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IB Physics Higher Level · Topic B.1

B.1 Thermal energy transfers

94 free, high-quality past-paper-style questions on Thermal energy transfers for IB Physics HL, each with a full mark scheme. Covers: Net radiation exchange; Kelvin and Celsius temperatures; Internal energy and temperature; and more. For when you have finished the past papers, or want to practise one topic at a time.B.1 at Standard Level →Read the B.1 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.428 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked

1B-1A-13Net radiation exchange Paper 1AEasy1 mark

A black-body sphere at 500 K is placed in surroundings at 300 K. σ = 5.67 × 10−8 W m−2 K−4.

What is the net power lost per unit area of the sphere?

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Marking pointMarkNotes
Step 1The sphere radiates σT⁴ = 5.67 × 10−8 × 500⁴ = 3544 W m−2.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2It also absorbs radiation from the surroundings: σTs⁴ = 5.67 × 10−8 × 300⁴ = 459 W m−2.—
Step 3Net loss = σ(T⁴ − Ts⁴) = 3544 − 459 = 3.1 × 10³ W m−2.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the sphere emits 3544 W m−2 but absorbs 459 W m−2 from its surroundings.
  • BThis is the emitted power alone, ignoring the radiation absorbed from the surroundings.
  • CThis is the power absorbed from the surroundings, not the net loss.
  • DThis uses σ(T − Ts)⁴, which is not how the fourth powers combine.
2B-1A-32Kelvin and Celsius temperatures Paper 1AEasy1 mark

The temperature of a sample of gas is raised from 18 °C to 63 °C.

Which row gives the change in temperature of the gas and the ratio (final average kinetic energy of the particles) / (initial average kinetic energy of the particles)?

Change in temperatureRatio of average kinetic energies
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Step 1A change in temperature is the same in kelvin as in degrees Celsius: 63 − 18 = 45 K.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The average kinetic energy is proportional to the kelvin temperature, Ek = (3/2)kBT; the temperatures are 291 K and 336 K.—
Step 3Ratio = 336/291 = 1.15.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis is √(336/291) = 1.07, the ratio of the rms speeds, not of the kinetic energies.
  • BCorrect: ΔT = 45 K and Ek ∝ T gives 336/291 = 1.15.
  • CThis uses the Celsius temperatures, 63/18 = 3.5; kinetic energy is proportional to the absolute temperature.
  • DThis adds 273 to the temperature change; the conversion applies to a temperature, not to a difference of temperatures.
3B-1A-33Internal energy and temperature Paper 1AEasy1 mark

A steel block of mass 4.0 kg at 30 °C is placed in contact with a steel nail of mass 0.050 kg at 90 °C. The two objects are isolated from their surroundings.

Which row identifies the object with the greater internal energy before contact, and the direction of the net transfer of thermal energy?

Greater internal energyNet thermal energy transfer
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Marking pointMarkNotes
Step 1Internal energy is the total random kinetic energy plus the total intermolecular potential energy of all the particles, so it depends on the number of particles as well as on the temperature. The block has 80 times as many particles, so its internal energy is greater.—All 2 steps must be completed — there is no mark for a part-answer.
Step 2Temperature measures the average kinetic energy per particle; the net transfer of thermal energy is determined by the temperature difference, from the nail (90 °C) to the block (30 °C), until the temperatures are equal.✓ 1Answer B

Answer: B  ·  2 stages of work, one mark

Every option, and why

  • AThis assumes energy flows from the object with more internal energy; the direction is set by the temperature difference.
  • BCorrect: the block has more internal energy, but the nail is at the higher temperature.
  • CThis identifies internal energy with temperature: the hotter nail has more energy per particle but far fewer particles.
  • DThis combines both errors — internal energy taken to be set by temperature alone, and energy flowing towards the higher temperature.
4B-1A-35Conduction Paper 1AEasy1 mark

One end of a long ceramic rod is held in a flame. After some time the other end of the rod becomes warm.

Which statement best explains how thermal energy is transferred along the rod?

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Step 1In a solid the particles vibrate about fixed positions; at the hot end their mean kinetic energy is greater. Through collisions (interactions) with their less energetic neighbours they pass on kinetic energy, so energy moves down the rod while the particles stay in place.✓ 1Answer A

Answer: A  ·  1 stage of work, one mark

Every option, and why

  • ACorrect: conduction arises from the difference in the kinetic energy of neighbouring particles.
  • BThis describes bulk movement of particles, which is convection in a fluid; the particles of a solid remain in fixed positions.
  • CThis describes radiation. Energy reaching the cold end by radiation through an opaque ceramic is negligible.
  • DTemperature is linked to the kinetic energy of the particles; conduction is the transfer of kinetic energy, not potential energy.
5B-1A-61Convection Paper 1AEasy1 mark

In a cold store, the cooling unit is mounted just below the ceiling. Air that passes over the cold pipes of the unit is cooled.

