gas.
Describe the arrangement and motion of the particles in a liquid and in a gas.
Explain why the hands feel cold as the alcohol evaporates.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 liquid: particles are close together, touching, and not in a regular pattern | ✓ 1 | |
| B1 liquid: particles move around each other | ✓ 1 | |
| B1 gas: particles are far apart and move quickly in random directions | ✓ 1 | |
| Part (b) | ||
| E1 the fastest, most energetic particles escape from the surface of the liquid | ✓ 1 | |
| E1 so the average kinetic energy of the particles left behind falls, and energy is taken from the skin | ✓ 1 | |
Exam tip: In a particle explanation, link a change in energy to what the particles are doing.
In a sports-science laboratory, a technician studies water in a beaker. Which description of the particles in a liquid is correct?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: liquid particles are close together (touching) but have no regular pattern and can slide past each other. A describes no real state (liquid particles are not far apart). B describes a solid. D describes a gas. | ✓ 1 | Answer C |
Answer: C
Common error: Do not describe liquid particles as 'far apart'. The key difference between a liquid and a gas is that liquid particles are almost touching.
Equal masses of copper (specific heat capacity 390 J/kg°C) and lead (specific heat capacity 130 J/kg°C) are used as heat sinks in a railway signalling box. The same quantity of energy is transferred to each block. How does the temperature rise of the lead compare with that of the copper?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: ΔT = E/(mc). E and m are the same, so ΔT is inversely proportional to c, and the ratio is 390/130 = 3.0 for lead to copper. B inverts the ratio. C assumes equal energy gives equal temperature rise. D is wrong because the masses are equal, so the mass cancels. | ✓ 1 | Answer A |
Answer: A
Common error: Equal energy does not mean equal temperature rise; the smaller the specific heat capacity, the larger the rise for the same mass.
Water in a swimming pool is a liquid.
Describe the arrangement and movement of the particles in the water. [2]
Explain why it is very difficult to compress the water. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Particles are close together / touching but in no regular pattern (random arrangement) | ✓ 1 | |
| Particles move around / slide past each other | ✓ 1 | |
| Part (b) | ||
| There is almost no space between the particles | ✓ 1 | |
| So they cannot be pushed much closer together | ✓ 1 | |
Common error: Liquid particles are not 'far apart'. They are as close together as in a solid but are free to move past each other.
A technician allows a sample of liquid wax to cool in a room and records its temperature every minute.
Time / min: 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
Temperature / °C: 80 | 71 | 64 | 58 | 58 | 58 | 58 | 53 | 49 | 46
State the freezing point of the wax. [1]
State the time interval during which the wax is freezing. [1]
Explain why the temperature stays constant while the wax freezes, even though energy is still transferred to the surroundings. [2]
Explain why the temperature falls by less each minute during the first 3 minutes. [1]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 58 °C | ✓ 1 | |
| Part (b) | ||
| From 3 minutes to 6 minutes (accept 3 to 6 min) | ✓ 1 | |
| Part (c) | ||
| Energy is released as the particles move into fixed positions / attractions between particles form | ✓ 1 | |
| This released energy balances the energy transferred to the surroundings, so the kinetic energy of the particles (temperature) does not change | ✓ 1 | |
| Part (d) | ||
| As the wax cools the temperature difference between the wax and the room decreases, so energy is transferred more slowly | ✓ 1 | |
Common error: During a change of state energy is still being transferred, but it changes the arrangement of the particles rather than their kinetic energy.
A sealed aerosol can of fixed volume is left in a hot room. The temperature of the gas inside rises. Which statement explains why the gas pressure increases?
| Marking point | Mark | Notes |
|---|---|---|
| Higher temperature means a higher average molecular speed, so collisions with the walls are more frequent and each collision has a greater force, which raises the pressure. The molecules do not change size or mass when heated, and the can is sealed so the number of molecules cannot increase. | ✓ 1 | Answer B |
Answer: B
Common error: Heating a gas does not make its molecules bigger, heavier or more numerous; only their speed changes.
A steel cube has sides of length 0.20 m and a mass of 64 kg. What is the density of the steel?
| Marking point | Mark | Notes |
|---|---|---|
| Volume = 0.20 × 0.20 × 0.20 = 0.0080 m³, so density = 64 ÷ 0.0080 = 8000 kg/m³. 320 kg/m³ comes from dividing by the side length only, 1600 kg/m³ from dividing by a face area, and 107 kg/m³ from using 3 × 0.20 as the volume. | ✓ 1 | Answer D |
Answer: D
Common error: Volume of a cube is side cubed, not side times three.
A volcanologist collects a rock sample. The rock has a mass of 0.84 kg and a volume of 3.0 × 10−4 m³.
Calculate the density of the rock.
A second sample of the same rock has a volume of 1.5 × 10−3 m³. Calculate its mass.
A piece of pumice has a density of 640 kg/m³. The density of water is 1000 kg/m³. State whether pumice floats and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| density = mass ÷ volume = 0.84 ÷ 3.0 × 10−4 | ✓ 1 | |
| 2800 kg/m³ | ✓ 1 | |
| Part (b) | ||
| mass = density × volume = 2800 × 1.5 × 10−3 | ✓ 1 | |
| 4.2 kg | ✓ 1 | |
| Part (c) | ||
| floats because its density is less than that of water | ✓ 1 | |
Exam tip: Use the density found in the first part for any other sample of the same material.
A clay water jug is porous, so water seeps to its outer surface and evaporates. The water inside stays cooler than the surrounding air. Which statement explains why?
| Marking point | Mark | Notes |
|---|---|---|
| Correct: the fastest molecules are the ones able to leave the surface, so the average kinetic energy, and therefore the temperature, of those left behind falls. The first option is wrong because losing the slowest molecules would raise the average. The second is wrong because collisions with the walls do not slow all the molecules. The last is wrong because evaporation is the cause of the cooling. | ✓ 1 | Answer C |
Answer: C
Exam tip: For evaporation questions, say which molecules escape and what happens to the average kinetic energy of those that remain.
Volcanologists heat a sample of solid basalt rock in a furnace. At a certain temperature the rock begins to melt.
Describe how the arrangement of the particles changes when the rock melts.
The temperature of the rock stays constant while it melts, even though energy is still being supplied. Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| in the solid the particles are in a fixed, regular arrangement | ✓ 1 | |
| in the liquid the particles are still close together but irregularly arranged and able to move past each other | ✓ 1 | |
| Part (b) | ||
| the energy is used to overcome the forces between the particles | ✓ 1 | |
| so the average kinetic energy of the particles does not increase, so the temperature stays constant | ✓ 1 | |
Exam tip: During a change of state the energy supplied changes how the particles are arranged, not their average kinetic energy.
A clay jug holds 5.0 kg of water at 30 °C. Over one hour, evaporation from the outer surface of the jug removes 105 kJ of energy from the water. The specific heat capacity of water is 4200 J/kg °C.
Calculate the fall in temperature of the water.
State the final temperature of the water.
Suggest why an unglazed clay jug keeps water cooler than a glazed jug of the same size.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 change in thermal energy = mass × specific heat capacity × temperature change, rearranged | ✓ 1 | |
| M1 substitution of 105 000 ÷ (5.0 × 4200) | ✓ 1 | |
| A1 5.0 °C | ✓ 1 | |
| Part (b) | ||
| 25 °C (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| unglazed clay is porous, so water seeps to the outer surface | ✓ 1 | |
| giving a large wet surface from which water evaporates, removing energy from the water | ✓ 1 | |
Common error: Convert kJ to J before using specific heat capacity in J/kg °C.
Gallium melts at 30 °C, so a lump of it will melt in a warm hand. Which statement describes the particles as solid gallium melts?
| Marking point | Mark | Notes |
|---|---|---|
| In a liquid the particles remain in contact, so the spacing hardly changes, but they are no longer held in a regular pattern and can move past one another. Moving far apart and filling the container describes a gas, not a liquid. Particles in a solid vibrate and do not stop when the solid melts; they vibrate more and begin to move around. The particles themselves do not change size; it is the arrangement and movement that change. | ✓ 1 | Answer B |
Answer: B
Common error: Melting does not make the particles bigger or push them far apart; that only happens on boiling.
The water in a swimming pool is 2.5 m deep. The density of water is 1000 kg/m3 and the gravitational field strength is 10 N/kg. What is the pressure on the bottom of the pool due to the water alone?
| Marking point | Mark | Notes |
|---|---|---|
| Pressure due to a liquid = height × density × gravitational field strength = 2.5 × 1000 × 10 = 25000 Pa. The value 2500 Pa omits the gravitational field strength. The value 250 Pa uses 10 instead of 1000 for the density. The value 125000 Pa adds atmospheric pressure of about 100000 Pa, but the question asks for the pressure due to the water alone. | ✓ 1 | Answer C |
Answer: C
Common error: Remember to include g when using pressure = hρg; omitting it is a frequent slip.
Gallium is a metal that melts at 30 °C and boils at about 2400 °C.
Describe the arrangement and movement of the particles in solid gallium.
A sample of gallium is warmed from 20 °C to 40 °C. Describe what happens to the particles during this change.
Explain why gallium vapour fills the whole of a closed container.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| particles are in a regular pattern / fixed positions, closely packed | ✓ 1 | |
| particles vibrate about their fixed positions | ✓ 1 | |
| Part (b) | ||
| the regular arrangement is lost / particles are no longer in fixed positions | ✓ 1 | |
| particles stay close together but move past one another | ✓ 1 | |
| Part (c) | ||
| particles in a gas move randomly at high speed | ✓ 1 | |
| there are almost no forces holding the particles together, so they spread out to occupy the available volume | ✓ 1 | |
Exam tip: When describing states of matter, mention arrangement, spacing and movement separately.
A library shelf holds a stack of books of total mass 6.0 kg. The stack rests on a table with a flat face in contact. The gravitational field strength is 10 N/kg.
Calculate the weight of the stack.
The area in contact with the table is 0.048 m2. Calculate the pressure exerted on the table.
The stack is turned onto its side so that the area in contact is 0.020 m2. Calculate the new pressure on the table.
Explain why the pressure is greater in (c) than in (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 weight = 6.0 × 10 | ✓ 1 | |
| A1 60 N | ✓ 1 | |
| Part (b) | ||
| M1 pressure = force ÷ area = 60 ÷ 0.048 (ECF from (a)) | ✓ 1 | |
| A1 1250 Pa | ✓ 1 | |
| Part (c) | ||
| M1 60 ÷ 0.020 | ✓ 1 | |
| A1 3000 Pa | ✓ 1 | |
| Part (d) | ||
| the same force acts on a smaller area | ✓ 1 | |
Exam tip: Pressure depends on both force and area: the same force on a smaller area gives a larger pressure.
A hiker of mass 60 kg stands still on snow wearing a pair of snowshoes. Each snowshoe has an area of 0.20 m2 in contact with the snow. What is the pressure on the snow? (g = 10 N/kg)
| Marking point | Mark | Notes |
|---|---|---|
| Weight = 60 × 10 = 600 N. Both snowshoes support the hiker, so the area is 2 × 0.20 = 0.40 m2 and p = F / A = 600 / 0.40 = 1500 Pa. 150 Pa comes from using the mass instead of the weight. 3000 Pa uses the area of only one snowshoe. 120 Pa comes from multiplying force by area. | ✓ 1 | Answer B |
Answer: B
Common error: Pressure uses force (weight in newtons), not mass, and the total area in contact with the ground.
A walker of mass 80 kg crosses deep snow. (g = 10 N/kg)
Calculate the weight of the walker.
In ordinary boots, each boot has an area of 0.020 m2 in contact with the snow. Calculate the pressure on the snow when the walker stands on both boots.
The walker puts on snowshoes. Each snowshoe has an area of 0.25 m2. Calculate the pressure on the snow when standing on both snowshoes.
Explain why the snowshoes stop the walker sinking in as far.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 W = m g | ✓ 1 | |
| A1 W = 80 × 10 = 800 N | ✓ 1 | |
| Part (b) | ||
| M1 p = F / A with area 2 × 0.020 = 0.040 m2 | ✓ 1 | |
| A1 p = 800 / 0.040 = 20 000 Pa | ✓ 1 | |
| Part (c) | ||
| M1 p = 800 / (2 × 0.25) | ✓ 1 | |
| A1 p = 1600 Pa | ✓ 1 | |
| Part (d) | ||
| B1 the same weight acts over a larger area, so the pressure on the snow is smaller | ✓ 1 | |
Exam tip: To compare pressures in a 'why' question, say what stays the same (the force) and what changes (the area).
In a shop's stockroom a stack of 6 identical crates stands on the floor. Each crate has a mass of 25 kg. The base of the bottom crate measures 0.40 m by 0.30 m. (g = 10 N/kg)
Calculate the weight of the stack.
Calculate the pressure that the stack exerts on the floor.
The floor can safely withstand a pressure of 5000 Pa. The stack is placed on a pallet. Ignoring the weight of the pallet, calculate the minimum area of the pallet that must be in contact with the floor.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 total mass 6 × 25 = 150 kg, W = m g | ✓ 1 | |
| A1 W = 1500 N | ✓ 1 | |
| Part (b) | ||
| M1 p = F / A with A = 0.40 × 0.30 = 0.12 m2 | ✓ 1 | |
| A1 p = 1500 / 0.12 = 12 500 Pa | ✓ 1 | |
| Part (c) | ||
| M1 A = F / p | ✓ 1 | |
| A1 A = 1500 / 5000 = 0.30 m2 | ✓ 1 | |
Exam tip: Rearrange p = F / A to find area when the pressure is limited: A = F / p.
A sealed probe is lowered into a deep lake. At a depth of 40 m the water pressure on the top face of the probe is P. The probe is small. What is the water pressure on one of its vertical side faces at the same depth?
| Marking point | Mark | Notes |
|---|---|---|
| At one depth in a liquid the pressure is the same in every direction, so the side face experiences P. A pressure of zero ignores the sideways push of the liquid. Half of P wrongly makes pressure depend on the orientation of the surface. Greater than P wrongly adds an extra sideways effect; the water is not flowing and pressure depends only on depth and density. | ✓ 1 | Answer C |
Answer: C
Common error: Pressure in a liquid depends on depth and density, not on the direction of the surface it acts on.
In a coffee-roasting trial, 0.20 kg of beans of specific heat capacity 1800 J/kg °C are heated from 20 °C to 220 °C. How much energy is transferred to the beans?
| Marking point | Mark | Notes |
|---|---|---|
| Energy = 0.20 × 1800 × (220 − 20) = 0.20 × 1800 × 200 = 72000 J. 79200 J uses 220 °C as the temperature change instead of 200 °C. 360000 J leaves out the mass. 7200 J is a power-of-ten slip in the multiplication. | ✓ 1 | Answer D |
Answer: D
Common error: The temperature change is final minus initial temperature, not the final temperature.
A hot-water bottle holds 1.5 kg of water. The water starts at 70 °C and cools to 30 °C during the night. The specific heat capacity of water is 4200 J/kg °C.
State the equation linking energy transferred, mass, specific heat capacity and temperature change.
Calculate the energy transferred from the water to the surroundings.
Explain one reason why water is a suitable liquid for a hot-water bottle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ΔQ = m × c × Δθ | ✓ 1 | |
| Part (b) | ||
| temperature change = 70 − 30 = 40 °C | ✓ 1 | |
| substitution: 1.5 × 4200 × 40 | ✓ 1 | |
| 252000 J (accept 252 kJ) | ✓ 1 | |
| Part (c) | ||
| Accept any valid reason with an explanation, e.g. water has a high specific heat capacity | ✓ 1 | |
| so it releases a large amount of energy for each degree it cools, keeping the bottle warm for longer | ✓ 1 | |
Exam tip: Write the temperature change as a subtraction before substituting, so a method mark is safe even if the arithmetic slips.
A student in a robotics club measures a solid metal block. It measures 2.0 cm × 3.0 cm × 5.0 cm and has a mass of 81 g. What is the density of the metal?
| Marking point | Mark | Notes |
|---|---|---|
| Volume = 2.0 × 3.0 × 5.0 = 30 cm3, so density = 81 g ÷ 30 cm3 = 2.7 g/cm3 = 2700 kg/m3, because 1 g/cm3 is 1000 kg/m3. 2.7 kg/m3 keeps the number from g/cm3 but changes the unit without converting. 0.37 g/cm3 is volume ÷ mass, the wrong way up. 13.5 g/cm3 divides the mass by the area of only one face (2.0 × 3.0) instead of the full volume. | ✓ 1 | Answer C |
Answer: C
Common error: Always convert units before or after dividing: 1 g/cm³ = 1000 kg/m³, and density is mass divided by volume, never the reverse.
The robotics classroom at a school is a cuboid of length 8.0 m, width 5.0 m and height 3.0 m. The density of the air in the room is 1.2 kg/m3. The gravitational field strength is 10 N/kg.
Calculate the volume of air in the room.
Calculate the mass of the air in the room.
Calculate the weight of the air in the room.
The room is warmed while its windows are open. Explain why the mass of air in the room decreases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 volume = 8.0 × 5.0 × 3.0 | ✓ 1 | |
| A1 120 m3 | ✓ 1 | |
| Part (b) | ||
| M1 mass = density × volume = 1.2 × 120 | ✓ 1 | |
| A1 144 kg (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 weight = mass × g = 144 × 10 | ✓ 1 | |
| A1 1440 N (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| The air expands when warmed and some of it leaves through the windows | ✓ 1 | |
| The same volume of room now contains fewer molecules, so the density of the air is lower | ✓ 1 | |
Exam tip: Mass is measured in kg and weight in N. Find the mass first with density × volume, then multiply by g to get the weight.
A clothes steamer used by a fashion designer boils water to make steam. The steam is directed at a garment, where it turns back into liquid water.
Name the change of state that happens when steam turns into liquid water.
Describe the arrangement and movement of the molecules in steam compared with those in liquid water.
A damp cotton garment dries faster on a warm day than on a cold day. Explain why, in terms of molecules.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Condensation | ✓ 1 | |
| Part (b) | ||
| In steam the molecules are far apart | ✓ 1 | |
| In liquid water the molecules are close together and touching | ✓ 1 | |
| In steam the molecules move randomly and quickly, in liquid water they slide past each other more slowly | ✓ 1 | |
| Part (c) | ||
| Evaporation happens when the faster molecules at the surface escape from the liquid | ✓ 1 | |
| At a higher temperature more molecules have enough kinetic energy to escape, so the garment dries faster | ✓ 1 | |
Common error: Steam is not the visible white cloud you see; steam itself is an invisible gas. When describing gases, say the molecules are far apart, not that they are 'light' or 'spread out evenly'.
A solid metal cube has sides of length 2.0 cm and a mass of 64 g. What is the density of the metal?
| Marking point | Mark | Notes |
|---|---|---|
| Volume = 2.03 = 8.0 cm3, so density = 64 ÷ 8.0 = 8.0 g/cm3. 16 g/cm3 divides by a face area (4.0 cm2) instead of the volume. 32 g/cm3 divides by the side length only. 0.13 g/cm3 inverts the formula (volume ÷ mass). | ✓ 1 | Answer B |
Answer: B
Common error: Volume of a cube is side cubed, not side squared or side times two.
A hiker of mass 80 kg crosses deep snow. Each boot has a contact area of 0.020 m2. Snowshoes can be strapped on; with both snowshoes the total contact area is 0.25 m2. Ignore the mass of the snowshoes. Use g = 10 N/kg.
Calculate the weight of the hiker.
Calculate the pressure on the snow when the hiker stands on both boots without snowshoes.
Calculate the pressure on the snow when the hiker stands wearing both snowshoes.
Explain why wearing snowshoes stops the hiker sinking so far into the snow.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = mg | ✓ 1 | |
| 80 × 10 = 800 N | ✓ 1 | |
| Part (b) | ||
| p = F ÷ A with A = 0.040 m2 | ✓ 1 | |
| 800 ÷ 0.040 = 20000 Pa (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 800 ÷ 0.25 | ✓ 1 | |
| 3200 Pa (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| snowshoes give a larger contact area | ✓ 1 | |
| the same weight acts over a larger area so the pressure on the snow is smaller | ✓ 1 | |
Exam tip: In a pressure question, state clearly which force and which area you are using.
Gallium is a metal with a melting point of 30 °C and a boiling point of 2400 °C.
State the physical state of gallium at 20 °C and at 100 °C.
Describe the arrangement and the motion of the particles in liquid gallium.
Gallium vapour is held in a sealed container. Explain how the vapour exerts a pressure on the walls of the container.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| solid at 20 °C | ✓ 1 | |
| liquid at 100 °C | ✓ 1 | |
| Part (b) | ||
| particles close together / touching | ✓ 1 | |
| irregular (random) arrangement | ✓ 1 | |
| particles move around / slide past each other | ✓ 1 | |
| Part (c) | ||
| particles move randomly and collide with the walls | ✓ 1 | |
| each collision exerts a force; very many collisions give a steady force per unit area | ✓ 1 | |
Exam tip: When describing a state, mention both the spacing and the movement of the particles.
A beekeeper fills a jar of volume 540 cm3 with honey. The mass of the honey is 756 g.
Calculate the density of the honey in g/cm3.
Write your answer to (a) in kg/m3.
A bucket holds 14 kg of the honey. Use a density of 1400 kg/m3 to calculate the volume of honey in the bucket in m3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ρ = m ÷ V | ✓ 1 | |
| 756 ÷ 540 = 1.4 g/cm3 | ✓ 1 | |
| Part (b) | ||
| 1400 kg/m3 | ✓ 1 | |
| Part (c) | ||
| V = m ÷ ρ | ✓ 1 | |
| 14 ÷ 1400 = 0.010 m3 | ✓ 1 | |
Exam tip: Convert g/cm³ to kg/m³ by multiplying by 1000.
A gas is cooled to a temperature of −73 °C. What is this temperature on the kelvin scale?
| Marking point | Mark | Notes |
|---|---|---|
| T(K) = θ(°C) + 273 = −73 + 273 = 200 K. 346 K adds 73 instead of using −73. −346 K subtracts 273 from −73. 73 K ignores the negative sign and the 273 conversion. | ✓ 1 | Answer C |
Answer: C
Common error: Absolute temperatures on the kelvin scale can never be negative.
A crate stands on a floor and exerts a pressure p on it. An identical crate is placed on top of the first, directly above it. What is the pressure now exerted by the lower crate on the floor?
| Marking point | Mark | Notes |
|---|---|---|
| The force on the floor doubles because the weight doubles, but the contact area is unchanged, so the pressure doubles to 2p. The pressure does not stay at p because the force has changed. 4p would need both the force and the pressure to be squared. p/2 would need the force to halve. | ✓ 1 | Answer A |
Answer: A
Common error: Pressure depends on both force and area; check which of the two has changed.
A shopper of mass 60 kg stands on one narrow heel, which has a contact area of 1.5 cm2. Assume the whole weight is on this heel. Use g = 10 N/kg.
Write the contact area in m2.
Calculate the pressure on the floor under the heel.
The shop floor tiles are damaged by pressures above 2.0 × 106 Pa. State whether the tile is damaged by the heel, with a reason.
Calculate the smallest heel area, in cm2, that would not damage the tile.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1.5 × 10−4 m2 | ✓ 1 | |
| Part (b) | ||
| W = 60 × 10 = 600 N | ✓ 1 | |
| p = F ÷ A | ✓ 1 | |
| 600 ÷ 1.5 × 10−4 = 4000000 Pa (4.0 × 106 Pa) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| yes, because 4.0 × 106 Pa is greater than 2.0 × 106 Pa | ✓ 1 | |
| Part (d) | ||
| A = F ÷ p = 600 ÷ (2.0 × 106) = 3.0 × 10−4 m2 | ✓ 1 | |
| 3.0 cm2 | ✓ 1 | |
Common error: Convert cm² to m² by dividing by 10 000, not by 100.
In a study of how infections spread in a school, a classroom is modelled as a box 8.0 m long, 6.0 m wide and 3.0 m high. The density of air is 1.2 kg/m3. Use g = 10 N/kg.
Calculate the volume of the classroom.
Calculate the mass of air in the classroom.
Calculate the weight of the air in the classroom.
A ventilation system moves air at 0.40 m3 per second. Calculate the time taken to move a volume of air equal to the volume of the classroom.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = 8.0 × 6.0 × 3.0 | ✓ 1 | |
| 144 m3 | ✓ 1 | |
| Part (b) | ||
| m = ρV | ✓ 1 | |
| 1.2 × 144 = 172.8 kg (accept 173 kg) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 1728 N (accept 1730 N) (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| t = volume ÷ rate | ✓ 1 | |
| 144 ÷ 0.40 = 360 s (ECF from (a)) | ✓ 1 | |
Exam tip: Check that all three lengths are in metres before multiplying to get a volume.
A student measures a rectangular metal block. Its edges are 3.0 cm, 5.0 cm and 4.5 cm long and its mass is 135 g. What is the density of the metal?
| Marking point | Mark | Notes |
|---|---|---|
| Volume = 3.0 × 5.0 × 4.5 = 67.5 cm3, so density = mass ÷ volume = 135 ÷ 67.5 = 2.0 g/cm3. 0.50 g/cm3 comes from dividing volume by mass the wrong way round. 9.0 g/cm3 divides the mass by the area of one face (15 cm2) instead of the volume. 45 g/cm3 divides the mass by a single edge length. | ✓ 1 | Answer C |
Answer: C
Common error: Density needs the whole volume, found by multiplying all three edges, not one length or one area.
A sealed rigid glass vial contains air. The vial is warmed in a water bath. Which statement explains why the pressure of the air inside the vial rises?
| Marking point | Mark | Notes |
|---|---|---|
| Warming increases the average kinetic energy of the molecules, so they move faster. They collide with the walls more frequently and each collision exerts a greater force, so the pressure rises. Molecules do not grow when heated. The vial is sealed, so the number of molecules is fixed. The vial is rigid, so the molecules cannot move further apart on average, and fewer collisions would lower the pressure. | ✓ 1 | Answer A |
Answer: A
Common error: Heating a gas makes its molecules move faster; it does not make the molecules expand or increase in number.
A student measures the density of a metal block. The block is a cuboid. Using a ruler she finds its length is 6.0 cm, its width is 4.0 cm and its height is 2.5 cm. A balance shows that its mass is 162 g.
Calculate the volume of the block.
Calculate the density of the metal in g/cm3.
Convert your answer to (b) into kg/m3. (1 g/cm3 = 1000 kg/m3)
Suggest two ways the student could improve the accuracy of her measurement of the volume.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 volume = length × width × height = 6.0 × 4.0 × 2.5 | ✓ 1 | |
| A1 60 cm3 | ✓ 1 | |
| Part (b) | ||
| M1 density = mass ÷ volume = 162 ÷ 60 | ✓ 1 | |
| A1 2.7 g/cm3 (accept 2.70) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| A1 2700 kg/m3 (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| A valid improvement, for example measure each edge at several positions and take the average | ✓ 1 | |
| A second, different valid improvement, for example use vernier calipers or a micrometer for smaller resolution, or repeat the measurements | ✓ 1 | |
Exam tip: Show the substitution into density = mass ÷ volume and keep units consistent: grams with cm³, kilograms with m³.
A pharmacist warms 0.25 kg of a water-based solution from 20 °C to 37 °C using an electric heater of power 150 W. The specific heat capacity of the solution is 4200 J/kg °C. In parts (a) and (b), assume that all the energy from the heater is transferred to the solution.
Calculate the energy transferred to the solution.
Calculate the time for which the heater is switched on.
In practice the heater is on for longer than the time in (b). Suggest why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 temperature change = 37 − 20 = 17 °C | ✓ 1 | |
| M1 energy = mcΔθ = 0.25 × 4200 × 17 | ✓ 1 | |
| A1 17 850 J | ✓ 1 | |
| Part (b) | ||
| M1 time = energy ÷ power = 17 850 ÷ 150 | ✓ 1 | |
| A1 119 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Some energy is transferred to the surroundings or to the container, so more energy must be supplied | ✓ 1 | |
Exam tip: Write the temperature change as a separate step before substituting into energy = mass × specific heat capacity × temperature change.
Which statement describes the particles in a liquid?
| Marking point | Mark | Notes |
|---|---|---|
| In a liquid the particles are touching or very close, with no regular pattern, and they move around by sliding past one another, so the liquid flows but is hard to compress. The first description is a gas. The second description is a solid. The fourth description mixes the large separation of a gas with the fixed positions of a solid and matches no state of matter. | ✓ 1 | Answer C |
Answer: C
Common error: Do not say that particles in a liquid are 'further apart' than in a solid; they are only slightly further apart on average.
A technician investigating the spread of an infection keeps a liquid sample in a sealed glass vial.
Describe the arrangement and the movement of the particles in a liquid.
Explain why the liquid takes the shape of the vial but its volume does not change when it is poured into a different vial.
Explain why it is difficult to compress a liquid.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| particles are close together (touching) | ✓ 1 | |
| arranged irregularly, with no regular pattern | ✓ 1 | |
| particles move around and slide past each other | ✓ 1 | |
| Part (b) | ||
| the particles can slide past each other, so the liquid flows to fill the lower part of any container | ✓ 1 | |
| the particles stay close together, so the volume stays fixed | ✓ 1 | |
| Part (c) | ||
| the particles are already close together, so there is very little space between them | ✓ 1 | |
Exam tip: Describe both the arrangement and the movement of the particles; answers that give only one earn half the marks.
A retailer's delivery van has a total mass of 3200 kg when loaded. It stands on level ground on four identical tyres. The area of each tyre in contact with the road is 0.020 m2. The gravitational field strength is 10 N/kg.
Calculate the weight of the loaded van.
Calculate the total area of the tyres in contact with the road.
Calculate the pressure that the tyres exert on the road.
The van delivers all of its goods. State and explain how the pressure on the road changes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 weight = mass × gravitational field strength | ✓ 1 | |
| A1 3200 × 10 = 32 000 N | ✓ 1 | |
| Part (b) | ||
| 4 × 0.020 = 0.080 m2 | ✓ 1 | |
| Part (c) | ||
| M1 pressure = force ÷ area | ✓ 1 | |
| A1 32 000 ÷ 0.080 = 4.0 × 105 Pa (400 000 Pa) | ✓ 1 | |
| Part (d) | ||
| the pressure decreases | ✓ 1 | |
| the weight (force) is smaller but the contact area is unchanged | ✓ 1 | |
Common error: Use the area of all four tyres, not one, when finding the pressure on the road.
A cube of side 0.10 m has a mass of 2.7 kg. What is the density of the material?
| Marking point | Mark | Notes |
|---|---|---|
| The volume is 0.10 × 0.10 × 0.10 = 1.0 × 10−3 m3, so density = 2.7 ÷ 1.0 × 10−3 = 2.7 × 103 kg/m3. 2.7 × 10−3 comes from multiplying mass by volume. 2.7 × 102 comes from dividing by the area of one face (0.010 m2). 27 comes from dividing by the side length only. | ✓ 1 | Answer D |
Answer: D
Exam tip: Cube the side length to find the volume of a cube, and check that your density is sensible for the material.
A hot-water bottle contains 1.5 kg of water at 70 °C. The water cools to 30 °C. The specific heat capacity of water is 4200 J/kg °C.
State what is meant by specific heat capacity.
Calculate the energy transferred from the water as it cools.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the energy needed to raise the temperature of 1 kg | ✓ 1 | |
| of a substance by 1 °C | ✓ 1 | |
| Part (b) | ||
| M1 temperature change = 40 °C | ✓ 1 | |
| M1 energy = mass × specific heat capacity × temperature change | ✓ 1 | |
| A1 2.52 × 105 J | ✓ 1 | |
Exam tip: Use the temperature change, not the final temperature, in ΔE = m c Δθ.
A delivery van has a mass of 3200 kg. It stands on four tyres. The area of each tyre in contact with the road is 0.020 m2. The gravitational field strength is 10 N/kg.
Calculate the weight of the van.
Calculate the pressure that the tyres exert on the road.
Explain why fitting wider tyres reduces the pressure on the road.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 weight = mass × gravitational field strength | ✓ 1 | |
| A1 32 000 N | ✓ 1 | |
| Part (b) | ||
| M1 total area = 4 × 0.020 = 0.080 m2 | ✓ 1 | |
| M1 pressure = force ÷ area | ✓ 1 | |
| A1 4.0 × 105 Pa , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| the weight (force) stays the same but the contact area increases | ✓ 1 | |
| pressure = force ÷ area, so pressure decreases | ✓ 1 | |
Common error: Use the total contact area of all four tyres, not the area of one tyre, when the whole weight is shared.
A wooden bar from a xylophone is a cuboid of length 0.30 m, width 0.040 m and thickness 0.020 m. Its mass is 0.168 kg.
Calculate the volume of the bar.
Calculate the density of the wood.
The density of water is 1000 kg/m3. State whether the bar floats in water and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 volume = length × width × thickness | ✓ 1 | |
| A1 2.4 × 10−4 m3 | ✓ 1 | |
| Part (b) | ||
| M1 density = mass ÷ volume | ✓ 1 | |
| A1 700 kg/m3 , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| it floats | ✓ 1 | |
| its density is less than the density of water | ✓ 1 | |
Common error: Keep all lengths in metres before multiplying; mixing cm and m gives a volume that is wrong by a large factor.
A wooden crate of mass 60 kg stands on a floor. Its base is a square of side 0.50 m. Take g = 10 N/kg. What pressure does the crate exert on the floor?
| Marking point | Mark | Notes |
|---|---|---|
| Weight = 60 × 10 = 600 N. Area = 0.50 × 0.50 = 0.25 m². Pressure = 600 ÷ 0.25 = 2400 Pa. 240 Pa uses the mass in kg instead of the weight in newtons. 150 Pa multiplies force by area instead of dividing. 600 Pa is the weight itself, with the area ignored. | ✓ 1 | Answer C |
Answer: C
Common error: Pressure needs the force (weight) in newtons, not the mass in kilograms.
A classroom is a cuboid 7.5 m long, 6.0 m wide and 3.0 m high. The air in it has a density of 1.2 kg/m3. Use g = 10 N/kg.
Calculate the volume of the classroom.
Calculate the mass of the air in the classroom.
Calculate the weight of the air in the classroom.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 volume = length × width × height | ✓ 1 | |
| A1 135 m3 | ✓ 1 | |
| Part (b) | ||
| M1 mass = density × volume | ✓ 1 | |
| A1 162 kg (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 weight = mass × g | ✓ 1 | |
| A1 1620 N (ECF from (b)) | ✓ 1 | |
Exam tip: Rearrange density = mass ÷ volume to mass = density × volume before substituting.
A nurse warms 0.25 kg of vaccine solution from 4 °C to 20 °C in a water bath. The specific heat capacity of the solution is 4000 J/kg °C.
Calculate the energy needed to warm the solution.
The heater in the bath transfers energy to the solution at a useful rate of 80 W. Calculate the time taken.
Suggest why the time taken in practice is longer than your answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 ΔE = m c Δθ | ✓ 1 | |
| M1 0.25 × 4000 × 16 | ✓ 1 | |
| A1 16 000 J | ✓ 1 | |
| Part (b) | ||
| M1 time = energy ÷ power | ✓ 1 | |
| A1 200 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| energy is also transferred to the surroundings / container (so useful power is less than 80 W) | ✓ 1 | |
Exam tip: The temperature change is final minus initial temperature: 20 − 4 = 16 °C, not 20 °C.
Hand sanitiser is a liquid that dries on the skin even though the skin is far below the boiling point of the liquid.
Name the process by which the liquid changes to a gas below its boiling point.
Explain why the skin feels cool as the sanitiser dries.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 evaporation | ✓ 1 | |
| Part (b) | ||
| only the molecules with the most kinetic energy escape from the surface | ✓ 1 | |
| so the average kinetic energy of the remaining liquid decreases | ✓ 1 | |
| so the temperature of the liquid falls and energy is transferred from the skin to it | ✓ 1 | |
Common error: Do not say the liquid 'boils' at room temperature; evaporation happens only at the surface and at any temperature.
Liquid nitrogen boils at −196 °C. Use 0 °C = 273 K. What is this temperature on the kelvin scale?
| Marking point | Mark | Notes |
|---|---|---|
| Kelvin = Celsius + 273 = −196 + 273 = 77 K, so 77 K is correct. 469 K comes from adding 196 instead of subtracting it. −469 K and −77 K are negative, which is impossible on the kelvin scale because 0 K is the lowest possible temperature. | ✓ 1 | Answer C |
Answer: C
Exam tip: A kelvin temperature can never be negative, so a negative answer tells you to check the conversion.
Gallium melts at 30 °C, so a small solid piece melts when held in a warm hand. Which statement describes the gallium particles after melting, compared with before?
| Marking point | Mark | Notes |
|---|---|---|
| In a liquid the particles stay close together, touching, but are no longer held in fixed positions so they slide past one another. Moving far apart describes a gas. Fixed positions with larger vibrations describes a warmer solid. Particles do not change size on melting. | ✓ 1 | Answer A |
Answer: A
Common error: Do not say the particles themselves melt or expand; only their arrangement and movement change.
A mountaineer records air temperatures on an expedition. Use 0 °C = 273 K.
The temperature at the summit is −23 °C. Convert this temperature to kelvin.
Explain how the average kinetic energy of the air molecules at the summit compares with that at base camp, where the temperature is higher.
The temperature at base camp is 290 K. Convert this temperature to °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 250 K | ✓ 1 | |
| Part (b) | ||
| The average kinetic energy is lower at the summit because the temperature is lower | ✓ 1 | |
| Part (c) | ||
| 17 °C | ✓ 1 | |
Common error: Subtract 273 to go from kelvin to °C; adding it is the most common slip.
Gallium is a metal with a melting point of 30 °C. Use 0 °C = 273 K.
Describe the arrangement and movement of the particles in solid gallium.
Describe how the arrangement and movement of the particles change when gallium melts.
Convert the melting point of gallium to kelvin.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Particles are in a regular pattern (lattice) and close together | ✓ 1 | |
| They vibrate about fixed positions | ✓ 1 | |
| Part (b) | ||
| Particles stay close together/touching but the regular pattern is lost | ✓ 1 | |
| Particles can move/slide past each other | ✓ 1 | |
| Part (c) | ||
| 303 K | ✓ 1 | |
Common error: Say that liquid particles are 'close together but free to move past each other'; 'far apart' describes a gas.
A sealed rigid glass jar contains air at a pressure of 100 kPa and a temperature of 27 °C. The jar is warmed to 87 °C. Its volume does not change.
What is the new pressure of the air in the jar?
| Marking point | Mark | Notes |
|---|---|---|
| First convert both temperatures to kelvin: T1 = 27 + 273 = 300 K and T2 = 87 + 273 = 360 K. At constant volume, p1/T1 = p2/T2, so p2 = 100 × 360 ÷ 300 = 120 kPa. Option D uses the Celsius temperatures (100 × 87 ÷ 27). Option B turns the kelvin ratio upside down. Option A turns the Celsius ratio upside down. | ✓ 1 | Answer C |
Answer: C
Common error: Gas law calculations only work with kelvin temperatures. Using °C gives answers that are far too large or far too small.
A marathon runner sweats, and the sweat evaporates from her skin at about 33 °C. This cools her body.
Which statement explains why evaporation cools her?
| Marking point | Mark | Notes |
|---|---|---|
| Only the particles with the most kinetic energy can overcome the attractive forces and escape from the liquid surface. Losing these high-energy particles lowers the average kinetic energy of the remaining liquid, so its temperature falls. Energy is then transferred from the skin to the cooler sweat. Option A reverses which particles escape. Option B treats 'cold' as something that can be transferred. Option C confuses evaporation, which happens at any temperature, with boiling. | ✓ 1 | Answer D |
Answer: D
Exam tip: For evaporative cooling, link three steps: the fastest particles escape, the average kinetic energy of the rest falls, and so the temperature falls.
A physiotherapist treats a footballer's swollen ankle. She can use either a bag containing 200 g of crushed ice at 0 °C, or a bag containing 200 g of water at 0 °C.
State the temperature of the ice while it is melting.
Explain why the bag of ice cools the ankle for longer than the bag of water.
Suggest why the ice bag is wrapped in a thin towel before it is placed on the skin.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0 °C | ✓ 1 | |
| Part (b) | ||
| The ice absorbs a large amount of energy from the ankle to melt (latent heat of fusion), without any rise in temperature | ✓ 1 | |
| The water's temperature starts to rise as soon as it absorbs energy from the ankle, so it soon stops being much colder than the ankle | ✓ 1 | |
| So the ice bag absorbs more energy from the ankle before it warms to body temperature / it stays at 0 °C for longer | ✓ 1 | |
| Part (c) | ||
| The towel acts as a (thermal) insulator that reduces the rate of energy transfer from the skin, so the skin is not damaged by cooling too quickly / frostbite / ice burn | ✓ 1 | |
Exam tip: Both bags start at 0 °C, so saying only that 'ice is colder' earns nothing. The marks come from the energy absorbed during melting at constant temperature, compared with the water's temperature rising straight away.
In a factory, molten aluminium is poured into a mould to make a car engine part. An engineer records the temperature of the aluminium as it cools.
Time in s (temperature in °C): 0 (720), 30 (695), 60 (672), 90 (660), 120 (660), 150 (660), 180 (660), 210 (660), 240 (660), 270 (641), 300 (624)
State the melting point of aluminium.
Calculate the average rate at which the temperature falls between 0 s and 60 s. Give the unit.
Explain why the temperature stays constant between 90 s and 240 s, even though the aluminium is still transferring energy to the mould.
State the state of the aluminium at 300 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 660 °C | ✓ 1 | |
| Part (b) | ||
| (720 − 672) ÷ 60 | ✓ 1 | |
| 0.80 °C/s | ✓ 1 | |
| Part (c) | ||
| The aluminium is freezing / changing from liquid to solid | ✓ 1 | |
| Bonds form / particles become held in fixed positions, which releases energy | ✓ 1 | |
| The energy released replaces the energy transferred to the mould, so the kinetic energy of the particles / the temperature does not fall | ✓ 1 | |
| Part (d) | ||
| Solid | ✓ 1 | |
Common error: Use the readings for the stated time interval only. Including the flat section when calculating a cooling rate gives a value that is too small.
A refrigerator in a medical clinic stores vaccines. A liquid called a coolant flows through pipes inside the storage compartment, where it turns into a gas. The gas is then pumped to pipes at the back of the refrigerator, where it turns back into a liquid.
State the change of state of the coolant in the pipes inside the storage compartment.
Explain how this change of state keeps the vaccines cold.
State the change of state of the coolant at the back of the refrigerator and what happens to the energy.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Evaporation / boiling / liquid to gas | ✓ 1 | |
| Part (b) | ||
| Changing from liquid to gas needs energy (latent heat) to separate the particles / overcome the forces between them | ✓ 1 | |
| This energy is absorbed from the air and vaccines inside the compartment | ✓ 1 | |
| So the thermal (internal) energy of the contents decreases and their temperature falls / is kept low | ✓ 1 | |
| Part (c) | ||
| Condensation / gas to liquid | ✓ 1 | |
| Energy (latent heat) is released and transferred to the surrounding air of the room | ✓ 1 | |
Exam tip: Changes of state work in pairs. A coolant absorbs energy when it evaporates and releases the same energy when it condenses. Always say where the energy comes from and where it goes.
A hospital drip bag contains a solution of density 1000 kg/m3. For the solution to flow into a patient's vein, the pressure of the liquid at the needle must be 6000 Pa greater than the pressure at the top of the liquid in the bag.
What is the minimum height of the liquid surface in the bag above the needle?
(g = 10 N/kg)
| Marking point | Mark | Notes |
|---|---|---|
| Pressure difference = ρ × g × h, so h = p ÷ (ρ × g) = 6000 ÷ (1000 × 10) = 0.60 m. Option C leaves out g, option B inverts the equation (ρg ÷ p), and option D multiplies all three quantities together. | ✓ 1 | Answer A |
Answer: A
Exam tip: Rearrange p = ρ × g × h before substituting. Then check that the answer is sensible: a drip bag hung a few metres up would be impractical.
An ice skater has a mass of 52 kg. Each skate blade touches the ice along a strip 260 mm long and 1.0 mm wide.
(g = 10 N/kg)
Calculate the weight of the skater.
The skater balances on one blade. Calculate the pressure that the blade exerts on the ice. Give your answer in Pa.
The skater then glides with her weight shared equally between both blades. State the new pressure on the ice.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = m × g = 52 × 10 | ✓ 1 | |
| W = 520 N | ✓ 1 | |
| Part (b) | ||
| area = 260 × 1.0 = 260 mm2 = 2.6 ×10−4 m2 | ✓ 1 | |
| p = F ÷ A = 520 ÷ 2.6 ×10−4 | ✓ 1 | |
| p = 2.0 ×106 Pa | ✓ 1 | |
| Part (c) | ||
| 1.0 ×106 Pa (the area is doubled, so the pressure is halved) | ✓ 1 | |
Common error: 1 mm² = 1 ×10⁻⁶ m², not 1 ×10⁻³ m². Squaring the length conversion is easy to forget.
A remotely operated underwater vehicle carries a camera behind a flat window. The vehicle inspects a pipeline on the seabed at a depth of 2500 m. The density of seawater is 1030 kg/m3.
(g = 10 N/kg)
State the equation linking pressure difference, height, density and gravitational field strength.
Calculate the pressure due to the seawater at a depth of 2500 m. Ignore atmospheric pressure.
The camera window has an area of 0.012 m2. Calculate the force that the seawater exerts on the window.
A second, identical window faces sideways at the same depth. Explain why the force on it is the same as the force on the first window.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| pressure difference = height × density × gravitational field strength (p = h × ρ × g) | ✓ 1 | |
| Part (b) | ||
| p = 2500 × 1030 × 10 | ✓ 1 | |
| p = 2.6 ×107 Pa (2.575 ×107 Pa) | ✓ 1 | |
| Part (c) | ||
| F = p × A = 2.575 ×107 × 0.012 | ✓ 1 | |
| F = 3.1 ×105 N | ✓ 1 | |
| Part (d) | ||
| pressure at a given depth in a liquid acts equally in all directions, so the pressure on both windows is the same | ✓ 1 | |
| the windows have the same area, so force = pressure × area is the same | ✓ 1 | |
Exam tip: When calculating the force, use the unrounded pressure value. Rounding too early can shift the final significant figure.
A water tower supplies an office building. The surface of the water in the tower is 32 m above the ground. The density of water is 1000 kg/m3.
(g = 10 N/kg)
A tap is on a floor 20 m above the ground. Calculate the pressure difference between the water surface in the tower and the water at this tap.
A coffee machine is connected to the same supply. It needs a pressure difference of at least 50 kPa between the water surface in the tower and the water at the machine. Calculate the greatest height above the ground at which the machine could be installed.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| height difference = 32 − 20 = 12 m | ✓ 1 | |
| p = h × ρ × g = 12 × 1000 × 10 | ✓ 1 | |
| p = 1.2 ×105 Pa | ✓ 1 | |
| Part (b) | ||
| h = p ÷ (ρ × g) = 50 000 ÷ (1000 × 10) | ✓ 1 | |
| h = 5.0 m below the water surface | ✓ 1 | |
| greatest height = 32 − 5.0 = 27 m | ✓ 1 | |
Common error: In p = h × ρ × g, h is the depth below the liquid surface, not the height above the ground. Subtract to find the vertical distance first.
During a flight, the air pressure inside a passenger aircraft cabin is 80 kPa. The air pressure outside the aircraft is 25 kPa. A cabin door has an area of 1.8 m2.
Calculate the pressure difference across the door.
Calculate the resultant force on the door due to this pressure difference.
The door is slightly larger than its frame and must be pulled inwards to open. Explain why passengers cannot open it during the flight.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 80 − 25 = 55 kPa (55 000 Pa) | ✓ 1 | |
| Part (b) | ||
| F = p × A | ✓ 1 | |
| F = 55 000 × 1.8 | ✓ 1 | |
| F = 99 000 N (9.9 ×104 N) | ✓ 1 | |
| Part (c) | ||
| the resultant force acts outwards, pressing the door against its frame | ✓ 1 | |
| pulling the door inwards would need a force of about 99 000 N, far more than a person can exert | ✓ 1 | |
Common error: Use the pressure difference to find the resultant force. Using 80 kPa alone ignores the air pushing on the outside of the door.
A cuboid fuel tank in an aircraft wing has a flat horizontal base of area 8.0 m2. It contains jet fuel to a depth of 1.5 m. The density of jet fuel is 800 kg/m3.
(g = 10 N/kg)
Calculate the mass of fuel in the tank.
Use the equation for pressure difference in a liquid to calculate the pressure that the fuel exerts on the base of the tank.
Show that dividing the weight of the fuel by the area of the base gives the same pressure as your answer to the previous part.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| volume = 8.0 × 1.5 = 12 m3 | ✓ 1 | |
| m = ρ × V = 800 × 12 = 9600 kg | ✓ 1 | |
| Part (b) | ||
| p = h × ρ × g = 1.5 × 800 × 10 | ✓ 1 | |
| p = 12000 Pa (1.2 × 104 Pa) | ✓ 1 | |
| Part (c) | ||
| weight = m × g = 9600 × 10 = 96000 N | ✓ 1 | |
| p = F ÷ A = 96000 ÷ 8.0 | ✓ 1 | |
| = 12000 Pa (1.2 × 104 Pa), the same as the previous answer | ✓ 1 | |
Exam tip: For a straight-sided container, weight ÷ base area and h × ρ × g always agree. This makes a quick way to check a liquid-pressure answer.
A diver swims down from a depth of 2.0 m to a depth of 12.0 m in seawater of density 1030 kg/m3.
What is the increase in pressure on the diver?
(g = 10 N/kg)
| Marking point | Mark | Notes |
|---|---|---|
| The change in depth is 12.0 − 2.0 = 10.0 m. Pressure difference = h × ρ × g = 10.0 × 1030 × 10 = 103 000 Pa. Option A leaves out g. Option C uses the final depth of 12.0 m instead of the change in depth. Option D adds the two depths together. | ✓ 1 | Answer B |
Answer: B
Exam tip: In p = hρg, h is the vertical distance between the two points being compared. Read carefully to see whether the question asks for a pressure or a pressure difference.
An engineer tests a new electric car of mass 1440 kg. The car rests on four identical tyres on a flat road. The weight of the car is shared equally between the four tyres.
(g = 10 N/kg)
Calculate the weight of the car.
The average pressure exerted by the tyres on the road is 2.4 × 105 Pa.
Calculate the area of contact between one tyre and the road. Give your answer in cm2.
Farm tractors that drive on soft, wet fields have very wide tyres.
Explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = m × g = 1440 × 10 | ✓ 1 | |
| W = 14 400 N | ✓ 1 | |
| Part (b) | ||
| A = F ÷ p, or rearrangement used | ✓ 1 | |
| total area = 14 400 ÷ (2.4 × 105) = 0.060 m2 OR force on one tyre = 14 400 ÷ 4 = 3600 N | ✓ 1 | |
| area of one tyre = 0.060 ÷ 4 = 0.015 m2 OR 3600 ÷ (2.4 × 105) = 0.015 m2 | ✓ 1 | |
| = 150 cm2 | ✓ 1 | |
| Part (c) | ||
| wide tyres give a larger area of contact with the ground | ✓ 1 | |
| so for the same weight the pressure on the soil is smaller and the tractor is less likely to sink in | ✓ 1 | |
Exam tip: The weight is shared by all four tyres. Either find the total contact area and divide by four, or find the force on one tyre first. To convert m² to cm², multiply by 10 000, not by 100.
A research submarine has a circular viewing window of area 0.080 m2. The submarine dives to a depth of 250 m in seawater.
Density of seawater = 1025 kg/m3
(g = 10 N/kg)
Calculate the pressure due to the seawater at a depth of 250 m.
Calculate the force exerted by the seawater on the window.
The submarine later operates at the same depth in a freshwater lake.
State how the force on the window changes.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = h × ρ × g = 250 × 1025 × 10 | ✓ 1 | |
| p = 2 562 500 Pa (accept 2.56 × 106 Pa or 2.6 × 106 Pa) | ✓ 1 | |
| Part (b) | ||
| F = p × A = 2 562 500 × 0.080 | ✓ 1 | |
| F = 205 000 N (2.05 × 105 N). Also accept any answer from 2.0 × 105 N to 2.1 × 105 N, for example 208 000 N from using p = 2.6 × 106 Pa. Allow the answer carried forward from an incorrect pressure in the first part. | ✓ 1 | |
| Part (c) | ||
| It decreases, because fresh water has a lower density, so the pressure at the same depth is lower | ✓ 1 | |
Common error: Rearranging p = F/A gives F = p × A. Dividing the pressure by the area instead gives an answer that is far too large and in the wrong unit.
In a hospital, a bag of saline solution is hung above a patient's arm. The solution flows down a tube and into a vein through a needle. The flow happens only if the pressure of the liquid at the needle is greater than the blood pressure in the vein.
Density of saline solution = 1020 kg/m3
(g = 10 N/kg)
State the equation linking pressure difference, height, density and gravitational field strength.
The pressure of the saline at the needle must be at least 9.2 kPa greater than the pressure at the surface of the liquid in the bag.
Calculate the minimum height of the liquid surface in the bag above the needle.
The nurse raises the bag higher above the patient's arm.
Explain why the saline then flows into the vein faster.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| pressure difference = height × density × gravitational field strength or p = h × ρ × g | ✓ 1 | |
| Part (b) | ||
| h = p ÷ (ρ × g) | ✓ 1 | |
| h = 9200 ÷ (1020 × 10) | ✓ 1 | |
| h = 0.90 m | ✓ 1 | |
| Part (c) | ||
| the greater height increases the pressure of the saline at the needle | ✓ 1 | |
| so the pressure difference between the saline and the blood is larger, which pushes liquid in faster | ✓ 1 | |
Common error: Convert kPa to Pa before using p = hρg. Leaving 9.2 kPa as 9.2 gives a height 1000 times too small.
A drone company 3D-prints a body panel from a plastic of density 1.24 g/cm3. The outside of the panel encloses a total volume of 60 cm3.
Calculate the mass of the panel if it were printed completely solid.
To save weight, the panel is printed with a hollow inside. It has a mass of 46.5 g.
Calculate the volume of plastic used.
Calculate the average density of the hollow panel.
The drone may land on a lake. The density of water is 1.00 g/cm3.
Suggest whether the sealed hollow panel would float on water. Give a reason for your answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| m = ρ × V = 1.24 × 60 | ✓ 1 | |
| m = 74.4 g ≈ 74 g | ✓ 1 | |
| Part (b) | ||
| V = m ÷ ρ = 46.5 ÷ 1.24 | ✓ 1 | |
| V = 37.5 cm3 | ✓ 1 | |
| Part (c) | ||
| average density = total mass ÷ total volume = 46.5 ÷ 60 | ✓ 1 | |
| = 0.775 g/cm3 ≈ 0.78 g/cm3 | ✓ 1 | |
| Part (d) | ||
| it would float | ✓ 1 | |
| because its average density (0.78 g/cm3) is less than the density of water | ✓ 1 | |
Exam tip: For a hollow object, use the total mass and the total outside volume to find the average density. This average decides whether a sealed object floats.
A sealed rigid tank of propellant gas on a satellite is heated by sunlight from 27 °C to 327 °C.
Which statement describes the average kinetic energy of the gas molecules?
| Marking point | Mark | Notes |
|---|---|---|
| The average kinetic energy of gas molecules is proportional to the Kelvin temperature. 27 °C = 300 K and 327 °C = 600 K, so the Kelvin temperature doubles and the average kinetic energy doubles. The factor of 12 comes from wrongly using Celsius values (327 ÷ 27). The factor of 1.4 (√2) describes the change in average speed, not kinetic energy. The kinetic energy of each molecule depends on temperature, not on how many molecules there are. | ✓ 1 | Answer B |
Answer: B
Exam tip: Any proportionality involving gas temperature uses kelvin. Convert both temperatures before you compare them.
An aerosol can of deodorant carries the warning 'Do not expose to temperatures above 50 °C'. At 20 °C the gas inside the can is at a pressure of 350 kPa. The volume of the can does not change.
Calculate the pressure of the gas in the can at 50 °C.
Explain, in terms of molecules, why the pressure in the can increases when it is heated.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Converts temperatures: 293 K and 323 K | ✓ 1 | |
| p2 = 350 × 323 ÷ 293 | ✓ 1 | |
| p2 = 386 kPa | ✓ 1 | |
| Part (b) | ||
| Molecules gain kinetic energy / move faster | ✓ 1 | |
| They collide with the walls of the can more often | ✓ 1 | |
| And with more force in each collision, so the force per unit area increases | ✓ 1 | |
Common error: Using 350 × 50 ÷ 20 gives 875 kPa, which is far too large. The equation p₁/T₁ = p₂/T₂ only works with Kelvin temperatures.
A cyclist uses a hand pump to inflate a tyre. The barrel of the pump has a uniform cross-section. When the plunger is fully pulled out, the column of air in the barrel is 30 cm long and at a pressure of 100 kPa. The valve into the tyre opens when the air in the barrel reaches 400 kPa. The plunger is pushed in slowly, so the temperature of the air stays constant.
Explain, in terms of molecules, why the pressure of the air increases as the plunger is pushed in.
Calculate the length of the air column when the valve opens.
Calculate the distance the plunger moves before the valve opens.
Suggest why the pump becomes warm if the plunger is pushed in quickly.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The same number of molecules is in a smaller volume, so the molecules collide with the walls more often (more collisions per second) | ✓ 1 | |
| This gives a greater force per unit area on the walls, so the pressure is greater | ✓ 1 | |
| Part (b) | ||
| The volume is proportional to the length, so 100 × 30 = 400 × L (or p1V1 = p2V2 used with volumes) | ✓ 1 | |
| L = 7.5 cm | ✓ 1 | |
| Part (c) | ||
| 30 − 7.5 = 22.5 cm | ✓ 1 | |
| Part (d) | ||
| Any one of the following : | ✓ 1 | |
| • work is done on the air, which increases the kinetic energy of the molecules and so raises the temperature | — | |
| • the compression is too quick for energy to be transferred to the surroundings, so the air heats up | — | |
| • friction between the plunger and the barrel transfers energy to the pump as thermal energy | — | |
Exam tip: If a container has a uniform cross-section, its volume is proportional to its length. You can then use lengths directly in p₁V₁ = p₂V₂.
A scuba diver breathes out an air bubble of volume 2.0 cm3 at a depth where the total pressure is 250 kPa. The bubble rises to the surface, where the pressure is 100 kPa.