Which statement best explains how the air throughout the store, including the air near the floor, is cooled?

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Step 1Air cooled at (almost) constant pressure contracts, so the same mass occupies a smaller volume and its density ρ = m/V increases.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The denser cool air sinks below the warmer, less dense air, which is pushed upwards to the cooling unit: a convection current circulates through the whole store.—
Step 3Air is a very poor conductor, so conduction through it is slow, and the pipes do not emit "cold"; the dominant mechanism is convection.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: cooling increases the density of the air, and the density difference drives a convection current — this is why cooling units are placed high and heaters low.
  • BThis reverses the effect of temperature on density: cooled air contracts and becomes denser, it does not expand.
  • CThis names the wrong mechanism: still air is a poor thermal conductor, so conduction alone would cool the store very slowly.
  • D"Cold radiation" does not exist: the net radiative transfer is from the warmer surroundings to the cold pipes, and it does not cool the air (air is almost transparent to it).
6B-1A-62Evaporation and boiling Paper 1AEasy1 mark

A liquid can change into a gas by evaporation or by boiling.

Which statements about evaporation are correct?

I. It takes place only at the surface of the liquid.
II. It can take place at temperatures below the boiling point of the liquid.
III. When no energy is supplied to the liquid, it increases the mean kinetic energy of the molecules that remain in the liquid.

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Marking pointMarkNotes
Step 1I is correct: only molecules at the surface that have enough kinetic energy to overcome the attractions of their neighbours can escape. (In boiling, bubbles of vapour form throughout the liquid.)—All 3 steps must be completed — there is no mark for a part-answer.
Step 2II is correct: at any temperature some molecules have enough energy to escape, so evaporation happens below the boiling point.—
Step 3III is wrong: the molecules that escape are the most energetic ones, so the mean kinetic energy of those left behind decreases and the liquid cools.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis rejects II, as if evaporation needed the boiling point to be reached; puddles dry at room temperature.
  • BCorrect: evaporation is a surface process that occurs at any temperature, and it cools the liquid that remains.
  • CThis confuses evaporation with boiling (vapour forming throughout the liquid) and reverses the cooling effect.
  • DThis accepts III: removing the fastest molecules lowers, not raises, the mean kinetic energy of the rest.
7IBP2H-010-00-01Kelvin and Celsius temperatures Paper 1AEasy1 mark

A biological sample is taken out of liquid nitrogen at −196 °C and allowed to warm to a laboratory temperature of 20 °C. What is the change in temperature of the sample?

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A temperature difference has the same size in kelvin as in degrees Celsius, because both scales use the same interval. ΔT = 20 − (−196) = 216 °C = 216 K. Converting both values first gives the same result: 293 K − 77 K = 216 K. 176 K treats −196 as +196. 370 K adds the kelvin temperatures. 489 K converts only one of the two temperatures to kelvin.✓ 1Answer B

Answer: B

Exam tip: Add 273 only when you need an absolute temperature, for example in Stefan–Boltzmann or Wien. For a temperature change, Δθ in °C equals ΔT in K.

8IBP2H-010-00-05Evaporation and boiling Paper 1AEasy1 mark

A block of pure ice at 0 °C is heated steadily and melts. Which statement describes the water molecules during melting?

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A pure substance changes phase at a constant temperature. Temperature measures the mean random kinetic energy of the particles, so the mean kinetic energy does not change. The energy supplied (the latent heat) does work against the intermolecular forces as the molecules leave their fixed lattice positions. Their mean potential energy therefore increases, so option B is correct. Options A, C and D wrongly assume the energy supplied must raise the kinetic energy and the temperature.✓ 1Answer B

Answer: B

Exam tip: Link 'constant temperature' directly to 'constant mean kinetic energy'. The energy supplied must then be increasing the potential energy.

9IBP2H-010-01-03Convection Paper 1AEasy1 mark

On board an orbiting space station, a candle flame is roughly spherical and hot gases do not rise away from it. What is the best explanation?

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Convection needs a buoyant force on less dense fluid. That force comes from a pressure difference in the fluid caused by its weight. In a freely falling space station the air has no apparent weight, so there is no such pressure difference. Hot, less dense gas is therefore not pushed upwards, and convection currents do not form. Option A is wrong because gravity at that height is still about 90% of its surface value. Options C and D are false: the cabin air is at normal pressure and it expands when heated.✓ 1Answer B

Answer: B

Exam tip: Convection requires both a change of density with temperature and a gravitational field, or an effective one, in the frame of the fluid.

10IBP2H-010-01-06Internal energy and temperature Paper 1AEasy1 mark

A block of gallium is placed on a warm hand and melts at a constant temperature of 30 °C. During melting, what happens to the mean random kinetic energy and the intermolecular potential energy of the gallium atoms?