Density of sea water = 1030 kg/m3. g = 10 N/kg.
Calculate the volume of the bubble when it reaches the surface.
State one assumption you made in part (a).
Calculate the depth of the diver.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V2 = 250 × 2.0 ÷ 100 | ✓ 1 | |
| V2 = 5.0 cm3 | ✓ 1 | |
| Part (b) | ||
| The temperature of the air in the bubble stays constant (OR no air leaves or enters the bubble) | ✓ 1 | |
| Part (c) | ||
| Pressure due to the water = 250 − 100 = 150 kPa | ✓ 1 | |
| h = p ÷ (ρg) = 150 000 ÷ (1030 × 10) | ✓ 1 | |
| h = 15 m (14.6 m) | ✓ 1 | |
Common error: The 250 kPa includes atmospheric pressure. Only the extra 150 kPa is caused by the water above the diver, so subtract 100 kPa before using p = ρgh.
Some vaccines are stored in ultra-cold freezers at −73 °C. A sample of air is sealed inside the freezer. Room temperature is 27 °C.
Convert both temperatures to kelvin.
Calculate the ratio
average kinetic energy of air molecules in the freezer ÷ average kinetic energy of air molecules at room temperature.
A student says that the molecules in the freezer have negative kinetic energy because the temperature is negative. Explain why the student is wrong.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| −73 °C = 200 K | ✓ 1 | |
| 27 °C = 300 K | ✓ 1 | |
| Part (b) | ||
| Average kinetic energy is proportional to Kelvin temperature | ✓ 1 | |
| Ratio = 200 ÷ 300 = 2/3 (0.67) | ✓ 1 | |
| Part (c) | ||
| The Celsius scale does not start at absolute zero; the Kelvin temperature is positive, so the molecules are still moving and have positive kinetic energy | ✓ 1 | |
Common error: Using the Celsius values gives a meaningless ratio of −73 ÷ 27. Kinetic energy can never be negative, so always work in kelvin.
A jeweller wants to check whether an irregular lump of yellow resin is real amber. She has an electronic balance, a measuring cylinder, water and some thin thread.
Describe how she could use the measuring cylinder to find the volume of the lump.
The lump has a mass of 27.6 g. The water level rises from 40.0 cm3 to 64.0 cm3.
Calculate the density of the lump in g/cm3.
Real amber floats in a strong salt solution that has a density of 1.20 g/cm3. Suggest whether this lump would float in the solution. Give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Partly fill the measuring cylinder with water and read the volume | ✓ 1 | |
| Lower the lump in on the thread until it is fully under the water, and read the new volume | ✓ 1 | |
| Volume of the lump = new reading − first reading | ✓ 1 | |
| Part (b) | ||
| Volume = 64.0 − 40.0 = 24.0 cm3 | ✓ 1 | |
| density = mass ÷ volume = 27.6 ÷ 24.0 | ✓ 1 | |
| = 1.15 g/cm3 | ✓ 1 | |
| Part (c) | ||
| It would float, because its density (1.15 g/cm3) is less than the density of the solution (1.20 g/cm3) | ✓ 1 | |
Exam tip: The volume of the object is the change in the water reading, not the final reading. Write the subtraction out in full so that you earn the mark for the method.
A biofuel plant stores vegetable oil in a tall vertical tank. The oil is 6.5 m deep. The density of the oil is 880 kg/m3.
g = 10 N/kg
Calculate the pressure difference between the surface of the oil and the bottom of the tank.
A second tank holds oil to the same depth but is twice as wide. State how the pressure at the bottom of the second tank compares with the pressure at the bottom of the first tank.
There is a drain hatch with an area of 0.050 m2 at the bottom of the first tank. The atmosphere pushes on both the oil surface and the outside of the hatch, so only the pressure caused by the oil gives a resultant force.
Calculate the resultant force of the oil on the hatch.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| pressure difference = height × density × g = 6.5 × 880 × 10 | ✓ 1 | |
| = 57 200 Pa | ✓ 1 | |
| Part (b) | ||
| It is the same, because the pressure depends only on depth (and density), not on width | ✓ 1 | |
| Part (c) | ||
| force = pressure × area = 57 200 × 0.050 | ✓ 1 | |
| = 2860 N (accept 2900 N) | ✓ 1 | |
Exam tip: Rearrange p = F/A to F = p × A when you are asked for a force. Carry the unrounded pressure forward from the earlier part.
A lorry driver's seat sits on a sealed air spring. When nobody is sitting on the seat, the air spring contains 1.8 × 10−3 m3 of air at a pressure of 120 kPa. When the driver sits down, the air is compressed to a volume of 1.2 × 10−3 m3. The temperature of the air does not change.
Explain, in terms of molecules, why the pressure of the air increases when its volume decreases at constant temperature.
Calculate the pressure of the air when the driver is sitting on the seat.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The same number of molecules is in a smaller volume, so they hit the walls more often | ✓ 1 | |
| The average speed of the molecules does not change, because the temperature is constant | ✓ 1 | |
| So there is a greater force on each unit area of wall, which means the pressure is greater | ✓ 1 | |
| Part (b) | ||
| p1V1 = p2V2 | ✓ 1 | |
| 120 × 1.8 × 10−3 = p2 × 1.2 × 10−3 | ✓ 1 | |
| p2 = 180 kPa | ✓ 1 | |
Common error: Do not say the molecules move faster when the gas is squashed at constant temperature. Their average speed stays the same. They simply hit the walls more often.
The plunger of a sealed syringe is pushed in slowly so that the volume of the trapped air is halved. The temperature of the air stays constant. Which statement explains why the pressure of the air increases?
| Marking point | Mark | Notes |
|---|---|---|
| At constant temperature the average speed of the molecules is unchanged. In a smaller volume each molecule reaches the walls more often, so the total force per unit area rises: C is correct. A is wrong because speed depends on temperature, which is constant. B is wrong because the force of an individual impact depends on molecular speed, which has not changed. D is wrong because the molecules themselves do not change size. | ✓ 1 | Answer C |
Answer: C
Common error: Pressure changes can come from more frequent collisions or from harder collisions. Decide which one the temperature change allows before choosing.
A metal heating block in a hospital steriliser has mass 2.0 kg and specific heat capacity 500 J/kg °C. The block receives 150 kJ of energy. What is the temperature rise of the block?
| Marking point | Mark | Notes |
|---|---|---|
| ΔT = E / (m c) = 150 000 J / (2.0 kg × 500 J/kg °C) = 150 °C, so B is correct. A uses 150 instead of 150 000 J (kJ not converted). C multiplies the energy by the mass and divides by c. D divides only by the mass and forgets c. | ✓ 1 | Answer B |
Answer: B
Exam tip: Write the energy in joules before using E = mcΔT. A temperature rise of a few hundred degrees from a tiny answer like 0.15 should prompt a unit check.
A jeweller says a ring is made of pure gold. The density of pure gold is 19.3 g/cm3. A student tests the claim. The ring has mass 31.2 g. It is lowered into a displacement can that is full of water, and 2.0 cm3 of water flows into a measuring cylinder.
Describe how the student should use the displacement can to find the volume of the ring.
Calculate the density of the ring in g/cm3.
The measuring cylinder reading could be wrong by up to 0.2 cm3. Calculate the largest possible density of the ring.
Evaluate the jeweller's claim.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Fill the can until water just reaches the spout and wait until dripping stops | ✓ 1 | |
| Lower the ring (on a thread) gently into the can and collect the overflow in a measuring cylinder | ✓ 1 | |
| The volume of water collected equals the volume of the ring | ✓ 1 | |
| Part (b) | ||
| ρ = m / V | ✓ 1 | |
| = 31.2 / 2.0 = 15.6 g/cm3 | ✓ 1 | |
| Part (c) | ||
| Smallest volume = 2.0 − 0.2 = 1.8 cm3 | ✓ 1 | |
| ρ = 31.2 / 1.8 | ✓ 1 | |
| = 17.3 g/cm3 | ✓ 1 | |
| Part (d) | ||
| Even the largest possible density (17.3 g/cm3) is less than 19.3 g/cm3 | ✓ 1 | |
| so the evidence does not support the claim that the ring is pure gold (it must contain a less dense material or hollow spaces) | ✓ 1 | |
Exam tip: To evaluate a claim, compare your calculated value with the reference value and state a clear conclusion that follows from the numbers.
An F1 car and driver have a total mass of 800 kg. The car slows from 50 m/s to 30 m/s using its brakes. Assume that all the lost kinetic energy is transferred to the four steel brake discs, which have a total mass of 6.4 kg. The specific heat capacity of steel is 500 J/kg °C.
Calculate the kinetic energy lost by the car.
Calculate the temperature rise of the brake discs.
Suggest why the real temperature rise of the discs is smaller than your answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Initial KE = ½ × 800 × 502 = 1000000 J and final KE = ½ × 800 × 302 = 360000 J | ✓ 1 | |
| Energy lost = 640000 J | ✓ 1 | |
| Part (b) | ||
| ΔE = m c Δθ so Δθ = ΔE / (m c) | ✓ 1 | |
| = 640000 / (6.4 × 500) (ECF from (a)) | ✓ 1 | |
| 200 °C | ✓ 1 | |
| Part (c) | ||
| Some of the energy is transferred to the surroundings (air) by heating, or some goes to the tyres/road | ✓ 1 | |
Common error: Calculate the kinetic energy at each speed and subtract. Subtracting the speeds first (20 m/s) before squaring gives the wrong energy change.
A beekeeper heats 0.50 kg of solid beeswax at 20 °C using a heater with a constant useful power of 400 W. The wax warms to its melting point of 62 °C and then melts completely at that temperature. Specific heat capacity of solid beeswax = 2500 J/kg °C. Specific latent heat of fusion of beeswax = 1.75 × 105 J/kg.
Calculate the energy needed to warm the wax from 20 °C to 62 °C.
Calculate the energy needed to melt the wax at 62 °C.
Calculate the total time the heater needs.
Explain why the temperature of the wax stays constant while it melts.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ΔE = m c Δθ = 0.50 × 2500 × 42 | ✓ 1 | |
| 52500 J | ✓ 1 | |
| Part (b) | ||
| ΔE = m L = 0.50 × 1.75 × 105 | ✓ 1 | |
| 87500 J | ✓ 1 | |
| Part (c) | ||
| Total energy = 52500 + 87500 = 140000 J (ECF from (a) and (b)) | ✓ 1 | |
| t = E / P = 140000 / 400 = 350 s | ✓ 1 | |
| Part (d) | ||
| The energy supplied is used to break the bonds between molecules (increasing potential energy) | ✓ 1 | |
| not to increase their average kinetic energy, so the temperature does not rise | ✓ 1 | |
Exam tip: In an explain question about melting, say what the energy does (overcomes attractive forces) and what it does not do (increase the average kinetic energy).
A smoke cell is viewed through a microscope. Bright specks of smoke are seen moving in small, jerky, random paths. Which statement gives the best explanation?
| Marking point | Mark | Notes |
|---|---|---|
| The air molecules are too small to see but move randomly and collide with the much larger smoke particles. At any instant more molecules hit one side than the other, giving an unbalanced force that keeps changing direction, so B is correct. A is wrong because air molecules are not at rest. C would give steady, directed motion rather than random jerks. D would give straight-line motion towards the lamp. | ✓ 1 | Answer B |
Answer: B
Exam tip: The smoke specks are what you see; the air molecules are the cause of their motion and cannot be seen.
A student measures the specific heat capacity of an aluminium block of mass 0.80 kg. The block has a hole for an electric heater and a hole for a thermometer. The heater has a power of 48 W and is switched on for 8.0 minutes. The temperature of the block rises from 19.0 °C to 43.0 °C.
Describe how the student should set up the block to get reliable results.
Calculate the energy supplied by the heater.
Calculate the specific heat capacity of the aluminium from these results. Give the unit.
The accepted value is 900 J/kg °C. Explain why the student's value is higher.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Place a little oil in the thermometer hole so there is good thermal contact | ✓ 1 | |
| Wrap the block in insulating material to reduce energy losses | ✓ 1 | |
| Part (b) | ||
| Time = 8.0 × 60 = 480 s | ✓ 1 | |
| Energy = P t = 48 × 480 = 23 040 J | ✓ 1 | |
| Part (c) | ||
| Δθ = 43.0 − 19.0 = 24.0 °C | ✓ 1 | |
| c = E / (m Δθ) = 23 040 / (0.80 × 24.0) (ECF from (b)) | ✓ 1 | |
| 1200 J/kg °C | ✓ 1 | |
| Part (d) | ||
| Some energy from the heater is transferred to the surroundings | ✓ 1 | |
| So the block's temperature rise is smaller than it should be, which makes the calculated c larger | ✓ 1 | |
Exam tip: Use only the energy that reaches the block; any energy lost makes the measured specific heat capacity too large.
An electric heater of power 300 W warms 150 kg of water in a fish-holding tank by 3.0 °C. The specific heat capacity of water is 4200 J/kg °C. Assume all the energy from the heater warms the water.
Calculate the energy needed to warm the water.
Calculate the time the heater takes, in minutes.
Explain why the heater would take longer in practice.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Energy = m c Δθ | ✓ 1 | |
| = 150 × 4200 × 3.0 | ✓ 1 | |
| 1.89 × 106 J (1 890 000 J) | ✓ 1 | |
| Part (b) | ||
| Time = energy ÷ power = 1 890 000 ÷ 300 (ECF from (a)) | ✓ 1 | |
| 6300 s | ✓ 1 | |
| 105 minutes | ✓ 1 | |
| Part (c) | ||
| Some energy is transferred to the surroundings (the air, tank walls) | ✓ 1 | |
| So more energy must be supplied in total to reach the same temperature rise | ✓ 1 | |
Exam tip: State the assumption you are asked to make; it tells you which energy losses to ignore.
The aluminium housing of a robot actuator absorbs 270 kJ of thermal energy while its temperature rises from 22 °C to 42 °C. The specific heat capacity of aluminium is 900 J/kg °C.
State the temperature rise of the housing.
Calculate the mass of the housing.
The housing is replaced by a copper one of the same mass, which has a specific heat capacity of 390 J/kg °C. It absorbs the same energy. State whether the temperature rise is larger or smaller and explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 20 °C | ✓ 1 | |
| Part (b) | ||
| m = E / (c Δθ) with E = 270 000 J | ✓ 1 | |
| = 270 000 / (900 × 20) | ✓ 1 | |
| 15 kg | ✓ 1 | |
| Part (c) | ||
| Larger | ✓ 1 | |
| Copper has a lower specific heat capacity so less energy is needed per kg for each °C, and the same energy gives a bigger rise | ✓ 1 | |
Exam tip: Rearrange m c Δθ = E before you substitute, and check the unit of energy is J.
A cup of coffee cools from 90 °C to 40 °C and stays liquid. Which statement describes what happens to the particles of the coffee?
| Marking point | Mark | Notes |
|---|---|---|
| A is correct: lower temperature means lower average kinetic energy, so slower motion, but the liquid has not frozen so the particles stay close and can still slide past each other. B describes a solid and particles never stop moving. C is wrong because particles do not change size on cooling. D is wrong because the average speed falls with temperature and a liquid does not expand to wide spacing on cooling. | ✓ 1 | Answer A |
Answer: A
Exam tip: Temperature is linked to the average kinetic energy of particles. A change of state only happens when the particle arrangement changes, not just because the temperature falls.
A tall marble column stands on level ground. It is a solid cylinder of uniform cross-section with a base area of 0.50 m2 and a height of 6.0 m. The density of the marble is 2700 kg/m3. Use g = 10 N/kg.
Calculate the mass of the column.
Calculate the pressure that the column exerts on the ground. Give your answer to 3 significant figures.
A second column of the same marble has a height of 9.0 m. Calculate the pressure it exerts on the ground using p = hρg.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Volume = area × height = 0.50 × 6.0 = 3.0 m3 | ✓ 1 | |
| m = ρV | ✓ 1 | |
| m = 2700 × 3.0 = 8100 kg | ✓ 1 | |
| Part (b) | ||
| W = mg = 8100 × 10 = 81000 N (ECF from (a)) | ✓ 1 | |
| p = F ÷ A | ✓ 1 | |
| p = 81000 ÷ 0.50 = 162000 Pa (1.62 × 105 Pa) | ✓ 1 | |
| Part (c) | ||
| Substitution p = 9.0 × 2700 × 10 | ✓ 1 | |
| p = 243000 Pa (2.43 × 105 Pa) | ✓ 1 | |
Exam tip: For a solid column of uniform cross-section, p = hρg gives the pressure on the base without needing the area or the mass.
A space probe carries an aluminium block of mass 2.5 kg that holds a sensor. The specific heat capacity of aluminium is 900 J/kg °C. Before a manoeuvre a heater raises the temperature of the block from −40 °C to 20 °C.
Calculate the thermal energy needed to warm the block.
The heater has a power of 150 W. Assume that all of its energy is transferred to the block. Calculate the time taken.
In practice the block takes longer than this time to warm up. Suggest one reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Q = mcΔθ | ✓ 1 | |
| Δθ = 60 °C | ✓ 1 | |
| Q = 2.5 × 900 × 60 = 135000 J | ✓ 1 | |
| Part (b) | ||
| t = Q ÷ P (using E = Pt) | ✓ 1 | |
| t = 135000 ÷ 150 = 900 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Some energy is transferred to the surroundings (by radiation or conduction to the probe) rather than to the block | ✓ 1 | |
Common error: Temperature change is final minus initial. Starting below 0 °C means the change is larger than the final temperature, not smaller.
A student investigates the specific heat capacity of green coffee beans before roasting. A 120 W electric heater is placed in a 0.50 kg sample of beans. The heater is switched on for 5.0 minutes and the temperature of the beans rises from 20.0 °C to 56.0 °C.
Calculate the electrical energy transferred by the heater.
Assume that all of this energy is transferred to the beans. Calculate the specific heat capacity of the beans.
The accepted value for this type of bean is 1800 J/kg °C. Explain why the student's value is higher.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Time converted to 300 s | ✓ 1 | |
| E = Pt = 120 × 300 = 36000 J | ✓ 1 | |
| Part (b) | ||
| c = Q ÷ (mΔθ) | ✓ 1 | |
| Δθ = 36.0 °C | ✓ 1 | |
| c = 36000 ÷ (0.50 × 36.0) = 2000 J/kg °C (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| Some energy from the heater is lost to the surroundings, so less than 36000 J reaches the beans | ✓ 1 | |
| The energy used in the calculation is too large, so the calculated c is too large | ✓ 1 | |
Exam tip: Convert minutes to seconds before using E = Pt. Energy losses make the energy gained by the substance less than the energy supplied.
A cup of hot coffee stands uncovered in a room. The coffee cools and its volume slowly decreases, although it is well below its boiling point.
Explain, in terms of particles, why the coffee cools as it evaporates. Describe also one way in which evaporation differs from boiling.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The particles in the liquid have a range of speeds (kinetic energies) | ✓ 1 | |
| Fast particles near the surface have enough energy to overcome the attraction of the other particles | ✓ 1 | |
| These particles escape from the surface into the air | ✓ 1 | |
| The escaping particles have more than the average kinetic energy | ✓ 1 | |
| So the average kinetic energy of the remaining particles falls and the temperature decreases | ✓ 1 | |
| Evaporation happens only at the surface and at any temperature, but boiling occurs throughout the liquid at a fixed temperature | ✓ 1 | |
Common error: Evaporation is not the same as boiling. Evaporation happens at the surface below the boiling point, and it cools the liquid left behind because the most energetic particles leave.
A sample-return capsule brings back an irregular rock from an asteroid. The rock has a mass of 0.84 kg. Its volume is found by displacement in a large measuring cylinder. The water level reads 500 cm3 before the rock is added and 800 cm3 after it is fully submerged.
Describe how the readings are used to find the volume of the rock.
Calculate the density of the rock in kg/m3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Volume of rock is the difference between the two readings | ✓ 1 | |
| 800 − 500 = 300 cm3 | ✓ 1 | |
| Part (b) | ||
| Convert volume: 300 cm3 = 3.0 × 10−4 m3 | ✓ 1 | |
| ρ = m ÷ V | ✓ 1 | |
| ρ = 0.84 ÷ 3.0 × 10−4 = 2800 kg/m3 | ✓ 1 | |
Common error: Convert cm³ to m³ by multiplying by 10⁻⁶, not 10⁻³, before using kg/m³.
Salol is heated until it is a liquid and then left to cool. The temperature stays constant at about 42 °C for several minutes while the salol solidifies. Which statement explains this?
| Marking point | Mark | Notes |
|---|---|---|
| Correct (D): as the particles settle into fixed positions their potential energy falls and this energy is transferred out, so the average kinetic energy, and therefore the temperature, stays constant. A describes the sloping sections of the curve, not the flat one. B is wrong: the salol is still hotter than the surroundings and loses energy. C is wrong: particles in a solid still vibrate. | ✓ 1 | Answer D |
Answer: D
Common error: A flat section on a cooling curve does not mean that energy transfer has stopped.
A market trader sells a ring and claims that it is pure gold. A student tests the claim. The ring has a mass of 18.0 g. The density of pure gold is 19.3 g/cm3.
Describe how the student can use a displacement can and a measuring cylinder to find the volume of the ring.
The volume of water collected is 1.5 cm3. Calculate the density of the ring. Give the unit.
Evaluate the trader's claim.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 fill the can until water just runs out of the spout, then lower the ring (on a thread) into the can and collect the displaced water in the measuring cylinder | ✓ 1 | |
| B1 the volume of water collected equals the volume of the ring | ✓ 1 | |
| Part (b) | ||
| C1 density = mass ÷ volume | ✓ 1 | |
| C1 18.0 ÷ 1.5 | ✓ 1 | |
| A1 12.0 g/cm3 | ✓ 1 | |
| Part (c) | ||
| E1 the density of the ring (12.0 g/cm3) is less than the density of pure gold (19.3 g/cm3), so the ring is not pure gold | ✓ 1 | |
| E1 it could be an alloy of gold with a less dense metal, or have a hollow space inside | ✓ 1 | |
Exam tip: In an evaluate question, compare your calculated value with the given value and then state a clear conclusion.
A student investigates how the temperature of salol changes as it cools. Salol is melted in a boiling tube by standing it in hot water. The tube is then removed and the temperature is recorded every 60 s.
| Time / s | 0 | 60 | 120 | 180 | 240 | 300 | 360 |
| Temperature / °C | 70.0 | 57.0 | 46.0 | 42.0 | 42.0 | 42.0 | 38.0 |
Suggest why the salol is melted in hot water rather than being heated directly with a Bunsen burner flame.
State the melting point of salol, using the data.
Explain why the temperature stays constant between 180 s and 300 s although the salol is still losing energy to the surroundings.
Calculate the mean rate of cooling of the salol between 0 s and 120 s.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| E1 the hot water heats the salol gently and evenly, so it does not overheat or catch fire | ✓ 1 | |
| Part (b) | ||
| B1 42 °C | ✓ 1 | |
| Part (c) | ||
| E1 the salol is changing state from liquid to solid | ✓ 1 | |
| E1 the particles are forming bonds and their potential energy decreases, releasing the energy that is lost, so their average kinetic energy does not change | ✓ 1 | |
| Part (d) | ||
| C1 (70.0 − 46.0) ÷ 120 | ✓ 1 | |
| A1 0.20 °C/s | ✓ 1 | |
Exam tip: For rate questions, use the change in temperature divided by the time taken, and quote the unit.
A Formula 1 car has 120 kg of fuel in its tank. The density of the fuel is 750 kg/m3. The car, driver and fuel have a total mass of 800 kg. Use g = 10 N/kg.
Calculate the volume of the fuel.
Calculate the total weight of the car, driver and fuel.
The four tyres have a total contact area of 0.080 m2. Calculate the average pressure that the tyres exert on the track. Use your answer to (b) or 8000 N.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| C1 volume = mass ÷ density | ✓ 1 | |
| C1 120 ÷ 750 | ✓ 1 | |
| A1 0.16 m3 | ✓ 1 | |
| Part (b) | ||
| B1 800 × 10 = 8000 N | ✓ 1 | |
| Part (c) | ||
| C1 pressure = force ÷ area | ✓ 1 | |
| C1 8000 ÷ 0.080 (ECF from (b)) | ✓ 1 | |
| A1 1.0 × 105 Pa | ✓ 1 | |
Exam tip: Weight is a force in newtons: mass in kg multiplied by g.
A hospital heat pack contains 0.50 kg of water at 60 °C. The water cools to 40 °C while warming a patient's shoulder. Specific heat capacity of water = 4200 J/kg °C.
Calculate the energy transferred from the water as it cools.
A different pack contains 0.80 kg of a material with a specific heat capacity of 1750 J/kg °C. Calculate the fall in temperature of this material when it transfers the same energy of 42 000 J.
Suggest why water is a good choice of material for a heat pack.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| C1 ΔQ = mcΔθ | ✓ 1 | |
| C1 0.50 × 4200 × 20 | ✓ 1 | |
| A1 42 000 J | ✓ 1 | |
| Part (b) | ||
| C1 Δθ = ΔQ ÷ (mc) = 42 000 ÷ (0.80 × 1750) | ✓ 1 | |
| A1 30 °C | ✓ 1 | |
| Part (c) | ||
| E1 water has a high specific heat capacity, so it stores a large amount of energy per kilogram for each degree of cooling | ✓ 1 | |
Common error: The temperature change is the difference between the two temperatures, not either temperature on its own.
Two solid columns, X and Y, are made of the same marble and stand on level ground. Column X has height h and base area A. Column Y has height 2h and base area A/2. The pressure that X exerts on the ground is p. What is the pressure that Y exerts on the ground?
| Marking point | Mark | Notes |
|---|---|---|
| C is correct: for a uniform column, pressure = hρg, so doubling the height doubles the pressure (the base area does not matter). Check: the weight of Y is ρ(2h)(A/2)g = ρhAg, the same as X, but it acts over half the area, so the pressure is 2p. A would result from halving the area wrongly. B ignores the change in height. D doubles the height and also halves the area as if both effects multiplied separately. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a uniform vertical column of one material, pressure depends only on height, density and g, not on the cross-sectional area.
A solid marble column in a railway station is a vertical cylinder of height 4.0 m and cross-sectional area 0.36 m2. Marble has a density of 2700 kg/m3. Use g = 10 N/kg.
Calculate the pressure that the column exerts on the floor. [3]
Calculate the weight of the column. Use the unrounded answer to (a); give your answer to 2 significant figures. [2]
A second column of the same marble and the same height has half the cross-sectional area. State the pressure it exerts on the floor and explain your answer. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use of p = hρg | ✓ 1 | |
| p = 4.0 × 2700 × 10 | ✓ 1 | |
| p = 108000 Pa (1.08 × 105 Pa) | ✓ 1 | |
| Part (b) | ||
| Use of force = pressure × area, F = 108000 × 0.36 (ECF from (a)) | ✓ 1 | |
| W = 38880 N = 3.9 × 104 N | ✓ 1 | |
| Part (c) | ||
| Same pressure as the first column (108000 Pa) | ✓ 1 | |
| Pressure depends only on height, density and g (the weight halves but so does the area) | ✓ 1 | |
Exam tip: For a uniform column, either p = hρg or p = W/A works; use whichever gives the shortest route from the data given.
A technician fills a tall tank with an unknown liquid for buoyancy tests in a sports-science laboratory. A pressure sensor measures the pressure p due to the liquid alone at different depths h.
Depth h / m: 0.5 | 1.0 | 1.5 | 2.0 | 2.5
Pressure p / kPa: 6.3 | 12.6 | 18.9 | 25.2 | 31.5
Use g = 10 N/kg.
State the relationship between p and h shown by the data. [1]
Determine the gradient of a graph of p (y-axis) against h (x-axis), using the first and last data points. Give the unit. [2]
The gradient is equal to ρg. Determine the density of the liquid in kg/m3. [2]
The tank is emptied and refilled with water (density 1000 kg/m3). State how the gradient of the graph changes. [1]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Pressure is directly proportional to depth | ✓ 1 | |
| Part (b) | ||
| Gradient = change in p / change in h = (31.5 − 6.3)/(2.5 − 0.5) | ✓ 1 | |
| 12.6 kPa/m (accept 12600 Pa/m) | ✓ 1 | |
| Part (c) | ||
| ρ = gradient/g = 12600/10 (kPa converted to Pa) | ✓ 1 | |
| ρ = 1260 kg/m3 (1.26 × 103 kg/m3) | ✓ 1 | |
| Part (d) | ||
| The gradient is smaller (10 kPa/m, since ρ is smaller) | ✓ 1 | |
Common error: Convert kPa to Pa before dividing by g, otherwise the density comes out 1000 times too small.
A physiotherapy research team tests a copper block as a model for a heat pack. The block has mass 2.5 kg and starts at 18 °C. An electric heater of power 150 W is switched on for 6.0 minutes and all the energy transferred goes into the block. The specific heat capacity of copper is 390 J/kg°C.
Calculate the energy transferred to the block. [2]
Calculate the temperature rise of the block. Give your answer to 2 significant figures. [3]
Calculate the final temperature of the block. Use the unrounded temperature rise from (b). [1]
Suggest why the real final temperature is lower than your answer to (c). [1]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use of E = Pt with t = 360 s | ✓ 1 | |
| E = 150 × 360 = 54000 J | ✓ 1 | |
| Part (b) | ||
| Use of E = mcΔθ (rearranged for Δθ) | ✓ 1 | |
| Δθ = 54000/(2.5 × 390) | ✓ 1 | |
| Δθ = 55 °C (55.38...) | ✓ 1 | |
| Part (c) | ||
| 18 + 55.38 = 73.38 °C (73 °C) | ✓ 1 | |
| Part (d) | ||
| Some energy is transferred to the surroundings (not all goes into the block) | ✓ 1 | |
Common error: Time must be in seconds when using E = Pt; 6.0 minutes is 360 s, not 6.0 s.
A shot put used in athletics training is a solid metal sphere of mass 7.26 kg. A technician finds its volume with a displacement can and measures the volume of displaced water as 940 cm3.
Calculate the density of the metal in kg/m3. Give your answer to 2 significant figures. [3]
Describe how the technician uses the displacement can to find the volume of the shot put. [2]
A second shot put is made from the same metal but has twice the mass. State its density. [1]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use of ρ = m/V | ✓ 1 | |
| V = 940 × 10−6 m3 (940 cm3 converted to m3) | ✓ 1 | |
| ρ = 7.26/(940 × 10−6) = 7723.4 kg/m3 = 7.7 × 103 kg/m3 | ✓ 1 | |
| Part (b) | ||
| Fill the can until water just overflows from the spout, then lower the shot into the water | ✓ 1 | |
| Collect the displaced water in a measuring cylinder and read its volume | ✓ 1 | |
| Part (c) | ||
| The same density (7723.4 kg/m3), since the volume also doubles | ✓ 1 | |
Common error: 1 m³ is 10⁶ cm³, not 100 cm³. Converting 940 cm³ to 9.4 m³ or 0.94 m³ is a common slip.
A copper block of mass 0.50 kg is heated by a 60 W electric heater for 130 s. All the energy from the heater warms the block. The specific heat capacity of copper is 390 J/kg °C. What is the temperature rise of the block?
| Marking point | Mark | Notes |
|---|---|---|
| Energy supplied = 60 × 130 = 7800 J. Temperature rise = 7800 ÷ (0.50 × 390) = 40 °C. A rise of 20 °C comes from leaving out the mass, 10 °C from multiplying by the mass instead of dividing, and 0.31 °C from forgetting to multiply the power by the time. | ✓ 1 | Answer A |
Answer: A
Exam tip: Find the energy first (power × time) and then rearrange Q = mcΔθ for Δθ.
A bubble of volcanic gas in magma has a volume of 4.0 cm³ at a pressure of 300 kPa. The bubble rises to where the pressure is 100 kPa. The temperature stays constant. What is the new volume of the bubble?
| Marking point | Mark | Notes |
|---|---|---|
| At constant temperature pV is constant, so V = 4.0 × 300 ÷ 100 = 12 cm³. 1.3 cm³ comes from inverting the ratio, 8.0 cm³ from using the pressure difference instead of the ratio, and 36 cm³ from squaring the ratio. | ✓ 1 | Answer C |
Answer: C
Exam tip: If the pressure falls at constant temperature, the volume must rise by the same factor.
A theatre prop-maker is sold a ring described as pure gold. The ring is tested using a displacement can. The ring has a mass of 14.4 g. When the ring is lowered into the full can, the water that overflows has a volume of 0.90 cm³.
Describe how the overflowing water is used to find the volume of the ring.
Calculate the density of the ring in g/cm³.
Convert your answer to (b) into kg/m³.
Pure gold has a density of 19.3 g/cm³. The volume reading could be wrong by up to 0.05 cm³. Evaluate whether the ring could be pure gold. Give the highest possible density to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Place a measuring cylinder under the spout of the can | ✓ 1 | |
| Lower the ring in gently so water overflows into the cylinder | ✓ 1 | |
| Volume of the ring equals the volume of water collected | ✓ 1 | |
| Part (b) | ||
| density = mass ÷ volume = 14.4 ÷ 0.90 | ✓ 1 | |
| 16 g/cm³ | ✓ 1 | |
| Part (c) | ||
| 16 000 kg/m³ (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| smallest possible volume is 0.85 cm³ | ✓ 1 | |
| highest density = 14.4 ÷ 0.85 = 16.9 g/cm³ | ✓ 1 | |
| this is still below 19.3 g/cm³, so the ring cannot be pure gold | ✓ 1 | |
Exam tip: When asked to evaluate a claim, use the extreme values of the uncertainty to see if the conclusion still holds.
A theatre stage lift uses air trapped in a cylinder. The air has a volume of 2.0 × 10−3 m³ at a pressure of 150 kPa. The air is compressed slowly to a volume of 0.50 × 10−3 m³.
Calculate the new pressure of the air.
Explain, in terms of molecules, why the pressure increases as the volume decreases.
State one condition needed for the equation in (a) to apply.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p1V1 = p2V2 | ✓ 1 | |
| p2 = 150 × 2.0 ÷ 0.50 | ✓ 1 | |
| 600 kPa | ✓ 1 | |
| Part (b) | ||
| molecules are closer together, so collide with the walls more often | ✓ 1 | |
| greater force per unit area on the walls, so pressure rises | ✓ 1 | |
| Part (c) | ||
| the temperature of the air stays constant or the mass of air stays the same | ✓ 1 | |
Exam tip: Pressure and volume units only need to be consistent on each side, so kPa can be used without converting.
A ship's galley heats 2.0 kg of water from 15 °C to 100 °C using a 3.0 kW heater. The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy needed to heat the water.
Calculate the time the heater would take if all its energy went into the water.
In practice the heater takes 280 s. Calculate the energy wasted.
Calculate the percentage of the energy supplied that is wasted.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy = m c Δθ = 2.0 × 4200 × 85 | ✓ 1 | |
| 714 000 J | ✓ 1 | |
| Part (b) | ||
| time = energy ÷ power = 714 000 ÷ 3000 | ✓ 1 | |
| 238 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| energy supplied = 3000 × 280 = 840 000 J | ✓ 1 | |
| wasted = 840 000 − 714 000 = 126 000 J | ✓ 1 | |
| Part (d) | ||
| 126 000 ÷ 840 000 × 100 | ✓ 1 | |
| 15% (ECF from (c)) | ✓ 1 | |
Exam tip: Percentage wasted is the wasted energy divided by the total energy supplied.
A sealed syringe traps 60 cm3 of air at a pressure of 100 kPa. The plunger is pushed in slowly until the air occupies 24 cm3. The temperature of the air stays constant. What is the final pressure of the air?
| Marking point | Mark | Notes |
|---|---|---|
| At constant temperature pV is constant, so p = 100 × 60 ÷ 24 = 250 kPa. 40 kPa comes from using the volume ratio the wrong way up. 136 kPa comes from adding the volume change to the pressure. 6000 kPa comes from multiplying the pressure by the starting volume and stopping. | ✓ 1 | Answer D |
Answer: D
Exam tip: Compressing a gas at constant temperature raises its pressure, so check that your answer is larger than the starting pressure.
A sealed rigid steel canister of gas is heated. The mass of gas and its volume stay constant. Which statement explains why the gas pressure rises?
| Marking point | Mark | Notes |
|---|---|---|
| Heating raises the average kinetic energy, so molecules move faster, collide with the walls more frequently and with greater force, which raises the pressure. Molecules do not expand when heated, no new molecules are made because the mass of gas is fixed, and the mass of each molecule does not change. | ✓ 1 | Answer B |
Answer: B
Common error: The number of molecules and the size of molecules do not change when a gas is heated; only their speed does.
A copper block of mass 0.80 kg is at 21 °C. It is heated by a 150 W electric heater for 2.0 minutes. The specific heat capacity of copper is 390 J/kg °C. Assume all the electrical energy is transferred to the block.
Calculate the energy supplied by the heater.
Calculate the rise in temperature of the block. Give your answer to 3 significant figures.
Calculate the final temperature of the block. Either the rounded or the unrounded rise may be used.
In a real experiment the final temperature is lower than this. Suggest why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = power × time, with time = 120 s | ✓ 1 | |
| A1 18 000 J | ✓ 1 | |
| Part (b) | ||
| M1 change in thermal energy = mass × specific heat capacity × temperature change, rearranged | ✓ 1 | |
| M1 substitution of 18 000 ÷ (0.80 × 390) | ✓ 1 | |
| A1 57.7 °C (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| A1 78.7 °C (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| some energy is transferred to the surroundings (by heating) | ✓ 1 | |
Common error: Convert minutes to seconds before using energy = power × time.
A student measures the specific heat capacity of an aluminium block of mass 1.0 kg. A 12 V heater, carrying a current of 4.0 A, is placed in the block and switched on for 5.0 minutes (300 s). The temperature of the block rises by 15.0 °C.
Calculate the energy supplied by the heater.
Calculate the specific heat capacity of aluminium from these results.
The accepted value is 900 J/kg °C. Explain why the student's value is higher.
Suggest one change to the experiment to improve the result.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = voltage × current × time | ✓ 1 | |
| A1 14 400 J | ✓ 1 | |
| Part (b) | ||
| M1 change in thermal energy = mass × specific heat capacity × temperature change, rearranged | ✓ 1 | |
| M1 substitution of 14 400 ÷ (1.0 × 15.0) (ECF from (a)) | ✓ 1 | |
| A1 960 J/kg °C | ✓ 1 | |
| Part (c) | ||
| some energy is transferred to the surroundings | ✓ 1 | |
| so the temperature rise is smaller than it would be with no loss, so the calculated value is too large | ✓ 1 | |
| Part (d) | ||
| wrap the block in insulation (lagging) / place the block on an insulating mat | ✓ 1 | |
Exam tip: In a heating practical, energy lost to the surroundings makes the measured temperature rise too small, so the calculated specific heat capacity comes out too large.
A crater lake contains 2.0 × 109 kg of water. A volcanic vent supplies energy to the lake at a constant rate of 2.8 × 106 W. The specific heat capacity of water is 4200 J/kg °C. Assume all the energy from the vent heats the water and no energy is lost.
Calculate the energy needed to raise the temperature of the lake by 0.50 °C.
Calculate the time taken for the vent to supply this energy.
Suggest why the real time is longer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 change in thermal energy = mass × specific heat capacity × temperature change | ✓ 1 | |
| M1 substitution of 2.0 × 109 × 4200 × 0.50 | ✓ 1 | |
| A1 4.2 × 1012 J | ✓ 1 | |
| Part (b) | ||
| M1 time = energy ÷ power | ✓ 1 | |
| A1 1.5 × 106 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| some energy is transferred to the surroundings (air, rocks) / evaporation from the surface | ✓ 1 | |
Exam tip: Write powers of ten carefully; 2.0 × 10⁹ × 4200 × 0.50 should give 4.2 × 10¹², not 4.2 × 10¹¹.
A hot-water bottle contains 1.5 kg of water at 80 °C. The water cools to 50 °C. The specific heat capacity of water is 4200 J/kg °C. How much thermal energy does the water transfer to the surroundings?
| Marking point | Mark | Notes |
|---|---|---|
| Energy = mass × specific heat capacity × temperature change = 1.5 × 4200 × 30 = 189000 J. The value 126000 J leaves out the mass. The value 504000 J uses the starting temperature of 80 °C instead of the change in temperature. The value 819000 J adds the two temperatures to get 130 °C instead of subtracting them. | ✓ 1 | Answer D |
Answer: D
Exam tip: Use the temperature change (final minus initial), never one of the temperatures on its own.
A sealed aerosol can of constant volume is warmed in sunlight. Which statement explains why the pressure of the gas inside the can increases?
| Marking point | Mark | Notes |
|---|---|---|
| Higher temperature means a greater average molecular speed, so collisions with the walls are more frequent and each collision exerts a larger force; the total force per unit area rises. Molecules do not expand when heated. The can is sealed, so the number of molecules cannot increase. The mass of each molecule does not change with temperature. | ✓ 1 | Answer A |
Answer: A
Exam tip: Link pressure to both the number of collisions per second and the force of each collision.
A hob heats 1.5 kg of water in an aluminium saucepan of mass 0.80 kg. The temperature of the water and pan rises from 20 °C to 70 °C. The specific heat capacity of water is 4200 J/kg °C and of aluminium is 900 J/kg °C.
Calculate the thermal energy transferred to the water.
Calculate the thermal energy transferred to the saucepan.
The hob supplies energy to the pan at a constant rate of 1800 W. Calculate the time taken, assuming all the energy goes to the water and saucepan.
Suggest why the actual time taken is longer than your answer to (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = 1.5 × 4200 × 50 | ✓ 1 | |
| A1 315000 J | ✓ 1 | |
| Part (b) | ||
| M1 energy = 0.80 × 900 × 50 | ✓ 1 | |
| A1 36000 J | ✓ 1 | |
| Part (c) | ||
| M1 time = (315000 + 36000) ÷ 1800 (ECF from (a) and (b)) | ✓ 1 | |
| A1 195 s | ✓ 1 | |
| Part (d) | ||
| some energy is transferred to the surroundings (by heat loss from the pan, or by evaporation) | ✓ 1 | |
Exam tip: Add the energy needed by the pan and by the water before dividing by the power.
A hot-water bottle contains 2.0 kg of water at 70 °C. Overnight the water cools to 35 °C. The specific heat capacity of water is 4200 J/kg °C.
Calculate the thermal energy transferred by the water as it cools.
The cooling takes 8.0 hours. Calculate the average power transferred to the surroundings. Give your answer to 3 significant figures. Use 294000 J.
Explain why water is a good liquid to use in a hot-water bottle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = 2.0 × 4200 × 35 | ✓ 1 | |
| A1 294000 J | ✓ 1 | |
| Part (b) | ||
| M1 power = energy ÷ time | ✓ 1 | |
| M1 time = 8.0 × 3600 = 28800 s | ✓ 1 | |
| A1 10.2 W | ✓ 1 | |
| Part (c) | ||
| water has a high specific heat capacity, so it stores a lot of thermal energy per kilogram for each degree of temperature change | ✓ 1 | |
Common error: Convert hours to seconds before using power = energy ÷ time.
A wind turbine stands on a concrete foundation block. The block is a cuboid measuring 8.0 m by 8.0 m by 2.5 m. The density of the concrete is 2400 kg/m3. The gravitational field strength is 10 N/kg.
Calculate the volume of the block.
Calculate the mass of the block.
The block rests on its 8.0 m by 8.0 m face. Calculate the pressure it exerts on the ground.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 8.0 × 8.0 × 2.5 | ✓ 1 | |
| A1 160 m3 | ✓ 1 | |
| Part (b) | ||
| M1 mass = density × volume = 2400 × 160 (ECF from (a)) | ✓ 1 | |
| A1 384000 kg | ✓ 1 | |
| Part (c) | ||
| M1 weight = 384000 × 10 = 3.84 × 106 N (ECF from (b)) | ✓ 1 | |
| M1 pressure = force ÷ area = 3.84 × 106 ÷ 64 | ✓ 1 | |
| A1 60000 Pa | ✓ 1 | |
Exam tip: Convert mass to weight with W = mg before using pressure = force ÷ area.
A swimming pool measures 12 m by 6.0 m and the water is 1.5 m deep, so its volume is 108 m3. The density of water is 1000 kg/m3 and its specific heat capacity is 4200 J/kg °C.
Calculate the mass of water in the pool.
Calculate the thermal energy needed to raise the temperature of the water by 2.0 °C. Give your answer in standard form to 3 significant figures.
A heater supplies energy to the water at a rate of 150 kW. Calculate the time needed to raise the temperature by 2.0 °C. Use the unrounded energy from (b).
Suggest one reason why the pool takes longer than this to warm up.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 mass = density × volume = 1000 × 108 | ✓ 1 | |
| A1 108000 kg | ✓ 1 | |
| Part (b) | ||
| M1 energy = mass × specific heat capacity × temperature change | ✓ 1 | |
| M1 108000 × 4200 × 2.0 (ECF from (a)) | ✓ 1 | |
| A1 9.07 × 108 J | ✓ 1 | |
| Part (c) | ||
| M1 time = energy ÷ power = 9.072 × 108 ÷ 150000 | ✓ 1 | |
| A1 6048 s | ✓ 1 | |
| Part (d) | ||
| energy is lost from the surface by evaporation, or conduction and convection to the air or ground | ✓ 1 | |
Common error: Power is in kilowatts here; change 150 kW to 150000 W before dividing.
A sealed sample of helium in a laboratory is heated from 27 °C to 327 °C. Use 0 °C = 273 K. What happens to the average kinetic energy of the helium atoms?
| Marking point | Mark | Notes |
|---|---|---|
| Average kinetic energy is proportional to the kelvin temperature. The temperatures are 300 K and 600 K, so the energy doubles. Staying the same ignores the temperature rise. A factor of about 12 comes from dividing the Celsius values (327 / 27). Quadrupling comes from squaring the factor, as if kinetic energy depended on temperature squared. | ✓ 1 | Answer D |
Answer: D
Common error: Never take ratios of Celsius temperatures; convert to kelvin first.
A solar thermal store holds 500 kg of water. The water is heated by 12 °C. The specific heat capacity of water is 4200 J/kg °C. How much energy is transferred to the water?
| Marking point | Mark | Notes |
|---|---|---|
| Energy = m c Δθ = 500 × 4200 × 12 = 25 200 000 J, which is 2.5 × 107 J. 2.1 × 106 J leaves out the temperature change. 5.0 × 104 J leaves out the mass. 6.0 × 103 J leaves out the specific heat capacity. | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the equation, substitute all three quantities, and check that none of the data has been left unused.
A hot-water bottle contains 1.2 kg of water at 65 °C. During the night the water cools to 35 °C. The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy transferred from the water as it cools.
The water takes 6.0 hours to cool. Using the unrounded value from (a), calculate the average rate at which energy is transferred.
Suggest why water is a good liquid to use in a hot-water bottle.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 ΔQ = m c Δθ | ✓ 1 | |
| M1 ΔQ = 1.2 × 4200 × 30 | ✓ 1 | |
| A1 151 200 J | ✓ 1 | |
| Part (b) | ||
| M1 P = E / t = 151 200 / (6.0 × 3600) | ✓ 1 | |
| A1 P = 7.0 W | ✓ 1 | |
| Part (c) | ||
| B1 water has a high specific heat capacity, so it stores a lot of energy for each kilogram per degree of temperature change | ✓ 1 | |
Exam tip: Convert hours to seconds (× 3600) before dividing energy in joules by time to obtain watts.
A sneeze releases a spray of tiny liquid droplets into the air in a ward. A hospital physicist uses the particle model to describe the droplets and the air around them.
Describe the arrangement and movement of the molecules in a liquid and in a gas.
Explain why a gas is much easier to compress than a liquid.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 liquid molecules are close together | ✓ 1 | |
| B1 liquid molecules move randomly and slide past each other | ✓ 1 | |
| B1 gas molecules are far apart | ✓ 1 | |
| B1 gas molecules move rapidly in random directions | ✓ 1 | |
| Part (b) | ||
| B1 gas molecules are far apart, so there is a lot of empty space to move into | ✓ 1 | |
| B1 liquid molecules are already touching, so they cannot be pushed much closer | ✓ 1 | |
Exam tip: Describe both the spacing and the motion of particles when a question asks about the arrangement and movement.
A solar water heater at a community building heats 120 kg of water from 15 °C to 45 °C. The specific heat capacity of water is 4200 J/kg °C. The solar panels transfer energy to the water at an average rate of 3000 W.
Calculate the energy needed to heat the water.
Calculate the time needed, in minutes, if all the energy from the panels is transferred to the water. Use the unrounded value from (a).
In practice the water takes longer to heat. Suggest one reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 ΔQ = m c Δθ | ✓ 1 | |
| M1 ΔQ = 120 × 4200 × 30 | ✓ 1 | |
| A1 15 120 000 J | ✓ 1 | |
| Part (b) | ||
| M1 t = E / P = 15 120 000 / 3000 | ✓ 1 | |
| M1 t = 5040 s and divide by 60 | ✓ 1 | |
| A1 84 minutes | ✓ 1 | |
| Part (c) | ||
| B1 some energy is lost to the surroundings or not all the energy reaching the panels is absorbed by the water | ✓ 1 | |
Exam tip: Write the time in the unit the question asks for; dividing by 60 turns seconds into minutes.
A student heats a 1.0 kg block of metal, for use in a thermal store, with a 50 W electric heater for 6.0 minutes. The temperature of the block rises from 20.0 °C to 38.0 °C.
Calculate the electrical energy supplied to the heater.
Use this energy to calculate the specific heat capacity of the metal.
The true specific heat capacity of the metal is lower than the student's value. Explain how this could happen.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 E = P t = 50 × 6.0 × 60 | ✓ 1 | |
| A1 E = 18 000 J | ✓ 1 | |
| Part (b) | ||
| M1 c = E / (m Δθ) = 18 000 / (1.0 × 18.0) | ✓ 1 | |
| A1 c = 1000 J/kg °C | ✓ 1 | |
| Part (c) | ||
| B1 some of the supplied energy is lost to the surroundings | ✓ 1 | |
| B1 so less energy than 18 000 J raised the block's temperature, and using the full value makes c appear too large | ✓ 1 | |
Common error: Time must be in seconds in E = P t; 6.0 minutes is 360 s.
Two uniform solid columns are made of the same marble and stand upright on level ground. Column X is 2.0 m tall with a base area of 0.50 m2. Column Y is 4.0 m tall with a base area of 0.25 m2. What is the ratio (pressure under X) : (pressure under Y)?
| Marking point | Mark | Notes |
|---|---|---|
| Each column has volume 1.0 m3 (2.0 × 0.50 and 4.0 × 0.25), so the weights are equal. Y has half the base area, so its pressure is double that of X, giving 1 : 2. This also follows from p = hρg, since Y is twice as tall. 1 : 1 forgets that the areas differ. 1 : 4 applies both the height factor and the area factor, counting the same effect twice. 2 : 1 turns the ratio upside down. | ✓ 1 | Answer B |
Answer: B
Exam tip: For a uniform upright column, pressure at the base is hρg, so only the height and the density matter.
An electric kettle of power 800 W heats 1.2 kg of water from 15 °C to 75 °C to fill a hot-water bottle. Assume all the electrical energy heats the water. The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy needed to heat the water.
Calculate the time taken. Use the unrounded answer to (a).
Suggest why the real time taken is longer than your answer to (b).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| temperature change = 75 − 15 = 60 °C | ✓ 1 | |
| substitution: 1.2 × 4200 × 60 | ✓ 1 | |
| 302400 J | ✓ 1 | |
| Part (b) | ||
| time = energy ÷ power = 302400 ÷ 800 | ✓ 1 | |
| 378 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| some energy heats the kettle itself or is transferred to the surroundings, so more energy must be supplied | ✓ 1 | |
Common error: Energy in joules divided by power in watts gives seconds; do not convert to minutes before dividing.
Liquid salol is left to cool in a test tube and its temperature is recorded. The temperature stays constant at about 42 °C while the salol solidifies, even though it is still transferring energy to the surroundings. Which statement explains this?
| Marking point | Mark | Notes |
|---|---|---|
| While the salol solidifies the molecules form fixed positions, so their potential energy falls and this supplies the energy transferred to the surroundings. Temperature depends on average kinetic energy, which does not change, so the reading is constant. Saying the molecules stop moving is wrong because the particles in a solid still vibrate and the temperature is still well above absolute zero. Kinetic energy falling with potential energy rising would make the temperature drop, not stay constant. The thermometer reads the temperature of the whole mixture of solid and liquid, so a limitation of the thermometer is not the cause. | ✓ 1 | Answer A |
Answer: A
Common error: Do not say that the molecules 'stop moving' or that no energy is lost during a change of state. Energy is still transferred out; it comes from potential energy, not kinetic energy.
A fisheries research submersible descends to a depth of 30 m in sea water of density 1030 kg/m3. The gravitational field strength is 10 N/kg. What is the pressure due to the water alone at this depth?
| Marking point | Mark | Notes |
|---|---|---|
| Pressure due to a liquid = depth × density × g = 30 × 1030 × 10 = 309 000 Pa = 309 kPa. 31 kPa leaves out the factor of g. 410 kPa adds atmospheric pressure (about 100 kPa), which gives the total pressure, not the pressure due to the water alone. 10.3 kPa is density × g with no depth, which would be the pressure at a depth of 1 m. | ✓ 1 | Answer B |
Answer: B
Exam tip: Read whether the question asks for the pressure due to the liquid or the total pressure. The total includes atmospheric pressure on top of h × ρ × g.
A student in a fashion-technology workshop wants to find the density of a small rectangular metal block that will be used as a weight in a garment hem.
Describe how the student can find the density of the block using a ruler and a balance.
The block measures 5.20 cm × 3.10 cm × 1.80 cm and has a mass of 78.3 g. Calculate the density of the block in g/cm3. Give your answer to 3 significant figures.
Suggest one way the student can reduce the uncertainty in the value of the volume.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Measure the length, width and height of the block with a ruler (repeating and averaging if needed) | ✓ 1 | |
| Multiply length × width × height to find the volume | ✓ 1 | |
| Measure the mass of the block using a balance | ✓ 1 | |
| Calculate the density using density = mass ÷ volume | ✓ 1 | |
| Part (b) | ||
| M1 volume = 5.20 × 3.10 × 1.80 = 29.0 cm3 (29.016) | ✓ 1 | |
| M1 density = 78.3 ÷ volume | ✓ 1 | |
| A1 density = 2.70 g/cm3 | ✓ 1 | |
| Part (c) | ||
| Measure each dimension at several places along the block and use the mean, or use a measuring instrument with a smaller scale division such as vernier calipers | ✓ 1 | |
Exam tip: When a calculation needs several lengths multiplied together, write the unrounded volume on your calculator and round only the final density.
Salol is a solid that melts at a low temperature. A student heated salol until it was a clear liquid at 70 °C and then let it cool in air while recording its temperature every 2 minutes.
Time (min): 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Temperature (°C): 70, 61, 53, 47, 43, 42, 42, 42, 42, 38, 33
State the melting point of salol from the data.
Determine the length of time during which the temperature stayed constant.
Explain, in terms of the molecules, why the temperature stays constant during this time even though the salol is still transferring energy to the surroundings.
Calculate the mean rate of cooling, in °C per minute, over the first 6 minutes. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 42 °C | ✓ 1 | |
| Part (b) | ||
| M1 identifies the constant readings from 10 min to 16 min | ✓ 1 | |
| A1 6 minutes | ✓ 1 | |
| Part (c) | ||
| The salol is changing from liquid to solid (molecules form fixed positions / bonds) | ✓ 1 | |
| The potential energy of the molecules decreases and this supplies the energy transferred out | ✓ 1 | |
| The average kinetic energy of the molecules does not change, so the temperature stays constant | ✓ 1 | |
| Part (d) | ||
| M1 temperature fall = 70 − 47 = 23 °C, divided by 6 min | ✓ 1 | |
| A1 3.83 °C per minute | ✓ 1 | |
Common error: A flat section on a cooling curve is not 'no energy transfer'. Energy is still lost, but it comes from the potential energy of the molecules, not their kinetic energy.
A fish hatchery heats the water in a tank using an electric heater of power 1.4 kW. The tank contains 200 kg of water. The specific heat capacity of water is 4200 J/kg °C. Assume all the energy from the heater warms the water.
Calculate the energy needed to raise the temperature of the water by 3.0 °C.
Calculate the time the heater takes to supply this energy.
In practice, the heater takes longer than this. Suggest why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = mass × specific heat capacity × temperature change | ✓ 1 | |
| M1 200 × 4200 × 3.0 | ✓ 1 | |
| A1 2 520 000 J (2.52 × 106 J) | ✓ 1 | |
| Part (b) | ||
| M1 time = energy ÷ power = 2 520 000 ÷ 1400 (ECF from (a)) | ✓ 1 | |
| A1 1800 s | ✓ 1 | |
| Part (c) | ||
| Some energy is transferred to the surroundings (the air, the tank walls) | ✓ 1 | |
Exam tip: Convert power to watts (1.4 kW = 1400 W) before using time = energy ÷ power, so that the answer comes out in seconds.
A fixed mass of gas is trapped in a cylinder by a piston. The piston is pushed in slowly so that the volume halves while the temperature stays constant. Which statement explains why the pressure of the gas increases?
| Marking point | Mark | Notes |
|---|---|---|
| With the same number of molecules in half the volume, each unit area of wall is hit twice as often, so the average force per unit area (pressure) doubles. Molecule speed depends on temperature, which is constant, so the first option is wrong. Molecules do not change size. Collisions between molecules do not produce pressure on the walls. | ✓ 1 | Answer D |
Answer: D
Common error: Pressure comes from molecules hitting the container walls, not from molecules hitting each other.
A liquid of mass 2.0 kg is heated. An energy transfer of 8400 J raises its temperature by 2.0 °C. What is the specific heat capacity of the liquid?
| Marking point | Mark | Notes |
|---|---|---|
| c = E ÷ (m × ΔT) = 8400 ÷ (2.0 × 2.0) = 2100 J/kg °C. 4200 divides by only one of mass or temperature change. 16800 multiplies the energy by one of them. 33600 multiplies the energy by both. | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the equation and substitute all three quantities before you calculate.
A copper block of mass 0.50 kg is heated by an electric heater rated at 100 W. The heater is switched on for 90 s and the temperature of the block rises from 18 °C to 48 °C. The specific heat capacity of copper is 390 J/kg °C.