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The temperature is constant, so the mean random kinetic energy of the atoms does not change. The energy absorbed is used to overcome the bonds between atoms, which increases their potential energy. The internal energy therefore increases even though the temperature stays the same. Option D would mean no energy is absorbed, which contradicts the need for latent heat.✓ 1Answer C

Answer: C

Exam tip: Temperature measures only the mean random kinetic energy. Internal energy also includes intermolecular potential energy.

11IBP2H-010-01-08Black-body spectra Paper 1AEasy1 mark

The argon in a sealed light bulb is heated from 27 °C to 327 °C. By what factor does the mean random kinetic energy of the argon atoms increase?

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Marking pointMarkNotes
The mean kinetic energy is \(\frac{3}{2}k_BT\), where T is the kelvin temperature. The temperature rises from 300 K to 600 K, so the mean kinetic energy doubles. The value 12 uses Celsius temperatures. The value 1.4 is the factor for the rms speed, which is proportional to \(\sqrt{T}\). The value 4 wrongly squares the temperature ratio.✓ 1Answer B

Answer: B

Common error: Always convert to kelvin before using any proportionality involving temperature.

12IBP3H-B1-20r17-01Evaporation and boiling Paper 1AEasy1 mark

Which statement correctly distinguishes evaporation of a liquid from boiling?

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Option A is correct: surface molecules with enough kinetic energy escape at any temperature, while boiling needs vapour bubbles to form throughout the liquid, which happens at one temperature for a given pressure. B is wrong because boiling also needs a continuous energy supply (latent heat). C swaps the two processes. D swaps the temperature conditions of the two processes.✓ 1Answer A

Answer: A

Common error: Boiling is not 'more evaporation': it is bubble formation inside the liquid at a fixed temperature, and it needs a continuous energy supply.

13IBP3H-B1-20r17-02Evaporation and boiling Paper 1AEasy1 mark

A hospital autoclave sterilises instruments using water at a pressure greater than atmospheric pressure. The water in the autoclave boils at a temperature above 100 °C. What is the reason for this?

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Option A is correct: boiling begins when the vapour pressure in a bubble can match the pressure outside, so raising the external pressure raises the boiling point. B is wrong because latent heat is not what sets the boiling temperature. C is wrong because the pressure does not slow the molecules; the molecules' average kinetic energy depends on temperature. D is wrong because pressure does not remove dissolved air and the effect is not about surface evaporation.✓ 1Answer A

Answer: A

Exam tip: Link boiling to pressure: boiling occurs when saturated vapour pressure equals the external pressure.

14IBP3H-B1-20r17-04Molecular model and density Paper 1AEasy1 mark

Air passes through an F1 brake duct at constant pressure and its temperature rises from 300 K to 450 K. What is the value of (final density) ÷ (initial density) of the air? Treat the air as an ideal gas.

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At constant pressure ρ = pM/(RT) so ρ is inversely proportional to the absolute temperature: ρf/ρi = 300/450 = 0.667, option B. A inverts the ratio. C squares the ratio (0.6672). D squares the inverted ratio.✓ 1Answer B

Answer: B

Exam tip: For a fixed mass at constant pressure the volume is proportional to T, so the density is proportional to 1/T; always use kelvin.

15IBP3H-B1-20r23-02Molecular model and density Paper 1AEasy1 mark

A fixed mass of an ideal gas in a flexible bag on a space probe is heated at constant pressure until its thermodynamic temperature doubles. The initial density of the gas is ρ. What is the final density of the gas?

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At constant pressure the volume is proportional to the thermodynamic temperature, so the volume doubles. The mass is fixed, so the density halves. The option stating no change wrongly assumes density depends only on mass, the option 2ρ inverts the proportionality, and the option ρ/4 wrongly squares the effect of the temperature change.✓ 1Answer A

Answer: A

Exam tip: For a fixed mass, density is inversely proportional to volume; work out what happens to the volume first.

16IBP3H-B1-20r23-03Convection Paper 1AEasy1 mark

Air near the heater of a coffee roaster rises through the roasting chamber and cooler air flows in to replace it. Which statement best explains why the heated air rises?

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Heating makes the air expand at roughly constant pressure, so its density falls. The denser surrounding air exerts a larger upthrust, so the warm air rises and sets up a convection current. The molecules themselves do not change mass, heat is not a substance with a preferred direction, and the description of energy passing between fixed neighbours is conduction, not convection.✓ 1Answer D

Answer: D

Common error: In convection it is the density of the fluid that changes; the mass of each molecule stays the same.