Calculate the energy gained by the block.
Calculate the energy supplied by the heater.
Calculate the percentage of the energy supplied that was transferred to the block.
Suggest why the block gained less energy than the heater supplied.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| temperature rise = 30 °C | ✓ 1 | |
| E = mcΔT = 0.50 × 390 × 30 | ✓ 1 | |
| 5850 J | ✓ 1 | |
| Part (b) | ||
| E = Pt | ✓ 1 | |
| 100 × 90 = 9000 J | ✓ 1 | |
| Part (c) | ||
| efficiency = useful energy ÷ total energy | ✓ 1 | |
| 5850 ÷ 9000 × 100 = 65% (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| energy was transferred (by heating) to the surroundings / to the heater itself | ✓ 1 | |
Common error: The temperature rise is final minus initial temperature, not the final temperature.
A solid is heated at a constant rate. When it is melting, its temperature stays constant even though energy is still being supplied. What explains this?
| Marking point | Mark | Notes |
|---|---|---|
| During melting, the energy supplied increases the potential energy of the particles as the bonds between them are broken. The average kinetic energy, and so the temperature, stays the same. The solid does continue to absorb energy. Particles never stop moving above absolute zero. Energy loss to the surroundings would not explain a steady temperature if the heater continues to supply energy at a constant rate. | ✓ 1 | Answer D |
Answer: D
Exam tip: State that the temperature is constant because the energy changes the arrangement of the particles, not their speed.
Wind with a density of 1.2 kg/m3 blows at 8.0 m/s at right angles through a circular area of 50 m2. What mass of air passes through the area each second?
| Marking point | Mark | Notes |
|---|---|---|
| The volume of air passing each second is 50 × 8.0 = 400 m3, so the mass is 1.2 × 400 = 480 kg. 60 kg multiplies density by area only. 400 kg is the volume, leaving out the density. 330 kg divides the volume by the density instead of multiplying. | ✓ 1 | Answer B |
Answer: B
Common error: Mass = density × volume; do not stop at the volume.
A solar panel supplies 60 W to an electric heater inside a copper block of mass 1.5 kg. The specific heat capacity of copper is 400 J/kg °C. The block starts at 22 °C and the heater is on for 5.0 minutes. Assume all the energy from the heater is transferred to the block.
Calculate the energy supplied to the block.
Calculate the final temperature of the block.
On a cloudy day the panel supplies only 40 W. Calculate the time needed to supply the same energy to the block.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| t = 300 s | ✓ 1 | |
| E = Pt = 60 × 300 = 18000 J | ✓ 1 | |
| Part (b) | ||
| ΔT = E ÷ (mc) | ✓ 1 | |
| 18000 ÷ (1.5 × 400) = 30 °C | ✓ 1 | |
| final temperature = 52 °C (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| t = E ÷ P | ✓ 1 | |
| 18000 ÷ 40 = 450 s (ECF from (a)) | ✓ 1 | |
Common error: Convert minutes to seconds before using E = Pt.
A supermarket sells olive oil in bottles of volume 2.0 litres (1 litre = 1.0 × 10−3 m3). The density of the oil is 900 kg/m3. Twelve bottles are packed in a crate of mass 2.4 kg. Ignore the mass of the bottles themselves. The base of the crate measures 0.30 m by 0.40 m. Use g = 10 N/kg.
Calculate the mass of oil in one bottle.
Calculate the weight of the full crate.
Calculate the pressure the crate exerts on the shelf it stands on.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = 2.0 × 10−3 m3 | ✓ 1 | |
| m = ρV | ✓ 1 | |
| 900 × 2.0 × 10−3 = 1.8 kg | ✓ 1 | |
| Part (b) | ||
| total mass = 12 × 1.8 + 2.4 = 24.0 kg | ✓ 1 | |
| weight = 24.0 × 10 = 240 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| area = 0.30 × 0.40 = 0.12 m2 | ✓ 1 | |
| p = F ÷ A | ✓ 1 | |
| 240 ÷ 0.12 = 2000 Pa (ECF from (b)) | ✓ 1 | |
Common error: Remember to add the mass of the crate to the mass of its contents.
Hydrogen made using electricity from a wind farm is stored in a tank of volume 0.50 m3 at a pressure of 2.4 × 106 Pa. The temperature of the gas is constant.
State one condition, other than constant temperature, needed for pV = constant to apply to the hydrogen.
Calculate the volume the same hydrogen would occupy at an atmospheric pressure of 1.0 × 105 Pa.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the mass (amount) of gas stays the same / no leaks | ✓ 1 | |
| Part (b) | ||
| p1V1 = p2V2 | ✓ 1 | |
| 2.4 × 106 × 0.50 = 1.0 × 105 × V2 | ✓ 1 | |
| 12 m3 | ✓ 1 | |
Exam tip: Use the same pressure unit on both sides of p₁V₁ = p₂V₂.
A jeweller checks a ring of mass 9.6 g. A measuring cylinder holds 20.0 cm3 of water. When the ring is fully submerged, the reading rises to 20.5 cm3. Pure gold has a density of 19.3 g/cm3.
Calculate the volume of the ring.
Calculate the density of the ring in g/cm3.
Write your answer to (b) in kg/m3.
Each reading on the measuring cylinder is uncertain by ±0.1 cm3. Explain why this method cannot show reliably that the ring is pure gold.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 20.5 − 20.0 = 0.5 cm3 | ✓ 1 | |
| Part (b) | ||
| ρ = m ÷ V | ✓ 1 | |
| 9.6 ÷ 0.5 = 19.2 g/cm3 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 19200 kg/m3 | ✓ 1 | |
| Part (d) | ||
| uncertainty in the volume of the ring could be up to ±0.2 cm3, a large fraction of 0.5 cm3 | ✓ 1 | |
| so the density could lie between about 14 and 32 g/cm3, too wide to compare 19.2 with 19.3 | ✓ 1 | |
Exam tip: Volume by displacement is the difference between two readings, so both uncertainties add.
A woman of mass 54 kg stands on one foot with all her weight on the heel of a shoe. The pressure under the heel is 1.2 × 106 Pa. The gravitational field strength g = 10 N/kg.
Calculate the weight of the woman.
Calculate the area of the heel in m2.
Give your answer to (b) in cm2.
The woman changes into flat shoes. With all her weight on one foot, the area of the sole in contact with the floor is 180 cm2. Calculate the pressure under the sole in Pa.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 weight = mass × g = 54 × 10 | ✓ 1 | |
| A1 540 N | ✓ 1 | |
| Part (b) | ||
| M1 area = force ÷ pressure | ✓ 1 | |
| M1 540 ÷ 1.2 × 106 | ✓ 1 | |
| A1 4.5 × 10−4 m2 (0.00045 m2) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| A1 4.5 cm2 (since 1 m2 = 10 000 cm2) (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 area = 180 ÷ 10 000 = 0.018 m2 | ✓ 1 | |
| M1 pressure = force ÷ area = 540 ÷ 0.018 | ✓ 1 | |
| A1 30 000 Pa (3.0 × 104 Pa) | ✓ 1 | |
Common error: Convert areas from cm² to m² by dividing by 10 000, not by 100, before using pressure = force ÷ area in pascals.
A student heats a sample of solid wax with a heater of constant power and records its temperature over time.
Time / min: 0, 2, 4, 5, 7, 9, 11, 13
Temperature / °C: 30, 42, 54, 60, 60, 60, 68, 76
State the melting point of the wax.
Explain, in terms of particles, why the temperature stays constant while the wax is melting.
Describe the arrangement and the motion of the particles in liquid wax.
Calculate the rate of temperature rise of the liquid wax between 11 min and 13 min.
For the solid wax between 0 min and 4 min the temperature rose at 6.0 °C/min. Suggest one reason why the liquid wax warms more slowly.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 60 °C | ✓ 1 | |
| Part (b) | ||
| The energy supplied is used to weaken or break the bonds (overcome the forces) between particles | ✓ 1 | |
| It does not increase the average kinetic energy of the particles, so the temperature does not rise | ✓ 1 | |
| Part (c) | ||
| Particles are close together (touching) in an irregular arrangement | ✓ 1 | |
| Particles move around and slide past each other | ✓ 1 | |
| Part (d) | ||
| M1 rate = change in temperature ÷ time = (76 − 68) ÷ (13 − 11) | ✓ 1 | |
| A1 4.0 °C/min | ✓ 1 | |
| Part (e) | ||
| Liquid wax has a larger specific heat capacity, or more energy is transferred to the surroundings at the higher temperature | ✓ 1 | |
Common error: During a change of state the energy supplied changes the arrangement of the particles, not their average kinetic energy, so the temperature stays constant.
A pharmacist measures the density of a liquid medicine. An empty measuring cylinder has a mass of 112.0 g. After 50.0 cm3 of the medicine is poured in, the cylinder and medicine have a mass of 173.0 g.
Calculate the mass of the medicine.
Calculate the density of the medicine in g/cm3.
Give the density in kg/m3. (1 g/cm3 = 1000 kg/m3)
Describe how the pharmacist should read the volume from the measuring cylinder scale to be accurate.
One dose of the medicine is 15 cm3. Calculate the mass of one dose in g.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 173.0 − 112.0 = 61.0 g | ✓ 1 | |
| Part (b) | ||
| M1 density = mass ÷ volume = 61.0 ÷ 50.0 | ✓ 1 | |
| A1 1.22 g/cm3 (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| A1 1220 kg/m3 (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| Place the cylinder on a level surface and read the scale with the eye level with the bottom of the meniscus | ✓ 1 | |
| Part (e) | ||
| M1 mass = density × volume = 1.22 × 15 | ✓ 1 | |
| A1 18.3 g (ECF from (b)) | ✓ 1 | |
Exam tip: When a container is on the balance, subtract the mass of the empty container to find the mass of the liquid alone.
A gaming chair contains a sealed air cushion. The air in the cushion is at a constant temperature.
Explain, in terms of air molecules, how the air exerts a pressure on the walls of the cushion.
When a player sits on the chair, the air in the cushion is compressed. Explain why its pressure increases.
State the temperature of absolute zero in °C.
The air in the cushion warms to 87 °C. Convert this to kelvin.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Molecules move randomly and collide with the walls | ✓ 1 | |
| Each collision exerts a force on the wall, and the total force on each unit of area is the pressure | ✓ 1 | |
| Part (b) | ||
| The same molecules are in a smaller volume, so they hit the walls more often (more collisions each second on each unit of area) | ✓ 1 | |
| The average force on each unit of area is greater, so the pressure is greater | ✓ 1 | |
| Part (c) | ||
| A1 −273 °C | ✓ 1 | |
| Part (d) | ||
| A1 360 K | ✓ 1 | |
Exam tip: When explaining pressure changes, link the cause to the number and force of collisions with the walls in each second.
A museum in a shopping arcade displays a statue on a solid marble column. The column is a cuboid with a square base of side 0.40 m and a height of 2.5 m. The density of marble is 2700 kg/m3. The gravitational field strength is 10 N/kg. Do not round any intermediate value.
Calculate the volume of the column.
Calculate the mass of the column.
Calculate the pressure that the column exerts on the floor. Ignore the statue.
A second column is made of the same marble with the same height but a square base of side 0.80 m. Explain how the pressure it exerts on the floor compares with the first column.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 volume = 0.40 × 0.40 × 2.5 | ✓ 1 | |
| A1 0.40 m3 | ✓ 1 | |
| Part (b) | ||
| M1 mass = density × volume | ✓ 1 | |
| A1 2700 × 0.40 = 1080 kg | ✓ 1 | |
| Part (c) | ||
| M1 weight = 1080 × 10 = 10 800 N | ✓ 1 | |
| M1 pressure = force ÷ area with area = 0.16 m2 | ✓ 1 | |
| A1 10 800 ÷ 0.16 = 67 500 Pa (6.75 × 104 Pa) | ✓ 1 | |
| Part (d) | ||
| the weight is four times greater (volume is four times greater) | ✓ 1 | |
| the area is also four times greater, so the pressure is the same | ✓ 1 | |
Exam tip: Find the area of the base separately from the volume; the volume uses the height but the pressure uses the area in contact with the floor.
A solar water-heating system warms 100 kg of water from 18°C to 48°C. The specific heat capacity of water is 4200 J/kg°C.
Calculate the energy transferred to the water.
The collectors transfer energy to the water at an average rate of 1500 W. Calculate the time needed to warm the water. Give your answer in seconds.
Suggest why the time taken in practice is longer than the time calculated.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 change in thermal energy = mass × specific heat capacity × change in temperature | ✓ 1 | |
| M1 100 × 4200 × 30 | ✓ 1 | |
| A1 1.26 × 107 J (12 600 000 J) | ✓ 1 | |
| Part (b) | ||
| M1 time = energy ÷ power | ✓ 1 | |
| A1 12 600 000 ÷ 1500 = 8400 s (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| some energy is transferred to the surroundings (lost from the pipes and tank) | ✓ 1 | |
Exam tip: The change in temperature is the difference (30°C), not the final temperature.
A sealed steel cylinder of gas at constant volume has a pressure of 150 kPa at 27 °C. It is heated to 127 °C. What is the new pressure?
| Marking point | Mark | Notes |
|---|---|---|
| At constant volume pressure is proportional to the temperature in kelvin. 27 °C = 300 K and 127 °C = 400 K, so the new pressure is 150 × 400 ÷ 300 = 200 kPa. 706 kPa comes from using the Celsius values in the ratio. 250 kPa comes from adding the 100 °C temperature rise to the pressure. 113 kPa comes from inverting the ratio of temperatures. | ✓ 1 | Answer B |
Answer: B
Common error: Always convert temperatures to kelvin (add 273) before using pressure–temperature relationships.
Frost on a field starts to melt in the morning sun. The temperature of the ice and water stays at 0 °C while it melts. Which statement explains this?
| Marking point | Mark | Notes |
|---|---|---|
| During melting the energy transferred breaks the bonds holding the particles in fixed positions, increasing their potential energy; the average kinetic energy, and so the temperature, stays constant. The idea that the particles speed up is wrong because that would raise the temperature. Ice does transfer energy to its particles, so the claim of no energy reaching it is wrong. The particles in water have greater potential energy than those in ice, so the last statement is false. | ✓ 1 | Answer A |
Answer: A
Common error: Temperature measures average kinetic energy of the particles; during a change of state the energy goes into potential energy instead.
The air in a sealed aircraft tyre is heated by braking. Which statement explains why the pressure of the air rises?
| Marking point | Mark | Notes |
|---|---|---|
| A higher temperature means the molecules have a greater average speed, so each collision with the wall exerts a bigger force and collisions are more frequent, increasing pressure. Molecules do not expand, and collisions with each other do not cause pressure. The number of molecules in a sealed tyre does not change. The mass of each molecule does not change. | ✓ 1 | Answer C |
Answer: C
Exam tip: Link pressure to collisions with the walls: say both how often and how hard the molecules collide.
A jeweller claims that a ring is made of pure gold. The ring has a mass of 24 g. The density of pure gold is 19.3 g/cm3. A student uses a displacement can and a measuring cylinder to find the volume of the ring.
Describe how the student finds the volume of the ring.
The student collects 1.6 cm3 of water. Calculate the density of the ring.
Evaluate the jeweller's claim. Take into account that the measuring cylinder reads to ±0.1 cm3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| fill the can with water until it overflows from the spout, and let it stop dripping | ✓ 1 | |
| lower the ring gently into the can and collect the displaced water in the measuring cylinder | ✓ 1 | |
| volume of water collected = volume of the ring | ✓ 1 | |
| Part (b) | ||
| M1 density = mass ÷ volume | ✓ 1 | |
| A1 15 g/cm3 | ✓ 1 | |
| Part (c) | ||
| density of the ring (15 g/cm3) is less than 19.3 g/cm3 | ✓ 1 | |
| volume range 1.5 to 1.7 cm3 gives densities of about 14.1 to 16.0 g/cm3, still below 19.3 | ✓ 1 | |
| so the ring is not pure gold (it may be an alloy or hollow) | ✓ 1 | |
Exam tip: When evaluating a claim, compare your calculated value with the claimed value and test whether the measurement uncertainty could close the gap.
On a farm, an immersion heater of power 2.0 kW heats 40 kg of water in a drinking trough from 8 °C to 18 °C. The specific heat capacity of water is 4200 J/kg °C. Assume all the energy from the heater goes into the water.
Calculate the energy needed to heat the water.
Calculate the time taken for the heater to supply this energy.
Suggest why the actual time taken would be longer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 temperature change = 10 °C | ✓ 1 | |
| M1 energy = m c Δθ | ✓ 1 | |
| A1 1.68 × 106 J | ✓ 1 | |
| Part (b) | ||
| M1 time = energy ÷ power | ✓ 1 | |
| A1 840 s , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| some energy is transferred to the surroundings / to the trough, so more energy must be supplied | ✓ 1 | |
Exam tip: Convert kilowatts to watts before using time = energy ÷ power with energy in joules.
A student heats a 0.50 kg aluminium block with a 60 W electric heater for 5.0 minutes. The temperature of the block rises from 20 °C to 60 °C.
Calculate the energy supplied by the heater.
Calculate the specific heat capacity of the aluminium from these results.
The accepted value for aluminium is lower than the student's result. Explain why the student's result is too high.
Suggest one improvement to the experiment.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = power × time with time = 300 s | ✓ 1 | |
| A1 18 000 J | ✓ 1 | |
| Part (b) | ||
| M1 temperature change = 40 °C | ✓ 1 | |
| M1 c = E ÷ (m Δθ) | ✓ 1 | |
| A1 900 J/kg °C , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| some energy is transferred to the surroundings | ✓ 1 | |
| so the block gains less energy than 18 000 J, but the calculation assumes all of it goes into the block | ✓ 1 | |
| Part (d) | ||
| insulate the block (e.g. wrap it in lagging) / put a lid on it | ✓ 1 | |
Common error: Convert minutes to seconds when using energy = power × time.
A fixed mass of gas in a sealed rigid container is heated. Which statement explains why the gas pressure rises?
| Marking point | Mark | Notes |
|---|---|---|
| Heating raises the average kinetic energy, so molecules move faster. They strike the walls more frequently and each collision exerts a larger force, so the pressure rises. The molecules themselves do not expand and their mass does not change. In a rigid container the volume is fixed, so the molecules are not pushed closer together. | ✓ 1 | Answer D |
Answer: D
Common error: Explain pressure changes using collisions with the container walls, not collisions between molecules.
The air in a car tyre is at an absolute pressure of 2.20 × 105 Pa when its temperature is 17 °C. After a long journey the temperature of the air is 47 °C. Assume the volume of the tyre and the amount of air in it stay constant.
Convert both temperatures to kelvin.
Calculate the pressure of the air after the journey. Give your answer to 3 significant figures.
Explain, in terms of molecules, why the pressure of the air rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 17 °C = 290 K | ✓ 1 | |
| A1 47 °C = 320 K | ✓ 1 | |
| Part (b) | ||
| M1 use of p1 ÷ T1 = p2 ÷ T2 with temperatures in kelvin | ✓ 1 | |
| M1 p2 = 2.20 × 105 × 320 ÷ 290 | ✓ 1 | |
| A1 2.43 × 105 Pa | ✓ 1 | |
| Part (c) | ||
| molecules gain kinetic energy / move faster | ✓ 1 | |
| molecules hit the tyre wall more often | ✓ 1 | |
| each collision exerts a greater force, so greater force per unit area | ✓ 1 | |
Common error: Using temperatures in °C in a pressure–temperature calculation gives a wrong answer; convert to kelvin first.
A supermarket sells olive oil in bottles that each hold 750 cm3. The oil in a full bottle has a mass of 0.69 kg.
Calculate the density of the oil in kg/m3.
The density of water is 1000 kg/m3. State what the oil floating on water shows about its density.
A customer pours 1.5 kg of this oil into a jug that holds 1.5 × 10−3 m3. Determine whether all the oil fits in the jug. Use the density 920 kg/m3.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 convert 750 cm3 to 7.5 × 10−4 m3 | ✓ 1 | |
| M1 density = mass ÷ volume | ✓ 1 | |
| A1 920 kg/m3 | ✓ 1 | |
| Part (b) | ||
| the oil is less dense than water | ✓ 1 | |
| Part (c) | ||
| M1 volume = mass ÷ density = 1.5 ÷ 920 | ✓ 1 | |
| A1 1.63 × 10−3 m3 | ✓ 1 | |
| A1 this is greater than 1.5 × 10−3 m3, so the oil does not all fit | ✓ 1 | |
Common error: Convert cm³ to m³ by dividing by 10⁶, not by 100 or 1000.
In a demonstration for a vaccination clinic, a nurse seals the nozzle of a syringe so that it traps 24 cm3 of air at a pressure of 1.00 × 105 Pa. The plunger is pushed in slowly until the trapped air has a volume of 15 cm3.
State two quantities that must stay constant for the pressure × volume of the air to stay constant.
Calculate the final pressure of the trapped air.
Explain, in terms of molecules, why the pressure increases as the volume decreases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 temperature | ✓ 1 | |
| B1 mass (or amount or number of molecules) of gas | ✓ 1 | |
| Part (b) | ||
| M1 p1V1 = p2V2 | ✓ 1 | |
| M1 p2 = 1.00 × 105 × 24 ÷ 15 | ✓ 1 | |
| A1 1.60 × 105 Pa | ✓ 1 | |
| Part (c) | ||
| the same number of molecules is in a smaller volume, so they hit the walls more often | ✓ 1 | |
| greater force per unit area, so greater pressure | ✓ 1 | |
Exam tip: Cm³ can be used for both volumes in p₁V₁ = p₂V₂ because the units cancel.
A sealed rigid steel flask contains a fixed mass of gas at 27 °C. The flask is warmed to 87 °C. By what factor does the gas pressure increase? Use 0 °C = 273 K.
| Marking point | Mark | Notes |
|---|---|---|
| Volume and mass are constant, so pressure is proportional to kelvin temperature: 360 K ÷ 300 K = 1.2. The value 0.83 inverts the ratio. A factor of 1.0 wrongly assumes a fixed volume means a fixed pressure. 3.2 comes from dividing the Celsius temperatures 87 ÷ 27. | ✓ 1 | Answer D |
Answer: D
Common error: Pressure is proportional to temperature in kelvin, never in degrees Celsius.
A student measures the pressure of a fixed mass of gas in a rigid container at several temperatures in °C. The graph of pressure (vertical) against temperature in °C is a straight line with a positive gradient. The line is extended back to zero pressure. At what temperature does it meet the temperature axis?
| Marking point | Mark | Notes |
|---|---|---|
| Pressure is proportional to the kelvin temperature, so it would reach zero at 0 K, which is −273 °C. 0 °C is where the line crosses the pressure axis at a non-zero pressure. −100 °C and 273 °C are not related to absolute zero; 273 °C has the wrong sign. | ✓ 1 | Answer B |
Answer: B
Exam tip: Absolute zero is 0 K, which is −273 °C; the straight line is only extended back to it, since real gases liquefy first.
A sealed rigid steel canister contains a fixed mass of gas. At base camp the temperature is 17 °C and the gas pressure is 232 kPa. At the summit the temperature is −23 °C. Use 0 °C = 273 K.
Explain, in terms of gas molecules, why the pressure falls as the canister cools.
Calculate the pressure of the gas at the summit.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The molecules move more slowly / have less kinetic energy on average | ✓ 1 | |
| They hit the walls less often | ✓ 1 | |
| and with less force per collision, so the force per unit area is smaller | ✓ 1 | |
| Part (b) | ||
| M1 converts to kelvin: 290 K and 250 K | ✓ 1 | |
| M1 uses p1/T1 = p2/T2 with kelvin temperatures, e.g. 232 × 250 ÷ 290 | ✓ 1 | |
| A1 200 kPa | ✓ 1 | |
Exam tip: Always write the two kelvin temperatures before substituting into the gas law; it earns a method mark even if the arithmetic goes wrong.
A sealed pressure cooker contains air at 27 °C and a pressure of 100 kPa. Ignore any steam. The volume of the air stays constant. Use 0 °C = 273 K.
The cooker is heated until the air reaches 127 °C. Calculate the new pressure of the air. Give your answer to 3 significant figures.
A safety valve opens when the pressure of the air reaches 140 kPa. Calculate the temperature in °C at which the valve opens.
Suggest why a pressure cooker needs a safety valve.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 converts to kelvin: 300 K and 400 K | ✓ 1 | |
| M1 p2 = p1T2/T1 = 100 × 400 ÷ 300 | ✓ 1 | |
| A1 133 kPa (133.3 kPa before rounding) | ✓ 1 | |
| Part (b) | ||
| M1 T2 = T1p2/p1 = 300 × 140 ÷ 100 | ✓ 1 | |
| A1 420 K | ✓ 1 | |
| A1 147 °C | ✓ 1 | |
| Part (c) | ||
| To release gas/steam if the pressure becomes too high, so the cooker does not burst/explode | ✓ 1 | |
Exam tip: In a final answer to a temperature question, check whether the question asks for kelvin or °C.
A 0.040 kg sample of solid gallium at 20 °C is placed in warm water. The melting point of gallium is 30 °C. The specific heat capacity of solid gallium is 370 J/kg °C and its specific latent heat of fusion is 8.0 × 104 J/kg.
Calculate the energy needed to raise the temperature of the sample from 20 °C to 30 °C.
Explain why the temperature of the gallium stays at 30 °C while it melts although it is still being heated.
Calculate the energy needed to melt all the sample at 30 °C, and hence the total energy needed to melt the sample starting at 20 °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 ΔQ = mcΔθ | ✓ 1 | |
| M1 Δθ = 10 °C, e.g. 0.040 × 370 × 10 | ✓ 1 | |
| A1 148 J | ✓ 1 | |
| Part (b) | ||
| The energy supplied is used to break the bonds/overcome the forces between particles, not to increase their kinetic energy | ✓ 1 | |
| Part (c) | ||
| M1 Q = mL = 0.040 × 8.0 × 104 | ✓ 1 | |
| A1 3200 J to melt | ✓ 1 | |
| A1 total 3348 J (ECF from (a)) | ✓ 1 | |
Common error: The latent heat energy has no temperature change in it: use Q = mL, not mcΔθ, while the substance changes state.
A food scientist tests a jar of honey. The jar holds 350 cm3 of honey of mass 497 g. Use g = 10 N/kg.
Calculate the density of the honey in kg/m3.
The honey is 0.10 m deep. Calculate the pressure due to the honey at the base of the jar.
A second jar holds the same honey to a depth of 0.20 m. State how the pressure due to the honey at its base compares with that in the first jar.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 converts units: 0.497 kg and 3.50 × 10−4 m3 | ✓ 1 | |
| M1 ρ = m/V | ✓ 1 | |
| A1 1420 kg/m3 | ✓ 1 | |
| Part (b) | ||
| M1 p = hρg | ✓ 1 | |
| M1 0.10 × 1420 × 10 | ✓ 1 | |
| A1 1420 Pa (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| It is twice as large | ✓ 1 | |
Exam tip: Convert cm³ to m³ by dividing by 10⁶ before using density in kg/m³.
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated points about molecular motion or the graph; little or no link between temperature, collisions and pressure. |
| Level 2 (3–4) | Explains how increased temperature increases pressure via molecular speed and collisions, OR links the graph intercept to absolute zero with some molecular reasoning; reasoning partly linked. |
| Level 3 (5–6) | Coherent, linked explanation covering both the molecular cause of the pressure increase (speed, frequency and force of collisions) and how extrapolation to zero pressure at −273 °C (0 K) shows absolute zero. |
| Level 0 (0) | No rewardable content. |
| Level 1 (1–2) | Isolated statements about molecules hitting the walls, or about pressure inside or outside the bottle; little attempt to link temperature and pressure changes. |
| Level 2 (3–4) | Explains how either the temperature rise affects the pressure inside (via molecular motion) or how the outside pressure change acts on the bottle; some linking of ideas but the comparison of inside and outside pressure is incomplete. |
| Level 3 (5–6) | Coherent, linked explanation of both effects: the warming raises the inside pressure only a little (molecular reasoning, Kelvin temperatures or calculation) while the outside pressure rises more, so there is a resultant inward force that squashes the bottle. |
A 'pot-in-pot' cooler can keep vegetables fresh without electricity. It is made of a small clay pot placed inside a larger clay pot. The gap between them is filled with wet sand. Water slowly soaks through the porous outer pot and evaporates from its outside surface.
Explain, in terms of particles, how the cooler keeps the vegetables cool, and why it works better on a hot, dry, windy day than on a cool, humid, still day.
Explain how the cooler works and why it works better in hot, dry, windy conditions.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Water particles in the liquid have a range of kinetic energies | ✓ 1 | |
| The particles with the most energy near the surface escape / evaporate | ✓ 1 | |
| The average kinetic energy of the remaining water falls, so its temperature falls | ✓ 1 | |
| Energy is transferred from the sand, inner pot and vegetables to the cooler water | ✓ 1 | |
| Higher air temperature means more particles have enough energy to escape, so the evaporation rate is higher | ✓ 1 | |
| Dry, moving air carries escaped water particles away, so fewer return to the liquid and evaporation is faster than in humid, still air | ✓ 1 | |
Exam tip: For 6-mark explanations, build a logical chain from particle energies to temperature drop to energy transfer, then address every condition named in the question.
An arm for a delivery drone is 3D-printed from plastic with a partly hollow internal structure. The arm has an outside volume of 4.0 ×10−5 m3 and a mass of 0.030 kg. Solid plastic of this type has a density of 1250 kg/m3. The density of water is 1000 kg/m3. Assume the mass of the air inside the arm is negligible.
Calculate the average density of the drone arm.
The drone crashes into a lake. The arm is sealed, so water cannot get inside. Explain whether the arm will float.
Calculate the percentage of the arm's outside volume that is occupied by solid plastic.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ρ = m ÷ V = 0.030 ÷ 4.0 ×10−5 | ✓ 1 | |
| ρ = 750 kg/m3 | ✓ 1 | |
| Part (b) | ||
| the average density of the arm (750 kg/m3) is less than the density of water (1000 kg/m3) | ✓ 1 | |
| so the arm floats | ✓ 1 | |
| Part (c) | ||
| volume of plastic = m ÷ ρ = 0.030 ÷ 1250 = 2.4 ×10−5 m3 | ✓ 1 | |
| fraction = 2.4 ×10−5 ÷ 4.0 ×10−5 | ✓ 1 | |
| = 60 % | ✓ 1 | |
Exam tip: For a hollow object, use the total mass and the total outside volume to find the average density. Use the density of the material to find how much of that volume is actually solid.
An oil–water separator at a factory is an open tank. It contains a layer of oil 0.40 m deep floating on a layer of water 1.20 m deep.
Density of oil = 850 kg/m3
Density of water = 1000 kg/m3
Atmospheric pressure = 1.01 × 105 Pa
(g = 10 N/kg)
Calculate the pressure due to the oil at the boundary between the oil and the water.
Calculate the total pressure at the bottom of the tank.
A drain plug of area 2.0 × 10−3 m2 is fitted in the bottom of the tank. The outside of the plug is open to the atmosphere.
Calculate the resultant force on the plug.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = 0.40 × 850 × 10 | ✓ 1 | |
| p = 3400 Pa | ✓ 1 | |
| Part (b) | ||
| pressure due to water = 1.20 × 1000 × 10 = 12 000 Pa | ✓ 1 | |
| total = 1.01 × 105 + 3400 + 12 000 | ✓ 1 | |
| total = 116 400 Pa ≈ 1.16 × 105 Pa | ✓ 1 | |
| Part (c) | ||
| atmospheric pressure acts on both sides, so the pressure difference = 3400 + 12 000 = 15 400 Pa | ✓ 1 | |
| F = 15 400 × 2.0 × 10−3 = 30.8 N ≈ 31 N | ✓ 1 | |
Common error: For layered liquids, work out hρg for each layer with its own density and add the results. Do not use the total depth with just one density.
A swimmer uses a snorkel, a tube that lets her breathe air from above the surface while her face is under water. Air in her lungs is at atmospheric pressure.
Density of seawater = 1030 kg/m3
(g = 10 N/kg)
Explain why the pressure of the water on the swimmer's chest increases as she goes deeper.
Her chest muscles can breathe in against a maximum pressure difference of 4.0 kPa.
Calculate the greatest depth of her chest below the surface at which she can still breathe through the snorkel.
Explain why using a much longer snorkel would not allow her to breathe at a greater depth.
A scuba diver breathes from a tank of compressed air at a depth of 18 m.
Atmospheric pressure = 1.0 × 105 Pa
Calculate the total pressure on the diver.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| there is a greater depth / taller column of water above her chest | ✓ 1 | |
| so a greater weight of water acts on each unit area of her chest / pressure = hρg increases with h | ✓ 1 | |
| Part (b) | ||
| h = p ÷ (ρ × g) | ✓ 1 | |
| h = 4000 ÷ (1030 × 10) | ✓ 1 | |
| h = 0.39 m | ✓ 1 | |
| Part (c) | ||
| the snorkel, whatever its length, still supplies air at atmospheric pressure to her lungs, while the water pressure on her chest increases with depth | ✓ 1 | |
| at depths greater than about 0.39 m the pressure difference is more than 4.0 kPa, so her muscles cannot expand her chest | ✓ 1 | |
| Part (d) | ||
| pressure due to water = 18 × 1030 × 10 = 185 400 Pa | ✓ 1 | |
| total = 185 400 + 1.0 × 105 = 2.9 × 105 Pa | ✓ 1 | |
Exam tip: 'Total pressure' below a liquid surface means atmospheric pressure plus hρg. 'Pressure due to the water' means hρg alone.
A rigid tank on a Mars rover stores nitrogen. The tank has a volume of 0.050 m3. At night the nitrogen is at 290 K and its pressure is 2.4 MPa. During the day, sunlight raises the pressure of the nitrogen to 2.7 MPa.
Calculate the daytime temperature of the nitrogen in °C.
In the daytime, all the nitrogen is released into an inflatable shelter. The nitrogen expands at the daytime temperature from (a), which stays constant, until its pressure is 15 kPa.
Calculate the volume of the nitrogen in the shelter.
At night, the nitrogen in the shelter cools from the daytime temperature in (a) to 220 K. The volume of the shelter stays the same.
Calculate the new pressure of the nitrogen.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p1/T1 = p2/T2, so T2 = 290 × 2.7 ÷ 2.4 | ✓ 1 | |
| T2 = 326 K | ✓ 1 | |
| 326 − 273 = 53 °C | ✓ 1 | |
| Part (b) | ||
| p1V1 = p2V2, so V2 = 2.7 × 106 × 0.050 ÷ 15 000 | ✓ 1 | |
| V2 = 9.0 m3 | ✓ 1 | |
| Part (c) | ||
| p2 = 15 × 220 ÷ 326, using the temperature in kelvin; allow error carried forward from the kelvin value in (a) | ✓ 1 | |
| p2 = 10 kPa (10.1 kPa) | ✓ 1 | |
Exam tip: In a multi-step gas problem, decide for each step which quantity is constant: volume or temperature. Then use p₁/T₁ = p₂/T₂ or p₁V₁ = p₂V₂ for that step, with temperatures in kelvin.
A football is pumped up indoors at 20 °C to a pressure of 160 kPa. It is then used in a winter match where the air temperature is −5 °C. The match rules require the pressure in the ball to be between 150 kPa and 170 kPa. Assume the volume of the ball does not change.
Calculate the pressure in the ball during the match.
Evaluate whether the ball meets the match rules.
Calculate the minimum pressure the ball should be pumped to at 20 °C so that it meets the rules during the match.
Explain why using the temperatures in °C in the equation would give an impossible answer.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Converts temperatures: 293 K and 268 K | ✓ 1 | |
| p2 = 160 × 268 ÷ 293 | ✓ 1 | |
| p2 = 146 kPa | ✓ 1 | |
| Part (b) | ||
| 146 kPa is below 150 kPa, so the ball does not meet the rules | ✓ 1 | |
| Part (c) | ||
| p1 = 150 × 293 ÷ 268 | ✓ 1 | |
| p1 = 164 kPa | ✓ 1 | |
| Part (d) | ||
| Pressure is proportional to Kelvin temperature, not Celsius temperature; using −5 °C would give a negative pressure, which is impossible | ✓ 1 | |
Exam tip: When you are asked to 'evaluate', compare your calculated value with the given limits and state a clear conclusion.
A welding workshop stores argon gas in a rigid steel cylinder. At 17 °C the pressure of the argon is 1.50 × 106 Pa. The cylinder has a safety valve that opens when the pressure inside reaches 1.80 × 106 Pa. One day the cylinder is left close to a hot furnace.
Explain, in terms of molecules, why the pressure of the argon increases as it is heated.
Calculate the temperature, in °C, at which the safety valve opens. Assume that the volume of the cylinder does not change.
The valve has an area of 2.0 × 10−5 m2. Atmospheric pressure outside the cylinder is 1.0 × 105 Pa.
Calculate the resultant force on the valve at the moment it opens.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Heating increases the average kinetic energy of the molecules, so they move faster on average | ✓ 1 | |
| The molecules hit the walls more often, because the volume is fixed | ✓ 1 | |
| Each collision applies a greater force (or gives a greater change of momentum), so the force per unit area, and therefore the pressure, increases | ✓ 1 | |
| Part (b) | ||
| T1 = 17 + 273 = 290 K | ✓ 1 | |
| p1/T1 = p2/T2 | ✓ 1 | |
| T2 = 290 × (1.80 × 106) ÷ (1.50 × 106) = 348 K | ✓ 1 | |
| T2 = 348 − 273 = 75 °C | ✓ 1 | |
| Part (c) | ||
| Pressure difference = 1.80 × 106 − 1.0 × 105 = 1.70 × 106 Pa | ✓ 1 | |
| F = p × A = 1.70 × 106 × 2.0 × 10−5 | ✓ 1 | |
| F = 34 N | ✓ 1 | |
Exam tip: When a gas pushes on one side of a surface and air pushes on the other, use the difference in pressure to find the resultant force.
A hotel lobby has a decorative solid stone column standing upright on the floor. The column is 3.2 m tall and has a uniform cross-sectional area of 0.15 m2. The density of the stone is 2400 kg/m3.
g = 10 N/kg
Calculate the mass of the column.
Calculate the pressure that the column exerts on the floor.
A designer says that a thicker column of the same stone and the same height would put the same pressure on the floor. Explain why the designer is correct.
The floor tiles can safely withstand a pressure of 60 000 Pa. The column is to be placed on a thin metal base plate that spreads its weight over a larger area. Ignoring the weight of the plate, calculate the minimum area of the base plate.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Volume = 0.15 × 3.2 = 0.48 m3 | ✓ 1 | |
| mass = density × volume = 2400 × 0.48 = 1152 kg | ✓ 1 | |
| Part (b) | ||
| Weight = 1152 × 10 = 11 520 N | ✓ 1 | |
| p = F ÷ A = 11 520 ÷ 0.15 | ✓ 1 | |
| = 76 800 Pa | ✓ 1 | |
| Part (c) | ||
| A thicker column has a larger volume, so its weight increases in the same proportion as its base area (e.g. double the area gives double the weight), or weight = density × area × height × g | ✓ 1 | |
| Since p = F ÷ A, the increases in force and area cancel, so the pressure (= density × height × g) is unchanged | ✓ 1 | |
| Part (d) | ||
| A = F ÷ p = 11 520 ÷ 60 000 | ✓ 1 | |
| = 0.192 m2 | ✓ 1 | |
Exam tip: Pressure is calculated from weight in newtons, not from mass in kilograms. Multiply the mass by g = 10 N/kg before you divide by the area.
The temperature of the gas in a sealed, rigid container rises from 27 °C to 54 °C. What happens to the average kinetic energy of the gas molecules?
| Marking point | Mark | Notes |
|---|---|---|
| Average kinetic energy is proportional to the temperature in kelvin: 300 K to 327 K, a ratio of 327 / 300 = 1.09, so B is correct. A doubles because the Celsius temperature has been used. C squares the doubled Celsius ratio. D confuses fixed volume with fixed temperature. | ✓ 1 | Answer B |
Answer: B
Common error: Ratios of temperature only work in kelvin. Doubling the Celsius value does not double the kinetic energy.
A hospital steriliser heats 2.5 kg of water from 20 °C to 100 °C using a 3.0 kW electric heater switched on for 5.0 minutes. Specific heat capacity of water = 4200 J/kg °C. Specific latent heat of vaporisation of water = 2.26 × 106 J/kg.
Calculate the energy gained by the water.
Calculate the energy supplied by the heater in 5.0 minutes.
Calculate the efficiency of the heating process.
The heater stays on for a further 2.0 minutes. Assume the efficiency stays at 93% and all the useful energy boils water at 100 °C. Calculate the mass of water that boils away.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Temperature rise = 80 °C | ✓ 1 | |
| ΔE = m c Δθ = 2.5 × 4200 × 80 | ✓ 1 | |
| 840000 J | ✓ 1 | |
| Part (b) | ||
| E = P t = 3000 × 300 | ✓ 1 | |
| 900000 J | ✓ 1 | |
| Part (c) | ||
| Efficiency = useful energy / total energy (ECF from (a) and (b)) | ✓ 1 | |
| 840000 / 900000 = 0.93 (93%) | ✓ 1 | |
| Part (d) | ||
| Energy supplied = 3000 × 120 = 360000 J | ✓ 1 | |
| Useful energy = 0.93 × 360000 = 334800 J | ✓ 1 | |
| m = E / L = 334800 / 2.26 × 106 = 0.148 kg (0.15 kg to 2 s.f.) | ✓ 1 | |
Exam tip: Convert minutes to seconds and kilowatts to watts before using E = Pt. Keep the extra figures in the working until the final answer.
Two metal studs for a jacket, X and Y, each have a mass of 0.50 kg. Each is given 6000 J of thermal energy. The temperature of X rises by 3.0 °C and the temperature of Y rises by 12 °C. What is the value of (specific heat capacity of X) ÷ (specific heat capacity of Y)?
| Marking point | Mark | Notes |
|---|---|---|
| c = E / (m Δθ). With the same energy and mass, c is inversely proportional to the temperature rise. cX = 6000/(0.50 × 3.0) = 4000 J/kg °C and cY = 6000/(0.50 × 12) = 1000 J/kg °C, so the ratio is 4, option B. A inverts the ratio. C subtracts the temperature rises (12 − 3). D multiplies them (3 × 12). | ✓ 1 | Answer B |
Answer: B
Exam tip: For equal energy and equal mass, the material that warms up less has the greater specific heat capacity.
Two bridge piers stand on firm ground. Pier X is a solid concrete block with a uniform cross-section. Pier Y is a solid block of lighter concrete with a uniform cross-section. The density of Y is 0.75 times the density of X, Y is twice as tall as X and the base area of Y is half that of X. What is (pressure of Y on the ground) ÷ (pressure of X on the ground)?
| Marking point | Mark | Notes |
|---|---|---|
| For a uniform block p = hρg, so the base area cancels. The ratio is 2 × 0.75 = 1.5, so B. A uses density only and ignores the height. C ignores the change in density. D wrongly also multiplies by 2 for the halved base area, forgetting that the weight falls by the same factor. | ✓ 1 | Answer B |
Answer: B
Common error: For a uniform solid column the base area cancels out of p = hρg. Do not apply an area factor to a pressure that comes from weight ÷ area unless the weight stays the same.
A solid cube of side L and uniform density rests on a table. It is cut into 8 identical smaller cubes, each of side L/2. One small cube is placed on the table. Compared with the pressure of the original cube on the table, the pressure of the small cube is
| Marking point | Mark | Notes |
|---|---|---|
| Correct (D): the weight is 1/8 as great and the contact area is 1/4 as great, so p = W/A is (1/8) ÷ (1/4) = 1/2 of the original. A uses the change in weight only. B uses the change in area only. C wrongly assumes that pressure depends only on the material. | ✓ 1 | Answer D |
Answer: D
Exam tip: When both the force and the area change, work out the factor for each and then divide.
A beekeeper heats water to make sugar syrup. The water is in an aluminium saucepan. The saucepan has a mass of 0.40 kg and contains 1.5 kg of water. The temperature of both rises from 20 °C to 100 °C.
Specific heat capacity of water = 4200 J/kg °C. Specific heat capacity of aluminium = 900 J/kg °C.
Calculate the energy transferred to the water.
Calculate the energy transferred to the saucepan.
The cooker supplies energy at a constant rate of 2.0 kW. Assume that all of this energy heats the water and the saucepan. Calculate the time taken. Give your answer to 3 significant figures. Use your unrounded answers to (a) and (b), or the values given in the mark scheme.
Suggest why the actual time taken is longer than your answer to (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| C1 ΔQ = mcΔθ | ✓ 1 | |
| C1 1.5 × 4200 × 80 | ✓ 1 | |
| A1 504 000 J | ✓ 1 | |
| Part (b) | ||
| C1 0.40 × 900 × 80 | ✓ 1 | |
| A1 28 800 J | ✓ 1 | |
| Part (c) | ||
| C1 total energy = 504 000 + 28 800 = 532 800 J (ECF from (a), (b)) | ✓ 1 | |
| C1 time = energy ÷ power = 532 800 ÷ 2000 | ✓ 1 | |
| A1 266 s | ✓ 1 | |
| Part (d) | ||
| E1 some energy is transferred to the surroundings (by heating the air, or through the sides of the pan), so more energy must be supplied | ✓ 1 | |
Common error: Remember to include the saucepan as well as the water, because both increase in temperature.
A freediver's depth gauge also records the total pressure p at depth h below the surface of the sea.
Depth h / m: 0 | 3 | 6 | 9 | 12
Total pressure p / kPa: 101.0 | 131.9 | 162.8 | 193.7 | 224.6
Use g = 10 N/kg.
Explain why the pressure is not zero at a depth of 0 m. [1]
Determine the gradient of a graph of p against h using the first and last points. [2]
Determine the density of the sea water. [2]
Calculate the depth at which the total pressure is 250 kPa. Give your answer to 3 significant figures. [3]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The atmosphere above the water exerts pressure on the surface (101 kPa) | ✓ 1 | |
| Part (b) | ||
| Gradient = (224.6 − 101.0)/(12 − 0) | ✓ 1 | |
| 10.3 kPa/m (10300 Pa/m) | ✓ 1 | |
| Part (c) | ||
| ρ = gradient/g = 10300/10 | ✓ 1 | |
| ρ = 1030 kg/m3 | ✓ 1 | |
| Part (d) | ||
| Use of p = patm + hρg (pressure due to water = 250 − 101 = 149 kPa) | ✓ 1 | |
| h = 149000/(1030 × 10) = 14.466 m | ✓ 1 | |
| h = 14.5 m | ✓ 1 | |
Exam tip: Total pressure at depth is atmospheric pressure plus hρg. Subtract atmospheric pressure first, or your depth will be far too small.
A sealed aerosol can contains gas at a pressure of 1.5 × 105 Pa at 20 °C. The volume of the gas is fixed. Use 0 °C = 273 K.
Convert 20 °C into kelvin.
The can is warmed to 50 °C. Calculate the pressure of the gas. Give your answer to 3 significant figures.
The can bursts when the gas pressure reaches 2.4 × 105 Pa. Calculate the temperature, in °C, at which this happens. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 293 K | ✓ 1 | |
| Part (b) | ||
| use p1/T1 = p2/T2 with temperatures in kelvin | ✓ 1 | |
| p2 = 1.5 × 105 × 323 ÷ 293 | ✓ 1 | |
| 1.65 × 105 Pa | ✓ 1 | |
| Part (c) | ||
| T2 = 293 × 2.4 ÷ 1.5 | ✓ 1 | |
| 469 K (468.8 K) | ✓ 1 | |
| 196 °C | ✓ 1 | |
Common error: Gas law calculations with temperature must use kelvin; using °C gives a wrong ratio.
In a ship's engine room, 0.30 kg of water at 90 °C is poured into 0.20 kg of water at 15 °C in an insulated container. The specific heat capacity of water is 4200 J/kg °C.
Calculate the final temperature of the mixture, assuming no energy is lost.
The measured final temperature is 58 °C. Calculate the energy transferred to the surroundings.
Suggest one way to reduce the energy transferred to the surroundings.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy lost by hot water = energy gained by cold water | ✓ 1 | |
| 0.30 × 4200 × (90 − T) = 0.20 × 4200 × (T − 15) | ✓ 1 | |
| 27 − 0.3T = 0.2T − 3 (or equivalent rearrangement) | ✓ 1 | |
| T = 60 °C | ✓ 1 | |
| Part (b) | ||
| energy lost by hot water = 0.30 × 4200 × 32 = 40 320 J | ✓ 1 | |
| energy gained by cold water = 0.20 × 4200 × 43 = 36 120 J | ✓ 1 | |
| difference = 4200 J | ✓ 1 | |
| Part (c) | ||
| use a lid or add more insulation around the container | ✓ 1 | |
Common error: Energy lost by the hot water equals energy gained by the cold water only if nothing escapes to the surroundings.
A hob with a power of 2.4 kW is used for 3.0 minutes to heat an aluminium saucepan of mass 0.50 kg containing 1.2 kg of water. Only 75% of the energy supplied by the hob is transferred to the saucepan and water. The specific heat capacity of aluminium is 900 J/kg °C and of water is 4200 J/kg °C.
Calculate the energy supplied by the hob in 3.0 minutes.
Calculate the energy transferred to the saucepan and water.
Calculate the rise in temperature of the saucepan and water. Use 324000 J.
The initial temperature was 20 °C. Calculate the final temperature. Use the unrounded temperature rise.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 energy = power × time = 2400 × 180 | ✓ 1 | |
| A1 432000 J | ✓ 1 | |
| Part (b) | ||
| A1 0.75 × 432000 = 324000 J (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 energy = (0.50 × 900 + 1.2 × 4200) × temperature rise | ✓ 1 | |
| M1 total of 5490 J/°C used: temperature rise = 324000 ÷ 5490 | ✓ 1 | |
| A1 59.0 °C | ✓ 1 | |
| Part (d) | ||
| A1 79.0 °C (ECF from (c)) | ✓ 1 | |
Common error: The pan and the water both warm up, so add their separate mass × specific heat capacity terms before dividing.
A geothermal power station pumps water at a rate of 4.0 kg per second through hot rocks. The water enters at 20 °C and leaves at 85 °C. The specific heat capacity of water is 4200 J/kg °C.
Calculate the energy transferred to the water each second.
State the rate of energy transfer to the water in watts.
The generators convert 15% of this energy into electrical energy. Calculate the electrical power output in kW, to 3 significant figures.
The electricity is sold at €0.10 per kWh. Calculate the income from 1.0 hour of operation. Use the unrounded power, 163.8 kW.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 ΔQ = m c Δθ | ✓ 1 | |
| M1 ΔQ = 4.0 × 4200 × 65 | ✓ 1 | |
| A1 1 092 000 J | ✓ 1 | |
| Part (b) | ||
| A1 1 092 000 W (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 0.15 × 1 092 000 = 163 800 W | ✓ 1 | |
| A1 164 kW | ✓ 1 | |
| Part (d) | ||
| M1 energy = 163.8 kW × 1.0 h = 163.8 kWh | ✓ 1 | |
| A1 163.8 × 0.10 = €16.38 | ✓ 1 | |
Exam tip: Keep unrounded values in a multi-step calculation and round only the final answer.
A space probe has an aluminium instrument block of mass 3.0 kg at −35 °C. The specific heat capacity of aluminium is 900 J/kg °C. A heater of power 135 W is switched on for 20 minutes.
Calculate the electrical energy supplied by the heater.
Calculate the temperature rise of the block if all of this energy heats the block.
In fact only 85% of the energy supplied by the heater is transferred to the block. Calculate the final temperature of the block.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| time = 20 × 60 = 1200 s | ✓ 1 | |
| energy = 135 × 1200 = 162000 J | ✓ 1 | |
| Part (b) | ||
| Δθ = E ÷ (mc) = 162000 ÷ (3.0 × 900) | ✓ 1 | |
| 60 °C (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| useful energy = 0.85 × 162000 = 137700 J | ✓ 1 | |
| temperature rise = 137700 ÷ 2700 = 51 °C | ✓ 1 | |
| final temperature = −35 + 51 = 16 °C | ✓ 1 | |
Common error: Convert minutes to seconds before using power × time, and add the temperature rise to the starting temperature, which here is negative.
A robot in a packing plant uses a suction pad to lift a flat panel that hangs beneath the pad. A pump reduces the air pressure inside the pad to 40 kPa. The pressure of the air outside is 100 kPa. The area of the pad is 50 cm2. The gravitational field strength is 10 N/kg.
Calculate the resultant force from the air pressure holding the panel against the pad.
Calculate the greatest mass of panel that the pad can hold.
Explain, in terms of molecules, why the pressure inside the pad falls when the pump removes air.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 pressure difference = 100 − 40 = 60 kPa = 60 000 Pa | ✓ 1 | |
| M1 area = 50 × 10−4 = 5.0 × 10−3 m2 and force = pressure × area | ✓ 1 | |
| A1 300 N | ✓ 1 | |
| Part (b) | ||
| M1 weight = mass × g so mass = 300 ÷ 10 (ECF from (a)) | ✓ 1 | |
| A1 30 kg | ✓ 1 | |
| Part (c) | ||
| There are fewer molecules in the same volume | ✓ 1 | |
| So there are fewer collisions per second with each unit area of the pad | ✓ 1 | |
Exam tip: The force from pressure depends on the difference in pressure on the two sides. Subtract the pressures first, then multiply by area in m².
A brick of weight 24 N has faces of sizes 0.10 m × 0.20 m, 0.20 m × 0.05 m and 0.10 m × 0.05 m. The brick is placed on a table with each face in turn underneath. What is the greatest pressure the brick exerts on the table?
| Marking point | Mark | Notes |
|---|---|---|
| Pressure is greatest on the smallest face, area 0.10 × 0.05 = 0.0050 m2: p = 24 ÷ 0.0050 = 4800 Pa. 1200 Pa comes from the largest face (smallest pressure). 2400 Pa comes from the middle face. 0.12 Pa multiplies force by area instead of dividing. | ✓ 1 | Answer C |
Answer: C
Common error: Pressure is force divided by area, so the smallest contact area gives the greatest pressure.
A Formula 1 car of mass 780 kg stands on four identical tyres. The contact area between each tyre and the road is 0.0125 m2. Use g = 10 N/kg.
Calculate the weight of the car.
Calculate the pressure the car exerts on the road when stationary.
At high speed the air also pushes the car downwards with a force of 11700 N. The road surface must not be subjected to a pressure above 3.0 × 105 Pa. Calculate the smallest total contact area that keeps the pressure within this limit.
State whether the actual total contact area of 0.050 m2 is large enough, and give a reason.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| W = mg | ✓ 1 | |
| 780 × 10 = 7800 N | ✓ 1 | |
| Part (b) | ||
| total area = 4 × 0.0125 = 0.050 m2 | ✓ 1 | |
| p = F ÷ A | ✓ 1 | |
| 7800 ÷ 0.050 = 156000 Pa (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| total downward force = 7800 + 11700 = 19500 N | ✓ 1 | |
| A = F ÷ p | ✓ 1 | |
| 19500 ÷ 300000 = 0.065 m2 (ECF from (a)) | ✓ 1 | |
| Part (d) | ||
| no, because 0.050 m2 is smaller than 0.065 m2 | ✓ 1 | |
Exam tip: At speed, add every downward force before calculating the pressure on the surface.
A beekeeper melts beeswax in a heated tray. A block of solid wax of mass 0.40 kg at 20 °C is heated to its melting point of 62 °C. The specific heat capacity of solid wax is 2500 J/kg °C. The heater has a power of 700 W.
Calculate the temperature rise of the wax.
Calculate the energy needed to raise the wax to its melting point.
Calculate the time needed if all the energy from the heater is transferred to the wax.
In practice the heating takes 80 s. Calculate the efficiency of the heating.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 62 − 20 = 42 °C | ✓ 1 | |
| Part (b) | ||
| Q = mcΔT | ✓ 1 | |
| 0.40 × 2500 × 42 = 42000 J (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| t = Q ÷ P | ✓ 1 | |
| 42000 ÷ 700 = 60 s (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| energy supplied = 700 × 80 = 56000 J | ✓ 1 | |
| efficiency = useful energy ÷ total energy | ✓ 1 | |
| 42000 ÷ 56000 × 100 = 75% | ✓ 1 | |
Common error: Use the useful energy (the wax's gain) on top of the fraction, and the total energy supplied underneath.
Liquid X has a mass of 2.0 kg and liquid Y has a mass of 1.0 kg. Each liquid receives the same quantity of thermal energy from its own heater. The temperature of X rises three times as much as the temperature of Y. How does the specific heat capacity of X compare with that of Y?
| Marking point | Mark | Notes |
|---|---|---|
| From c = ΔE ÷ (mΔθ), for equal energy c is proportional to 1 ÷ (mΔθ). X has twice the mass and three times the temperature rise, so mΔθ is 6 times larger and c is 6 times smaller. 6 times larger inverts the relationship. 1.5 times larger makes c proportional to Δθ ÷ m. 3 times smaller ignores the different masses. | ✓ 1 | Answer B |
Answer: B
Common error: Specific heat capacity depends on mass as well as temperature change, so always include both in the comparison.
The air in a tyre of a delivery van is at a pressure of 2.40 × 105 Pa and a temperature of 17°C. After a long journey the temperature of the air in the tyre is 47°C. Assume that the volume of the air in the tyre does not change.
Convert both temperatures to kelvin.
Calculate the pressure of the air in the tyre after the journey. Use the unrounded values and give your answer to 3 significant figures.
Explain, in terms of molecules, why the pressure of the air increases when the temperature rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 17 + 273 = 290 K | ✓ 1 | |
| 47 + 273 = 320 K | ✓ 1 | |
| Part (b) | ||
| M1 p1 ÷ T1 = p2 ÷ T2 | ✓ 1 | |
| M1 p2 = 2.40 × 105 × 320 ÷ 290 | ✓ 1 | |
| A1 2.65 × 105 Pa (265 000 Pa) | ✓ 1 | |
| Part (c) | ||
| the molecules move faster (have greater average kinetic energy) | ✓ 1 | |
| they hit the walls more often and with greater force, so the pressure is greater | ✓ 1 | |
Common error: Using 17 and 47 directly in the equation gives a much larger answer; always use kelvin.
A student seals a fixed mass of gas in a rigid flask and measures its pressure at different temperatures. At 0 °C the pressure is 100.0 kPa. At 80 °C the pressure is 129.3 kPa.
Calculate the gradient of a straight-line graph of pressure against temperature in °C. Give your answer to 3 significant figures.
The straight line is extended downwards until the pressure is zero. Estimate the temperature at which this happens. Give your answer to 3 significant figures.
Explain, in terms of molecules, why the pressure of a gas would be zero at this temperature.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 gradient = (129.3 − 100.0) ÷ (80 − 0) | ✓ 1 | |
| A1 0.366 kPa/°C | ✓ 1 | |
| Part (b) | ||
| M1 fall in pressure to be removed = 100.0 kPa | ✓ 1 | |
| M1 temperature change = 100.0 ÷ gradient | ✓ 1 | |
| A1 −273 °C | ✓ 1 | |
| Part (c) | ||
| the molecules would have no (minimum) kinetic energy / stop moving | ✓ 1 | |
| so there would be no collisions with the walls and no force on them | ✓ 1 | |
Exam tip: Extrapolating a line to zero pressure gives an estimate of absolute zero; the answer should be close to −273 °C.
A constant-volume gas thermometer contains a fixed mass of gas. The gas pressure is 1.20 × 105 Pa when the bulb is in melting ice at 273 K. The bulb is then placed in liquid nitrogen and the pressure falls to 3.40 × 104 Pa.
State why the volume of the gas must be kept constant.
Calculate the temperature of the liquid nitrogen in kelvin. Give your answer to 2 significant figures.
Convert your answer to (b) to degrees Celsius.
The average kinetic energy of the nitrogen molecules is proportional to the temperature in kelvin. Use your answer to (b) to find the average kinetic energy of the molecules in liquid nitrogen as a fraction of their average kinetic energy at 273 K. Give your answer to 2 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| pressure depends on volume as well as temperature, so otherwise the change in pressure would not show the change in temperature alone | ✓ 1 | |
| Part (b) | ||
| M1 use of p1 ÷ T1 = p2 ÷ T2 | ✓ 1 | |
| M1 T2 = 273 × 3.40 × 104 ÷ 1.20 × 105 | ✓ 1 | |
| A1 77 K | ✓ 1 | |
| Part (c) | ||
| A1 −196 °C (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| M1 fraction = 77 ÷ 273 | ✓ 1 | |
| A1 0.28 (ECF from (b)) | ✓ 1 | |
Exam tip: Keep the temperature in kelvin in both the gas-law equation and any kinetic energy ratio.
A solar water heater has a panel of area 2.5 m2. The intensity of sunlight on the panel is 600 W/m2 and 40% of the incident power is transferred to the water. The tank holds 120 kg of water. The specific heat capacity of water is 4200 J/kg °C.
Calculate the power transferred to the water.
Calculate the energy needed to raise the temperature of the water from 15 °C to 40 °C.
Calculate the time needed, in seconds.
Convert your answer to (c) to hours. Give your answer to 2 significant figures.
Suggest one reason why the heating takes longer in practice.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 0.40 × 600 × 2.5 | ✓ 1 | |
| A1 600 W | ✓ 1 | |
| Part (b) | ||
| M1 ΔE = m c Δθ | ✓ 1 | |
| M1 120 × 4200 × 25 | ✓ 1 | |
| A1 1.26 × 107 J | ✓ 1 | |
| Part (c) | ||
| M1 time = energy ÷ power | ✓ 1 | |
| A1 21 000 s (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| A1 5.8 hours (ECF from (c)) | ✓ 1 | |
| Part (e) | ||
| energy is transferred from the tank or pipes to the surroundings / sunlight intensity falls when it is cloudy | ✓ 1 | |
Exam tip: Divide the energy by the power in watts to get seconds, then divide by 3600 for hours.
The pressure of dry air in a sealed section of telephone cable is monitored to detect leaks. The cable volume is fixed. At 15 °C the pressure is 144 kPa. Use 0 °C = 273 K.
Calculate the pressure expected at −7 °C if no gas has escaped.
At −7 °C an engineer reads 128 kPa. Use your answer to (a) to decide whether gas has leaked from the cable.
Calculate the temperature in °C at which a leak-free cable would give a reading of 128 kPa.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 converts to kelvin: 288 K and 266 K | ✓ 1 | |
| M1 p2 = 144 × 266 ÷ 288 | ✓ 1 | |
| A1 133 kPa | ✓ 1 | |
| Part (b) | ||
| The reading (128 kPa) is lower than the expected pressure (133 kPa) (ECF from (a)) | ✓ 1 | |
| so gas has leaked (escaped) | ✓ 1 | |
| Part (c) | ||
| M1 T2 = 288 × 128 ÷ 144 | ✓ 1 | |
| A1 256 K | ✓ 1 | |
| A1 −17 °C | ✓ 1 | |
Exam tip: For a 'decide' question, quote both numbers you are comparing and state the conclusion clearly.
An electronics technician uses a soldering iron to melt a bead of solid solder. While it is melting, the solder stays at a temperature of 183 °C.
Describe how the arrangement and movement of the particles change when the solder melts.
Explain why the temperature of the solder stays at 183 °C until all of it has melted, even though the iron keeps supplying energy.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| In the solid, the particles are close together in a regular arrangement | ✓ 1 | |
| In the solid, they vibrate about fixed positions; in the liquid, they can move around or slide past each other | ✓ 1 | |
| In the liquid, the particles are still close together but are arranged randomly | ✓ 1 | |
| Part (b) | ||
| The energy supplied is used to overcome or break the forces/bonds between the particles, or to increase the potential energy of the particles | ✓ 1 | |
| It does not increase the (average) kinetic energy of the particles, so the temperature does not rise | ✓ 1 | |
Common error: Particles in a liquid are not far apart. They are about as close together as in a solid. The difference is that they are arranged randomly and can move past each other.
A hotel uses an electric water boiler to make hot drinks. The boiler heats 0.75 kg of water from 20 °C to 100 °C.
Specific heat capacity of water = 4200 J/kg °C.
Calculate the energy needed to raise the temperature of the water from 20 °C to 100 °C.
The heater transfers energy to the water at a rate of 2000 W.
Calculate the time taken to heat the water to 100 °C. Assume no energy is lost to the surroundings.
The boiler stays switched on after the water reaches 100 °C. Its thermometer continues to read 100 °C.
Explain why the temperature does not rise above 100 °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ΔQ = m × c × ΔT | ✓ 1 | |
| ΔQ = 0.75 × 4200 × 80 | ✓ 1 | |
| ΔQ = 252 000 J | ✓ 1 | |
| Part (b) | ||
| time = energy ÷ power = 252 000 ÷ 2000 | ✓ 1 | |
| time = 126 s | ✓ 1 | |
| Part (c) | ||
| The water is boiling / changing state to steam | ✓ 1 | |
| The energy supplied is used to separate the particles / overcome the forces between them rather than to increase their kinetic energy | ✓ 1 | |
Common error: ΔT is the temperature change (80 °C), not the final temperature (100 °C).
An electric storage heater contains ceramic bricks. The bricks are heated overnight and give out thermal energy during the day. One brick has a mass of 2.4 kg. The specific heat capacity of the ceramic is 840 J/kg °C.
State what is meant by the specific heat capacity of a material.
Calculate the thermal energy needed to heat one brick from 20 °C to 520 °C.
The heating element supplies thermal energy to this brick at a rate of 800 W. Calculate the minimum time needed to heat the brick from 20 °C to 520 °C.
Suggest why the actual time is longer than your answer to the previous part.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| The energy needed to raise the temperature of 1 kg of the material by 1 °C (or 1 K) | ✓ 1 | |
| Part (b) | ||
| ΔT = 520 − 20 = 500 °C | ✓ 1 | |
| ΔQ = m × c × ΔT = 2.4 × 840 × 500 | ✓ 1 | |
| = 1 008 000 J (1.0 × 106 J) | ✓ 1 | |
| Part (c) | ||
| time = energy ÷ power = 1 008 000 ÷ 800 | ✓ 1 | |
| = 1260 s | ✓ 1 | |
| Part (d) | ||
| Some thermal energy is transferred to the surroundings or the heater casing while the brick is warming up | ✓ 1 | |
Exam tip: Always work out the temperature change first. Multiplying by the final temperature instead of by ΔT is a common way to lose a mark.
A student wants to find the specific heat capacity of sunflower oil. She has an immersion heater, a joulemeter, a beaker, a thermometer, an electronic balance and some insulating foam.
Describe an experiment she could do to find the specific heat capacity of the oil. Include the measurements she should take and how she should use them to calculate the specific heat capacity.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Use the electronic balance to find the mass of the oil, e.g. weigh the empty beaker and then the beaker with oil and subtract, or zero (tare) the balance with the empty beaker | ✓ 1 | |
| Wrap the beaker in insulating foam before heating, and place the immersion heater and the thermometer in the oil | ✓ 1 | |
| Record the initial temperature of the oil | ✓ 1 | |
| Connect the heater through the joulemeter and reset it to zero. Switch on and stir the oil. Switch off after a suitable temperature rise and record the highest temperature reached | ✓ 1 | |
| Read the energy supplied, E, from the joulemeter, and find the temperature rise ΔT = final temperature − initial temperature | ✓ 1 | |
| Calculate the specific heat capacity using c = E ÷ (m × ΔT), with the mass in kg | ✓ 1 | |
Exam tip: In a six-mark practical description, name each measuring instrument and say what it measures. Describe the steps in a sensible order, and end by saying how the results are used to calculate the answer.
In a hospital, a bag of saline solution hangs 1.5 m above a patient's arm. A tube joins the bag to a vein. The density of the saline solution is 1020 kg/m3. Use g = 10 N/kg.
Calculate the pressure due to the column of saline solution at the vein.
The pressure in the patient's vein is 3000 Pa above atmospheric pressure. Explain why saline solution flows into the vein.
The nurse lowers the bag until the flow just stops. Calculate the height of the bag above the vein when this happens.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = h ρ g | ✓ 1 | |
| = 1.5 × 1020 × 10 | ✓ 1 | |
| 15300 Pa | ✓ 1 | |
| Part (b) | ||
| The pressure due to the saline column (15300 Pa) is greater than the vein pressure (3000 Pa), so there is a resultant push into the vein | ✓ 1 | |
| Part (c) | ||
| h = p / (ρ g) | ✓ 1 | |
| = 3000 / (1020 × 10) | ✓ 1 | |
| 0.294 m (0.29 m to 2 s.f.) | ✓ 1 | |
Exam tip: Rearrange p = hρg before substituting. The height of the liquid column is the vertical height, not the length of the tube.
A team tests a fuel tank. A pressure sensor is lowered to different depths h in a liquid fuel. The pressure p due to the liquid alone is recorded.
h / m: 0.10, 0.20, 0.30, 0.40, 0.50
p / Pa: 760, 1500, 2230, 2990, 3720
Use g = 10 N/kg.
State the relationship between pressure and depth shown by the data.
Use the first and last readings to calculate the gradient of a graph of p against h.
The gradient equals ρg. Determine the density of the fuel.
Suggest why using the gradient is better than using one pair of readings.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Pressure increases in proportion to depth (p/h is approximately constant) | ✓ 1 | |
| Part (b) | ||
| Change in p = 3720 − 760 = 2960 Pa | ✓ 1 | |
| Change in h = 0.50 − 0.10 = 0.40 m | ✓ 1 | |
| Gradient = 7400 Pa/m | ✓ 1 | |
| Part (c) | ||
| ρ = gradient / g = 7400 / 10 (ECF from (b)) | ✓ 1 | |
| 740 kg/m3 | ✓ 1 | |
| Part (d) | ||
| It uses the pattern across several readings, so random errors in one reading have less effect | ✓ 1 | |
Exam tip: On a pressure against depth graph the gradient is ρg, so divide it by g to get density.
A smoke cell is placed under a microscope. Smoke particles are lit from the side and appear as bright specks.
Describe the motion of the bright specks.
Explain what causes this motion.
State what happens to the motion of the specks if the smoke cell is warmed.
State why the air molecules cannot be seen but the smoke particles can.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| They move in random directions | ✓ 1 | |
| With small, jerky, zigzag movements | ✓ 1 | |
| Part (b) | ||
| Air molecules are moving randomly and collide with the smoke particles | ✓ 1 | |
| More molecules hit one side of a particle than the other at any instant | ✓ 1 | |
| This gives an unbalanced force on the particle that keeps changing direction | ✓ 1 | |
| Part (c) | ||
| The motion becomes faster / more vigorous because the molecules move faster | ✓ 1 | |
| Part (d) | ||
| The air molecules are far smaller than the smoke particles | ✓ 1 | |
Common error: The smoke particles are not the molecules; they are much larger and are moved by collisions with the air molecules.
A sealed, rigid pneumatic cylinder in a robot arm contains air at a pressure of 2.4 × 105 Pa and a temperature of 27 °C. The cylinder warms to 87 °C. The volume of the air does not change.
Convert both temperatures to kelvin.
Calculate the new pressure of the air. Give your answer in Pa.
Explain, in terms of molecules, why the pressure rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 27 °C = 300 K | ✓ 1 | |
| 87 °C = 360 K | ✓ 1 | |
| Part (b) | ||
| p1/T1 = p2/T2 with temperatures in kelvin | ✓ 1 | |
| p2 = 2.4 × 105 × 360 / 300 (ECF from (a)) | ✓ 1 | |
| 2.88 × 105 Pa | ✓ 1 | |
| Part (c) | ||
| The molecules move faster as the temperature rises | ✓ 1 | |
| So they hit the walls more often and with greater force | ✓ 1 | |
| Each second there is a greater total force per unit area on the walls | ✓ 1 | |
Common error: Gas-law equations need temperatures in kelvin; using °C gives a wrong ratio.
A sealed tank of gas on a space probe is at 27 °C and a pressure of 2.0 × 105 Pa. Sunlight heats the tank to 177 °C. The volume of the tank does not change. What is the new pressure of the gas?
| Marking point | Mark | Notes |
|---|---|---|
| Convert to kelvin: 300 K and 450 K. At constant volume p/T is constant, so p = 2.0 × 105 × 450/300 = 3.0 × 105 Pa, so C. A inverts the temperature ratio (300/450). B uses the ratio of the Celsius values (177/27). D converts only the final temperature to kelvin (450/27). | ✓ 1 | Answer C |
Answer: C
Common error: Temperatures in gas law calculations must be in kelvin. Add 273 to each Celsius value before forming the ratio.
The air in a Formula 1 tyre can be treated as a fixed mass of gas at constant volume. At 27 °C the pressure of the air in the tyre is 1.40 × 105 Pa. After a few laps the temperature of the air is 87 °C.
Convert both temperatures to kelvin.
Calculate the pressure of the air in the tyre at 87 °C.
Explain, in terms of the air molecules, why the pressure increases as the temperature rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 300 K and 360 K | ✓ 1 | |
| Part (b) | ||
| C1 p1/T1 = p2/T2 | ✓ 1 | |
| C1 1.40 × 105 × 360 ÷ 300 | ✓ 1 | |
| A1 1.68 × 105 Pa (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| E1 the molecules move faster (higher average kinetic energy), so they hit the walls harder | ✓ 1 | |
| E1 and more often, so the force per unit area is greater | ✓ 1 | |
Common error: Temperatures must be in kelvin in gas law calculations: add 273 to the Celsius temperature.
A beekeeper uses a smoker. Its bellows contain 800 cm3 of air at atmospheric pressure, 1.0 × 105 Pa. The bellows are squeezed slowly until the air has a volume of 250 cm3. Assume that no air escapes.
State why the temperature of the air must stay the same for the calculation in (b) to be valid.
Calculate the pressure of the air in the bellows after squeezing.
Explain, in terms of molecules, why the pressure increases when the volume decreases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 pV = constant only applies to a fixed mass of gas at constant temperature | ✓ 1 | |
| Part (b) | ||
| C1 p1V1 = p2V2 | ✓ 1 | |
| C1 1.0 × 105 × 800 ÷ 250 | ✓ 1 | |
| A1 3.2 × 105 Pa | ✓ 1 | |
| Part (c) | ||
| E1 the molecules are in a smaller space, so they hit the walls more often | ✓ 1 | |
| E1 so the total force on each unit area of the wall is greater | ✓ 1 | |
Exam tip: Volumes in p₁V₁ = p₂V₂ may stay in cm³ if both are in the same unit.
A sealed gas cell on a space probe contains gas at 27 °C. Its volume cannot change. Sunlight raises the temperature of the gas to 177 °C. By what factor does the pressure of the gas increase?
| Marking point | Mark | Notes |
|---|---|---|
| B is correct: at constant volume pressure is proportional to the temperature in kelvin. 27 °C = 300 K and 177 °C = 450 K, so the factor is 450/300 = 1.5. A uses the ratio of the Celsius temperatures (177/27). C assumes pressure is unchanged because the volume is constant. D inverts the ratio. | ✓ 1 | Answer B |
Answer: B
Common error: Always convert to kelvin before using pressure-temperature relationships; ratios of Celsius temperatures are meaningless.
A sample of gas is sealed in a cylinder fitted with a piston in a sampling instrument on a space probe. The gas has an initial pressure of 1.8 × 105 Pa and an initial volume of 250 cm3. The piston is pushed in slowly until the volume is 90 cm3.
State the property of the gas that must stay constant, apart from its mass, for the final pressure to be calculated from pV = constant. [1]
Calculate the final pressure of the gas. [3]
Explain, in terms of gas molecules, why the pressure increases. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Temperature | ✓ 1 | |
| Part (b) | ||
| Use of p1V1 = p2V2 (volumes may be left in cm3) | ✓ 1 | |
| p2 = (1.8 × 105 × 250)/90 | ✓ 1 | |
| p2 = 500000 Pa (5.0 × 105 Pa) | ✓ 1 | |
| Part (c) | ||
| The same number of molecules is in a smaller volume, so they collide with the walls more often (more collisions per second per unit area) | ✓ 1 | |
| Greater force per unit area on the walls, so greater pressure | ✓ 1 | |
Exam tip: For pV = constant you may keep volumes in cm³ because the units cancel, but pressure must be in the same unit on both sides.
A sealed syringe contains 48 cm3 of air at a pressure of 1.0 × 105 Pa. The plunger is pushed in slowly until the volume of the air is 20 cm3.
Calculate the final pressure of the air, assuming the temperature stays constant.
Explain, in terms of molecules, why the pressure of the air increases.
State why the plunger is pushed in slowly.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 pressure × volume is constant, p1V1 = p2V2 | ✓ 1 | |
| M1 substitution of 1.0 × 105 × 48 = p2 × 20 | ✓ 1 | |
| A1 2.4 × 105 Pa | ✓ 1 | |
| Part (b) | ||
| the same number of molecules are in a smaller volume | ✓ 1 | |
| so the molecules collide with the walls and plunger more often | ✓ 1 | |
| so the average force on each unit area of the walls is greater | ✓ 1 | |
| Part (c) | ||
| so the temperature of the air stays constant (energy has time to transfer to the surroundings) | ✓ 1 | |
Exam tip: Link pressure to molecules in two steps: how often they collide with the walls, and how hard.
A diver releases an air bubble at a depth where the pressure is 3.0 × 105 Pa. The bubble has a volume of 12 cm3. The bubble rises to the surface of the pool, where the pressure is 1.0 × 105 Pa. Assume that the temperature of the air in the bubble stays constant.
State the condition, other than constant temperature, needed for pV to be constant for a fixed amount of gas.
Calculate the volume of the bubble at the surface.
Explain, in terms of molecules, why the pressure of the air in the bubble falls as its volume increases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the mass (amount) of gas stays the same, so no gas escapes or enters | ✓ 1 | |
| Part (b) | ||
| M1 p1V1 = p2V2 | ✓ 1 | |
| M1 3.0 × 105 × 12 = 1.0 × 105 × V2 | ✓ 1 | |
| A1 36 cm3 | ✓ 1 | |
| Part (c) | ||
| molecules travel further between collisions with the walls, so collisions are less frequent | ✓ 1 | |
| the force on each unit area of the wall is smaller, so pressure is lower | ✓ 1 | |
Exam tip: Write the gas law you are using before substituting numbers, and keep units for volume the same on both sides.
Helium gas is sealed in a container of fixed volume at 27 °C. The gas is then heated to 177 °C. Use 0 °C = 273 K.
Convert both temperatures to kelvin.
The average kinetic energy of the helium atoms is proportional to the kelvin temperature. Calculate the factor by which the average kinetic energy increases.
State what happens to the pressure of the gas and explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 300 K | ✓ 1 | |
| B1 450 K | ✓ 1 | |
| Part (b) | ||
| M1 450 / 300 (ECF from (a)) | ✓ 1 | |
| A1 1.5 | ✓ 1 | |
| Part (c) | ||
| B1 the pressure increases | ✓ 1 | |
| B1 the atoms move faster, so they hit the walls more often and with more force | ✓ 1 | |
Common error: Temperature in the kinetic model must be the kelvin temperature; the Celsius value is not proportional to energy.
Helium in a space probe is stored in a tank of volume 2.4 litres at a pressure of 6.0×105 Pa. A valve opens and the gas expands into the tank and an empty chamber, so that the total volume is 8.0 litres. The temperature stays constant. What is the final pressure?
| Marking point | Mark | Notes |
|---|---|---|
| At constant temperature pV is constant: p = 6.0×105 × 2.4 / 8.0 = 1.8×105 Pa. 2.0×106 Pa inverts the volume ratio. 4.2×105 Pa multiplies by the fraction of the new volume that was empty (5.6/8.0). 6.0×105 Pa assumes the pressure does not change when the volume does. | ✓ 1 | Answer A |
Answer: A
Common error: When volume increases at constant temperature, pressure must decrease; check your answer is smaller than the starting pressure.
A sealed steel canister on a coffee roaster contains air at 27 °C and a pressure of 1.0×105 Pa. The canister is heated to 327 °C. The volume of the canister does not change. Use 0 °C = 273 K.
State the initial and final temperatures in kelvin.
Calculate the final pressure of the air.
Explain, in terms of molecules, why the pressure rises as the air is heated.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| initial temperature = 300 K | ✓ 1 | |
| final temperature = 600 K | ✓ 1 | |
| Part (b) | ||
| p1/T1 = p2/T2, so p2 = 1.0×105 × 600 ÷ 300 | ✓ 1 | |
| 2.0×105 Pa (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| the molecules move faster (have more kinetic energy) | ✓ 1 | |
| they hit the walls more often | ✓ 1 | |
| and with greater force in each collision | ✓ 1 | |
Common error: Gas-law calculations need temperatures in kelvin; using °C gives a pressure that is far too small.
A fish-farm aerator pump compresses 90 cm3 of air at a pressure of 100 kPa into a smaller cylinder of volume 36 cm3. The compression is slow so that the temperature of the air stays constant. What is the final pressure of the air?
| Marking point | Mark | Notes |
|---|---|---|
| At constant temperature pV is constant, so p = 100 × 90 ÷ 36 = 250 kPa. 40 kPa uses the volume ratio the wrong way round. 100 kPa assumes that constant temperature means constant pressure. 154 kPa assumes the pressure rises by the same number of units as the volume falls (90 − 36 = 54), but pressure and volume are inversely proportional, not related by subtraction. | ✓ 1 | Answer D |
Answer: D
Exam tip: For a fixed mass of gas at constant temperature, write p₁V₁ = p₂V₂ first, then rearrange. A smaller volume must give a larger pressure, which checks your answer.
A robot arm uses a sealed canister of gas as a spring. The gas has a pressure of 300 kPa at a temperature of 27 °C. The volume of the canister cannot change. The canister is heated to 87 °C.
Convert 27 °C to a temperature in kelvin.
Calculate the pressure of the gas at 87 °C.
Explain, in terms of the gas molecules, why the pressure increases when the gas is heated.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 300 K | ✓ 1 | |
| Part (b) | ||
| M1 converts 87 °C to 360 K | ✓ 1 | |
| M1 p1/T1 = p2/T2, so p2 = 300 × 360 ÷ 300 (ECF from (a)) | ✓ 1 | |
| A1 360 kPa | ✓ 1 | |
| Part (c) | ||
| The molecules move faster / have more kinetic energy | ✓ 1 | |
| They hit the walls more often and with greater force, so the pressure is greater | ✓ 1 | |
Exam tip: Gas-law equations that use temperature need kelvin. Add 273 to the Celsius value before substituting; using 87 and 27 directly gives a wrong ratio.
A water tower holds water of density 1000 kg/m3. The surface of the water is 25 m above a tap on level ground. The surface of the water and the open tap are both exposed to the atmosphere. Use g = 10 N/kg.
Calculate the difference in pressure between the water at the tap and the water at the surface in the tower.
The tap contains a washer of area 4.0 cm2. Write this area in m2.
Calculate the force on the washer caused by this pressure difference.
A tap on a higher floor of a building is connected to the same tower. Explain why the force on its washer is smaller.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = hρg | ✓ 1 | |
| 25 × 1000 × 10 = 250000 Pa | ✓ 1 | |
| Part (b) | ||
| 4.0 × 10−4 m2 | ✓ 1 | |
| Part (c) | ||
| F = pA | ✓ 1 | |
| 250000 × 4.0 × 10−4 = 100 N (ECF from (a) and (b)) | ✓ 1 | |
| Part (d) | ||
| smaller height of water above this tap | ✓ 1 | |
| so smaller pressure difference (and smaller force for the same area) | ✓ 1 | |
Common error: 1 cm² is 1 × 10⁻⁴ m², not 1 × 10⁻² m².
The air in a Formula 1 tyre is sealed, and the volume of the air can be taken as constant. When the tyre is fitted, the air is at 27 °C and a pressure of 1.50 × 105 Pa. After a long stint the air in the tyre is at 87 °C.
Write both temperatures in kelvin.
Calculate the pressure of the air in the tyre after the stint.
Explain, in terms of the air molecules, why the pressure increases when the temperature rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 300 K | ✓ 1 | |
| 360 K | ✓ 1 | |
| Part (b) | ||
| p1 ÷ T1 = p2 ÷ T2 | ✓ 1 | |
| 1.50 × 105 × 360 ÷ 300 = 180000 Pa (1.80 × 105 Pa) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| molecules move faster / have more kinetic energy | ✓ 1 | |
| they hit the walls more often | ✓ 1 | |
| and with a greater force in each collision | ✓ 1 | |
Common error: Gas-law calculations need the temperature in kelvin, not °C.
In a hospital, a bag of saline solution is hung above a patient's arm so that the liquid flows down a tube into a vein. The saline has a density of 1000 kg/m3. The pressure in the vein is 4000 Pa above atmospheric pressure. Use g = 10 N/kg.
The surface of the saline is 1.2 m above the vein. Calculate the pressure due to the saline at the vein, above atmospheric pressure.
State and explain whether saline flows into the vein.
The bag is lowered to 0.30 m above the vein. Calculate the new pressure due to the saline and state what happens to the flow.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = hρg | ✓ 1 | |
| 1.2 × 1000 × 10 = 12000 Pa | ✓ 1 | |
| Part (b) | ||
| yes | ✓ 1 | |
| 12000 Pa is greater than the 4000 Pa in the vein (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 0.30 × 1000 × 10 = 3000 Pa | ✓ 1 | |
| less than 4000 Pa, so saline does not flow in (blood may flow back up the tube) | ✓ 1 | |
Exam tip: Pressure in a liquid depends on the vertical height, not on the length of the tube.
Water behind a hydroelectric dam has a density of 1000 kg/m3. A small inspection hatch on the dam wall has an area of 0.50 m2. Its centre is 20 m below the surface of the water. Assume the pressure over the whole hatch equals the pressure at a depth of 20 m, and ignore atmospheric pressure. Use g = 10 N/kg.
Calculate the pressure due to the water at the hatch.
Calculate the force on the hatch.
Explain why the wall of a dam is made thicker at the bottom than at the top.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = hρg | ✓ 1 | |
| 20 × 1000 × 10 = 200000 Pa | ✓ 1 | |
| Part (b) | ||
| F = pA | ✓ 1 | |
| 200000 × 0.50 = 100000 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| pressure increases with depth | ✓ 1 | |
| so the force on the wall is greater near the bottom, which needs a stronger wall | ✓ 1 | |
Exam tip: Pressure in a liquid increases with depth only; the width of the reservoir does not matter.
A sealed medical syringe traps 60 cm3 of air at a pressure of 1.0 × 105 Pa. The plunger is pushed in slowly until the air occupies 24 cm3. The temperature of the air stays constant. What is the new pressure of the air?
| Marking point | Mark | Notes |
|---|---|---|
| At constant temperature p1V1 = p2V2, so p2 = 1.0 × 105 × 60 ÷ 24 = 2.5 × 105 Pa. 4.0 × 104 Pa multiplies by 24/60 instead of 60/24, treating pressure as proportional to volume. 1.6 × 105 Pa adds a fraction of the volume removed (36/60) to the original pressure. 3.6 × 106 Pa multiplies the pressure by the volume change in cm3. | ✓ 1 | Answer D |
Answer: D
Exam tip: For a fixed mass of gas at constant temperature, pressure and volume are inversely proportional, so a smaller volume means a larger pressure.
An aerosol inhaler canister contains gas at a pressure of 2.4 × 105 Pa when its temperature is 27 °C. The canister is left in a hot car and its temperature rises to 57 °C. The volume of the gas stays constant.
Convert 27 °C to kelvin.
Calculate the pressure of the gas at 57 °C.
Explain, in terms of gas molecules, why the pressure rises as the temperature rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| A1 300 K | ✓ 1 | |
| Part (b) | ||
| M1 temperature at 57 °C = 330 K | ✓ 1 | |
| M1 p1/T1 = p2/T2, so 2.4 × 105 ÷ 300 = p2 ÷ 330 | ✓ 1 | |
| A1 2.64 × 105 Pa (accept 2.6 × 105 Pa) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| The molecules gain kinetic energy and move faster | ✓ 1 | |
| They collide with the walls more often and with greater force, so the pressure is larger | ✓ 1 | |
Common error: Temperatures in gas law calculations must be in kelvin; using °C gives a wrong ratio.
A technician in an epidemiology laboratory traps 60 cm3 of air at a pressure of 100 kPa in a sealed syringe. The plunger is pushed in slowly until the volume of the air is 24 cm3. The temperature of the air stays constant.
Calculate the new pressure of the air.
Explain, in terms of molecules, why the pressure of the air increases.
Explain why the plunger is pushed in slowly.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p1V1 = p2V2 | ✓ 1 | |
| M1 100 × 60 = p2 × 24 | ✓ 1 | |
| A1 250 kPa | ✓ 1 | |
| Part (b) | ||
| the molecules are in a smaller volume so they hit the walls more often each second | ✓ 1 | |
| so the force on each unit area of the walls is greater | ✓ 1 | |
| Part (c) | ||
| so that the temperature stays constant (energy can transfer to the surroundings) | ✓ 1 | |
Exam tip: Volumes in cm³ and pressures in kPa can be used directly in p₁V₁ = p₂V₂ as long as the units match on both sides.
A pumped-storage hydroelectric scheme stores water in a reservoir. At the dam, the water is 25 m deep. The density of water is 1000 kg/m3. The gravitational field strength is 10 N/kg. Ignore atmospheric pressure.
Calculate the pressure due to the water at the base of the dam.
Explain why the wall of a dam is made thicker at the base.
An inspection hatch at the base of the dam has an area of 0.80 m2. Calculate the force of the water on the hatch.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 pressure = height × density × gravitational field strength | ✓ 1 | |
| M1 25 × 1000 × 10 | ✓ 1 | |
| A1 2.5 × 105 Pa (250 000 Pa) | ✓ 1 | |
| Part (b) | ||
| pressure increases with depth, so the force on the wall is greater at the base | ✓ 1 | |
| Part (c) | ||
| M1 force = pressure × area | ✓ 1 | |
| A1 250 000 × 0.80 = 2.0 × 105 N (200 000 N) (ECF from (a)) | ✓ 1 | |
Exam tip: Pressure in a liquid depends on depth and density only, not on the width of the reservoir.
A sealed cylinder of gas is stored in an aircraft cargo hold. At 27 °C the pressure of the gas is 480 kPa. The volume of the cylinder is fixed.
State the temperature of absolute zero in °C.
The hold cools and the pressure of the gas falls. Explain, in terms of molecules, why the pressure falls.
The temperature falls to −33 °C. Calculate the new pressure.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| −273 °C | ✓ 1 | |
| Part (b) | ||
| the molecules have a lower average kinetic energy / move more slowly | ✓ 1 | |
| they hit the walls less often | ✓ 1 | |
| and with less force, so the pressure is smaller | ✓ 1 | |
| Part (c) | ||
| M1 temperatures in kelvin: 300 K and 240 K | ✓ 1 | |
| M1 p2 = p1 × T2 ÷ T1 | ✓ 1 | |
| A1 384 kPa | ✓ 1 | |
Common error: Pressure is proportional to the temperature in kelvin, not in °C.
A garden sprayer on a farm contains 2.4 litres of air at a pressure of 100 kPa. The pump compresses the air to 0.80 litres. The temperature of the air stays constant.
State one other condition needed for the pressure and volume of the gas to follow p1V1 = p2V2.
Calculate the pressure of the compressed air.
Explain, in terms of molecules, why the pressure rises when the volume decreases.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the mass (number of molecules) of gas stays the same / no air leaks | ✓ 1 | |
| Part (b) | ||
| M1 use of p1V1 = p2V2 | ✓ 1 | |
| M1 p2 = 100 × 2.4 ÷ 0.80 | ✓ 1 | |
| A1 300 kPa | ✓ 1 | |
| Part (c) | ||
| the same number of molecules in a smaller volume | ✓ 1 | |
| so the molecules hit the walls more often | ✓ 1 | |
| so the total force per unit area on the walls is greater | ✓ 1 | |
Exam tip: Pressure and volume can be in any units in p₁V₁ = p₂V₂ as long as the same units are used on both sides.
The air in a sealed rigid container is heated from 27 °C to 327 °C. Which statement about the average kinetic energy of the air molecules is correct?
| Marking point | Mark | Notes |
|---|---|---|
| Average kinetic energy is proportional to the temperature in kelvin. 27 °C = 300 K and 327 °C = 600 K, so the kinetic energy doubles. Dividing the Celsius values (327 ÷ 27 ≈ 12) is wrong because Celsius has no true zero. A factor of 4 would need the temperature in kelvin to quadruple. A factor of about 1.4 is the change in average molecular speed, not kinetic energy, since kinetic energy depends on speed squared. | ✓ 1 | Answer B |
Answer: B
Common error: Always convert to kelvin before comparing temperatures in kinetic theory; ratios of Celsius values have no physical meaning.
The water at the intake of a hydroelectric turbine is at a depth of 35 m below the surface of a reservoir. The density of water is 1000 kg/m3. Use g = 10 N/kg.
Calculate the pressure due to the water at this depth.
An inspection hatch of area 0.80 m2 is at this depth. Calculate the force of the water on the hatch.
Explain why the wall of the dam is made thicker at the bottom than at the top.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p = h ρ g | ✓ 1 | |
| M1 35 × 1000 × 10 | ✓ 1 | |
| A1 3.5 × 105 Pa | ✓ 1 | |
| Part (b) | ||
| M1 F = p A | ✓ 1 | |
| A1 2.8 × 105 N (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| pressure increases with depth | ✓ 1 | |
| so the force of the water on the wall is greater near the bottom | ✓ 1 | |
Exam tip: Pressure from a liquid depends only on depth, density and g, not on how wide the reservoir is.
A weather balloon carries a small radio transmitter used for telecommunications. At ground level the balloon contains 2.0 m3 of helium at a pressure of 100 kPa. At a high altitude the pressure of the helium is 25 kPa. Assume the temperature of the helium stays constant.
Calculate the volume of the balloon at high altitude.
Explain, in terms of helium molecules, why the pressure of the helium falls as the balloon expands.
In practice the air temperature falls as the balloon rises. State the effect of this on the volume compared with your answer to (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p1V1 = p2V2 | ✓ 1 | |
| M1 V2 = 100 × 2.0 ÷ 25 | ✓ 1 | |
| A1 8.0 m3 | ✓ 1 | |
| Part (b) | ||
| The molecules are further apart/travel further between collisions with the balloon wall, so collide less often | ✓ 1 | |
| so the force per unit area on the wall is smaller | ✓ 1 | |
| Part (c) | ||
| The volume would be smaller | ✓ 1 | |
Exam tip: For Boyle's law, p and V can be in any units provided both sides use the same ones.
A candle manufacturer tests a new wax. A 0.25 kg sample of solid wax is heated by a 50 W heater. The temperature stays at 54 °C for 900 s while the wax melts. The liquid wax then warms from 54 °C to 64 °C in 100 s.
Assume all the energy from the heater is transferred to the wax.
Calculate the energy transferred by the heater in 100 s.
Calculate the specific heat capacity of the liquid wax. Give the unit.
Calculate the energy needed to melt 1 kg of this wax.
In reality, some energy from the heater is transferred to the surroundings. Explain how this affects the value you calculated for the specific heat capacity of the liquid wax.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| E = P × t = 50 × 100 | ✓ 1 | |
| E = 5000 J | ✓ 1 | |
| Part (b) | ||
| c = ΔQ ÷ (m × ΔT) | ✓ 1 | |
| c = 5000 ÷ (0.25 × 10) | ✓ 1 | |
| c = 2000 J/kg °C (accept J/kg K) | ✓ 1 | |
| Part (c) | ||
| Energy to melt sample = 50 × 900 = 45 000 J | ✓ 1 | |
| Energy per kg = 45 000 ÷ 0.25 = 180 000 J/kg (accept 1.8 × 105 J/kg) | ✓ 1 | |
| Part (d) | ||
| Less than 5000 J actually went into the wax | ✓ 1 | |
| So the calculated value is too large / the true specific heat capacity is lower | ✓ 1 | |
Common error: Energy losses mean less energy reaches the sample, so a calculated value of c is an overestimate, not an underestimate.
The blood in a standing adult is at rest in a vertical artery for this question. Take the density of blood as 1060 kg/m3 and g = 10 N/kg. The pressure of the blood at the heart is 16000 Pa above atmospheric pressure. The feet are 1.30 m below the heart and the brain is 0.30 m above the heart.
Calculate the increase in pressure from the heart to the feet.
Calculate the pressure of the blood at the feet, as a pressure above atmospheric pressure.
Calculate the pressure of the blood at the brain, as a pressure above atmospheric pressure.
A person stands up very quickly and feels dizzy. Use your answers to explain why.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Δp = h ρ g = 1.30 × 1060 × 10 | ✓ 1 | |
| 13780 Pa | ✓ 1 | |
| Part (b) | ||
| 16000 + 13780 (ECF from (a)) | ✓ 1 | |
| 29780 Pa | ✓ 1 | |
| Part (c) | ||
| Pressure drop = 0.30 × 1060 × 10 = 3180 Pa | ✓ 1 | |
| Pressure at brain = 16000 − 3180 | ✓ 1 | |
| 12820 Pa | ✓ 1 | |
| Part (d) | ||
| Standing raises the brain further above the heart, so the pressure at the brain falls (compared with lying down) | ✓ 1 | |
| Less blood reaches the brain until the heart pressure increases to compensate | ✓ 1 | |
Common error: Pressure falls as you go up through a liquid and rises as you go down. Check whether to add or subtract ρgh for the point you are asked about.
A sealed rigid container used to transport fish food holds gas at a pressure of 1.5 × 105 Pa and a temperature of 7 °C. The volume of the gas stays constant.
Convert 7 °C to kelvin.
The pressure is doubled. Calculate the new temperature of the gas in kelvin.
Give the new temperature in °C.
A student says that doubling the pressure means doubling the Celsius temperature from 7 °C to 14 °C. Explain why this is wrong.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 280 K | ✓ 1 | |
| Part (b) | ||
| Pressure is proportional to kelvin temperature at constant volume (p1/T1 = p2/T2) | ✓ 1 | |
| T = 2 × 280 = 560 K (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| 560 − 273 = 287 °C | ✓ 1 | |
| Part (d) | ||
| Pressure is proportional to the temperature in kelvin, not in °C | ✓ 1 | |
| 14 °C is only 287 K, which is not double 280 K | ✓ 1 | |
Common error: Doubling the Celsius value does not double the kelvin value, so it does not double the pressure.
A fuel tank on a space probe contains gas of volume 30 litres at a pressure of 1.2 × 106 Pa and a temperature of 27 °C. A valve is opened and the gas expands into a second, empty tank of volume 50 litres that is connected to the first. The temperature stays at 27 °C. The gas then cools to −33 °C in the shade, with the valve still open. Assume the amount of gas is fixed.
Calculate the pressure of the gas immediately after it has expanded into both tanks.
Calculate the pressure after the gas has cooled to −33 °C. Use the unrounded value from (a).
Explain, using the particle model, why the pressure falls when the gas cools.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Total volume = 30 + 50 = 80 litres | ✓ 1 | |
| p1V1 = p2V2 | ✓ 1 | |
| p2 = 1.2 × 106 × 30 ÷ 80 = 450000 Pa | ✓ 1 | |
| Part (b) | ||
| Temperatures converted: 300 K and 240 K | ✓ 1 | |
| p ÷ T is constant at constant volume | ✓ 1 | |
| p = 450000 × 240 ÷ 300 = 360000 Pa (3.6 × 105 Pa) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| The molecules move more slowly (lower average kinetic energy) | ✓ 1 | |
| They hit the walls less often and with less force, so the force per unit area is smaller | ✓ 1 | |
Common error: Use kelvin for p ÷ T and T ÷ V relationships, and keep the same unit for volume on both sides of p₁V₁ = p₂V₂.
A sensor on the bed of a lake releases a gas bubble. The lake is 20 m deep and the density of the water is 1000 kg/m3. Atmospheric pressure is 1.0 × 105 Pa. The bubble has a volume of 2.0 cm3 at the bed. The bubble rises to the surface. Use g = 10 N/kg.
Calculate the pressure due to the water at the bed of the lake.
Calculate the total pressure on the bubble at the bed of the lake.
Calculate the volume of the bubble when it reaches the surface.
State two assumptions made in your calculation in (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p = hρg | ✓ 1 | |
| p = 20 × 1000 × 10 = 200000 Pa | ✓ 1 | |
| Part (b) | ||
| 200000 + 1.0 × 105 = 300000 Pa (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| p1V1 = p2V2 | ✓ 1 | |
| Surface pressure = 1.0 × 105 Pa | ✓ 1 | |
| V2 = 300000 × 2.0 ÷ 100000 = 6.0 cm3 (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| The temperature of the gas stays constant | ✓ 1 | |
| The mass (amount) of gas in the bubble stays constant (none dissolves or escapes) | ✓ 1 | |
Common error: Total pressure under water is atmospheric pressure plus the pressure due to the water. Using only hρg gives the wrong gas pressure.
A sealed rigid air tank in a train braking system contains air at a temperature of 15 °C and a pressure of 8.00 × 105 Pa. After a long braking period the pressure in the tank is 8.80 × 105 Pa. Assume the volume of the tank does not change.
Convert 15 °C to kelvin. [1]
Calculate the final temperature of the air in kelvin. [3]
Calculate the final temperature in °C. Use the unrounded answer to (b). [1]
Explain why the calculation in (b) must use the kelvin scale rather than the Celsius scale. [2]
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 288 K | ✓ 1 | |
| Part (b) | ||
| Use of p1/T1 = p2/T2 | ✓ 1 | |
| T2 = 288 × (8.80 × 105)/(8.00 × 105) | ✓ 1 | |
| T2 = 316.8 K (317 K) | ✓ 1 | |
| Part (c) | ||
| 316.8 − 273 = 43.8 °C (accept 44 °C) | ✓ 1 | |
| Part (d) | ||
| Pressure is proportional to the temperature in kelvin (not in °C) | ✓ 1 | |
| Zero kelvin is the temperature at which the pressure would be zero / particles would have no kinetic energy | ✓ 1 | |
Common error: Substituting Celsius temperatures into p/T = constant gives wrong answers; convert by adding 273 first and subtract 273 at the end.
A theatre fog machine heats 0.15 kg of water from 20 °C to 100 °C. Then 0.060 kg of this water is turned into steam at 100 °C. The heater has a power of 1200 W and all its energy is transferred to the water. The specific heat capacity of water is 4200 J/kg °C and the specific latent heat of vaporisation of water is 2.26 × 106 J/kg.
Calculate the energy needed to heat the water from 20 °C to 100 °C.
Calculate the energy needed to turn 0.060 kg of water into steam.
Calculate the time the heater is on for all of this.
Explain why turning water into steam needs far more energy per kilogram than raising its temperature by 1 °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy = m c Δθ = 0.15 × 4200 × 80 | ✓ 1 | |
| 50 400 J | ✓ 1 | |
| Part (b) | ||
| energy = m L = 0.060 × 2.26 × 106 | ✓ 1 | |
| 135 600 J | ✓ 1 | |
| Part (c) | ||
| total energy = 50 400 + 135 600 | ✓ 1 | |
| = 186 000 J (ECF from (a) and (b)) | ✓ 1 | |
| time = 186 000 ÷ 1200 = 155 s | ✓ 1 | |
| Part (d) | ||
| molecules must be completely separated, overcoming the forces between them | ✓ 1 | |
| this requires a large increase in potential energy, unlike the small kinetic energy rise when heating | ✓ 1 | |
Common error: Do not use mcΔθ for a change of state; use energy = mL because the temperature does not change.
A sealed steel cylinder of gas is used in a theatre special effect. The volume of the cylinder is fixed. At 15 °C the pressure of the gas is 2.40 × 105 Pa. During the effect the cylinder is heated to 87 °C.
Convert both temperatures to kelvin.
Calculate the pressure of the gas at 87 °C.
The cylinder has a safety valve that opens at 3.5 × 105 Pa. Deduce whether the valve opens during the effect.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 288 K | ✓ 1 | |
| 360 K | ✓ 1 | |
| Part (b) | ||
| M1 p1/T1 = p2/T2 | ✓ 1 | |
| M1 substitution of 2.40 × 105 ÷ 288 = p2 ÷ 360 (ECF from (a)) | ✓ 1 | |
| A1 3.0 × 105 Pa | ✓ 1 | |
| Part (c) | ||
| no, because 3.0 × 105 Pa is less than 3.5 × 105 Pa (ECF from (b)) | ✓ 1 | |
Common error: Temperatures in gas law calculations must be in kelvin; add 273 to the temperature in °C.
A copper block of mass 0.50 kg at 100 °C is placed in 0.40 kg of water at 20 °C inside a well-insulated container. Assume no energy is transferred to the container or the surroundings. The specific heat capacity of copper is 390 J/kg °C and that of water is 4200 J/kg °C.
State the energy relationship that allows the final temperature to be found.
Calculate the final temperature of the water and copper. Give your answer to 3 significant figures.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the energy lost by the copper equals the energy gained by the water | ✓ 1 | |
| Part (b) | ||
| M1 energy lost by copper = 0.50 × 390 × (100 − T) | ✓ 1 | |
| M1 energy gained by water = 0.40 × 4200 × (T − 20) | ✓ 1 | |
| M1 equates and rearranges, e.g. 53 100 = 1875 T | ✓ 1 | |
| A1 28.3 °C | ✓ 1 | |
Exam tip: Let the final temperature be T and write each temperature change as a positive number, (100 − T) for the copper and (T − 20) for the water.
Gallium has a melting point of 30 °C. A sample of solid gallium of mass 0.25 kg is at its melting point. The specific latent heat of fusion of gallium is 8.0 × 104 J/kg. The specific heat capacity of liquid gallium is 370 J/kg °C.
Explain why the temperature of the gallium stays constant while it melts, even though energy is being supplied.
Calculate the energy needed to melt all the gallium.
Calculate the energy needed to raise the temperature of the liquid gallium from 30 °C to 50 °C.
Calculate the total energy needed to melt the sample and then warm the liquid to 50 °C.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the energy supplied is used to overcome the forces between the particles / change their arrangement | ✓ 1 | |
| it does not increase the average kinetic energy of the particles, so temperature does not rise | ✓ 1 | |
| Part (b) | ||
| M1 energy = mass × specific latent heat = 0.25 × 8.0 × 104 | ✓ 1 | |
| A1 20000 J | ✓ 1 | |
| Part (c) | ||
| M1 energy = 0.25 × 370 × 20 | ✓ 1 | |
| A1 1850 J | ✓ 1 | |
| Part (d) | ||
| M1 20000 + 1850 (ECF from (b) and (c)) | ✓ 1 | |
| A1 21850 J | ✓ 1 | |
Exam tip: Use mass × specific latent heat for the change of state, and mass × specific heat capacity × temperature change for the warming; the two stages are added.
A sealed vaccine syringe contains 6.0 cm3 of air at a pressure of 100 kPa. With the outlet sealed, the plunger is pushed in until the air volume is 2.5 cm3. The temperature stays constant. What is the new pressure?
| Marking point | Mark | Notes |
|---|---|---|
| At constant temperature pV is constant, so p2 = 100 × 6.0 / 2.5 = 240 kPa. 42 kPa comes from inverting the volume ratio, so the pressure falls instead of rising. 600 kPa multiplies the pressure by the initial volume only. 350 kPa comes from multiplying the pressure by the difference in volumes. | ✓ 1 | Answer C |
Answer: C
Exam tip: When volume decreases at constant temperature, pressure must increase; use this to check the direction of your answer.
A constant-volume gas thermometer is used to find the temperature of liquid nitrogen. A fixed mass of gas is sealed in a bulb. At 27 °C the gas pressure is 150 kPa. When the bulb is placed in the liquid nitrogen the pressure falls to 38.5 kPa. Use 0 °C = 273 K.
Convert 27 °C to kelvin.
Calculate the temperature of the liquid nitrogen in kelvin.
Calculate this temperature in °C.
Explain, in terms of molecules, why the pressure of the gas falls as the bulb is cooled.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| B1 300 K | ✓ 1 | |
| Part (b) | ||
| M1 p1 / T1 = p2 / T2 | ✓ 1 | |
| M1 T2 = 300 × 38.5 / 150 (ECF from (a)) | ✓ 1 | |
| A1 T2 = 77 K | ✓ 1 | |
| Part (c) | ||
| A1 77 − 273 = −196 °C (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| B1 the molecules have a lower average kinetic energy, so move more slowly | ✓ 1 | |
| B1 they hit the walls less often and with less force, so the pressure is smaller | ✓ 1 | |
Common error: In gas-law calculations the temperature must be in kelvin; using 27 and the Celsius value of the cold temperature gives a meaningless result.
A small air bubble is released at the bottom of a lake at a depth of 20 m. Its volume there is 0.50 cm3. Atmospheric pressure is 100 kPa, the density of lake water is 1000 kg/m3 and the gravitational field strength is 10 N/kg. Assume the temperature of the air in the bubble stays constant as it rises to the surface.
Calculate the pressure due to the water at a depth of 20 m. Give your answer in kPa.
State the total pressure on the bubble at the bottom of the lake.
Calculate the volume of the bubble when it reaches the surface.
Explain, in terms of the air molecules, why the volume of the bubble increases as it rises.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 pressure = depth × density × g = 20 × 1000 × 10 | ✓ 1 | |
| A1 200 kPa (200 000 Pa) | ✓ 1 | |
| Part (b) | ||
| 300 kPa (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 p1V1 = p2V2 | ✓ 1 | |
| M1 300 × 0.50 = 100 × V2 (ECF from (b)) | ✓ 1 | |
| A1 1.5 cm3 | ✓ 1 | |
| Part (d) | ||
| The pressure of the water on the bubble decreases as it rises | ✓ 1 | |
| The molecules hit the inside of the bubble less often per unit area unless it expands, so the bubble expands until the pressures balance | ✓ 1 | |
Common error: The pressure on the bubble is the water pressure plus atmospheric pressure. Leaving out atmospheric pressure gives a volume that is far too large.
A student heats 0.100 kg of ice, initially at −20 °C, using a heater of power 420 W. Assume all the energy from the heater is transferred to the ice or water. The temperature is recorded:
Time / s: 0, 10, 90, 190
Temperature / °C: −20, 0, 0, 100
The temperature changes at a constant rate between readings, except where it stays at 0 °C.
State the time interval during which the ice is melting and give a reason from the data.
Use the data from 0 s to 10 s to calculate the specific heat capacity of ice.
Calculate the specific latent heat of fusion of ice.
Calculate the gradient of the temperature–time graph for the water section, then use it to find the specific heat capacity of water.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| from 10 s to 90 s | ✓ 1 | |
| temperature stays constant at 0 °C while energy is still supplied | ✓ 1 | |
| Part (b) | ||
| energy = Pt = 420 × 10 = 4200 J | ✓ 1 | |
| c = Q ÷ (mΔT) = 4200 ÷ (0.100 × 20) | ✓ 1 | |
| 2100 J/kg °C | ✓ 1 | |
| Part (c) | ||
| energy = 420 × 80 = 33600 J | ✓ 1 | |
| L = Q ÷ m | ✓ 1 | |
| 33600 ÷ 0.100 = 336000 J/kg (3.36 × 105 J/kg) | ✓ 1 | |
| Part (d) | ||
| gradient = 100 ÷ 100 = 1.0 °C/s | ✓ 1 | |
| energy transferred each second = mc × gradient, so 420 = 0.100 × c × 1.0 | ✓ 1 | |
| c = 4200 J/kg °C | ✓ 1 | |
Exam tip: On a heating curve, the flat sections show a change of state and the sloping sections show a change of temperature.
A scuba diver releases an air bubble at a depth of 30 m in a freshwater lake. At this depth the bubble has a volume of 0.80 cm3. The bubble then rises slowly to just below the surface. Atmospheric pressure is 1.0 × 105 Pa, the density of water is 1000 kg/m3 and g = 10 N/kg.
Calculate the pressure due to the water at a depth of 30 m.
Calculate the total pressure on the bubble at a depth of 30 m.
Calculate the volume of the bubble just below the surface, where the pressure is atmospheric pressure.
State two assumptions you made in answering (c).
Explain, in terms of gas molecules, why the pressure of the gas in the bubble falls as the bubble expands.
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 pressure = hρg = 30 × 1000 × 10 | ✓ 1 | |
| A1 300 000 Pa (3.0 × 105 Pa) | ✓ 1 | |
| Part (b) | ||
| A1 300 000 + 100 000 = 400 000 Pa (4.0 × 105 Pa) (ECF from (a)) | ✓ 1 | |
| Part (c) | ||
| M1 p1V1 = p2V2 | ✓ 1 | |
| M1 4.0 × 105 × 0.80 = 1.0 × 105 × V | ✓ 1 | |
| A1 3.2 cm3 (ECF from (b)) | ✓ 1 | |
| Part (d) | ||
| A valid assumption | ✓ 1 | |
| A second, different valid assumption . Valid assumptions include: the temperature of the gas stays constant; the mass of gas in the bubble does not change (none dissolves or escapes); the pressure inside the bubble equals the pressure of the water around it | ✓ 1 | |
| Part (e) | ||
| The molecules are further apart, so they travel further between collisions with the surface of the bubble, giving fewer collisions each second on each unit of area | ✓ 1 | |
| The average force on each unit of area is smaller, so the pressure is smaller | ✓ 1 | |
Exam tip: Pressure at depth in a liquid is hρg; add atmospheric pressure for the total pressure on an object under the water surface.
A sealed rigid container holds a fixed mass of gas at a pressure of 150 kPa and a temperature of 27°C. The gas is heated to 87°C. What is the new pressure?
| Marking point | Mark | Notes |
|---|---|---|
| The volume is constant, so pressure ÷ kelvin temperature is constant. The temperatures are 300 K and 360 K, so the new pressure = 150 × 360 ÷ 300 = 180 kPa. The value 125 kPa inverts the temperature ratio. The value 210 kPa adds the 60 degree rise to the pressure. The value 483 kPa uses Celsius temperatures (150 × 87 ÷ 27). | ✓ 1 | Answer B |
Answer: B
Exam tip: For any gas-law calculation, change every temperature to kelvin first.
A fish farm has a tank 20 m deep. A small air bubble of volume 2.0 cm3 is released at the bottom and rises to the surface. The density of water is 1000 kg/m3, atmospheric pressure is 1.0 × 105 Pa and g = 10 N/kg. Assume the temperature of the bubble stays constant.
Calculate the pressure due to the water at the bottom of the tank.
Calculate the total pressure on the bubble at the bottom of the tank.
Calculate the volume of the bubble at the surface.
In practice the water near the surface is warmer than at the bottom. Explain why the bubble would then have a volume greater than your answer to (c).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 pressure = height × density × g | ✓ 1 | |
| A1 2.0 × 105 Pa | ✓ 1 | |
| Part (b) | ||
| 3.0 × 105 Pa , ECF from (a) | ✓ 1 | |
| Part (c) | ||
| M1 use of p1V1 = p2V2 | ✓ 1 | |
| M1 V2 = 3.0 × 105 × 2.0 ÷ 1.0 × 105 | ✓ 1 | |
| A1 6.0 cm3 , ECF from (b) | ✓ 1 | |
| Part (d) | ||
| a higher temperature gives the molecules a greater average speed / kinetic energy | ✓ 1 | |
| they hit the walls harder, so the bubble expands until the pressure inside equals the pressure outside, giving a larger volume | ✓ 1 | |
Common error: The pressure on a bubble underwater is the water pressure plus atmospheric pressure, not the water pressure alone.
A syringe is sealed at its nozzle with 60 cm3 of air at atmospheric pressure, 100 kPa. A food technologist pushes the plunger in slowly until the trapped air has a volume of 24 cm3. The plunger has an area of 4.0 × 10−4 m2.
Calculate the pressure of the trapped air.
Calculate the force needed to hold the plunger in this position. Atmospheric pressure acts on the outside of the plunger.
State one assumption made in part (a).
| Marking point | Mark | Notes |
|---|---|---|
| Part (a) | ||
| M1 p1V1 = p2V2 | ✓ 1 | |
| M1 p2 = 100 × 60 ÷ 24 | ✓ 1 | |
| A1 250 kPa | ✓ 1 | |
| Part (b) | ||
| M1 difference in pressure = 250 − 100 = 150 kPa (ECF from (a)) | ✓ 1 | |
| M1 F = pA = 1.5 × 105 × 4.0 × 10−4 | ✓ 1 | |
| A1 60 N | ✓ 1 | |
| Part (c) | ||
| The temperature of the air stays constant (and no air escapes) | ✓ 1 | |
Common error: The force you apply balances only the pressure difference across the plunger, not the whole pressure of the trapped gas.
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