17IBP3H-B1-20r23-04Mixing and thermal equilibrium Paper 1AEasy1 mark

A drum contains 0.25 kg of coffee beans with a specific heat capacity of 1600 J kg−1 K−1. The beans are heated from 20 °C to 220 °C. What is the energy transferred to the beans?

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Energy = mcΔθ = 0.25 × 1600 × (220 − 20) = 8.0 × 104 J = 80 kJ. The 40 kJ option uses the mean temperature of 120 °C minus 20 °C instead of the temperature change, 88 kJ uses the final temperature 220 °C instead of the change, and 320 kJ leaves out the mass.✓ 1Answer B

Answer: B

Exam tip: Use the temperature change Δθ, not the final temperature; a change in °C equals the same change in kelvin.

18B-1A-01Stefan–Boltzmann law Paper 1AMedium1 mark

Two black bodies are spheres. Sphere X has radius r and temperature T. Sphere Y has radius 2r and temperature T/2.

What is the ratio (power radiated by X) / (power radiated by Y)?

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Marking pointMarkNotes
Step 1P = σAT⁴ with A = 4πr², so P ∝ r²T⁴.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2For Y: (2r)²(T/2)⁴ = 4 × (1/16) = 1/4 of the value for X.—
Step 3So PX/PY = 1/(1/4) = 4.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is the ratio the other way up, PY/PX.
  • BThe area and temperature effects would cancel only if Y were at T/√2, not T/2.
  • CCorrect: Y has 4 times the area but (1/2)⁴ = 1/16 of the emission per unit area, so it radiates a quarter as much.
  • DThis uses the fourth power of the temperature ratio but forgets that Y has four times the surface area.
19B-1A-02Luminosity in solar units Paper 1AMedium1 mark

A star has a luminosity of 25L⊙, where L⊙ is the luminosity of the Sun. The apparent brightness of the star at the Earth is 1.0 × 10−10 times the apparent brightness of the Sun at the Earth.

What is the distance of the star from the Earth?

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Marking pointMarkNotes
Step 1From b = L/4πd², the distance is d = √(L/4πb), so d ∝ √(L/b).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2The Sun is 1 AU from the Earth, so d/(1 AU) = √[(L/L⊙)/(b/b⊙)] = √(25/1.0 × 10−10) = √(2.5 × 1011).—
Step 3d = 5.0 × 105 AU (about 2.4 pc).✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis inverts the luminosity ratio: √[(1/25)/(1.0 × 10−10)] = 2.0 × 104 AU. A more luminous star must be further away to appear equally faint.
  • BThis ignores the difference in luminosity, as if the star were identical to the Sun: √(1/1.0 × 10−10) = 1.0 × 105 AU.
  • CCorrect: d ∝ √(L/b), so √(25/1.0 × 10−10) = 5.0 × 105 AU.
  • DThis uses an inverse (not inverse-square) law for the brightness, omitting the square root: 25/1.0 × 10−10 = 2.5 × 1011 AU.
20B-1A-17Specific latent heat Paper 1AMedium1 mark

Water in an electric kettle is kept boiling by a 2.20 kW heating element. 90 % of the energy supplied by the element is transferred to the water; the rest is lost to the surroundings. The graph shows how the mass m of water in the kettle varies with time t once it is boiling.

What is the specific latent heat of vaporization of water given by these data?

050100150200250300t / s0.900.951.001.051.101.151.201.25m / kg
Mass of water in the kettle against time (drawn to scale). The m axis does not start at zero.
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Step 1From the gradient: mass vaporized in 300 s = 1.200 − 0.930 = 0.270 kg, a rate of 9.0 × 10−4 kg s−1.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Power transferred to the water = 0.90 × 2200 = 1980 W.—
Step 3L = (energy per second)/(mass vaporized per second) = 1980 ÷ (9.0 × 10−4) = 2.2 × 106 J kg−1.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis divides the energy supplied in 300 s by the mass remaining (0.930 kg) instead of the mass vaporized: 1980 × 300/0.930 = 6.4 × 105 J kg−1.
  • BCorrect: 0.90 × 2200 W ÷ 9.0 × 10−4 kg s−1 = 2.2 × 106 J kg−1.
  • CThis ignores the energy lost: 2200 ÷ (9.0 × 10−4) = 2.4 × 106 J kg−1.
  • DThis divides by the efficiency instead of multiplying: (2200/0.90) ÷ (9.0 × 10−4) = 2.7 × 106 J kg−1.
21B-1A-22Mixing and thermal equilibrium Paper 1AMedium1 mark

A metal block of mass m and specific heat capacity c, at a temperature θ in degrees Celsius, is placed in a mass 2m of a liquid at 0 °C. The specific heat capacity of the liquid is 4c. No energy is transferred to the surroundings.

What is the final temperature, in degrees Celsius?

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Step 1Energy lost by the block = energy gained by the liquid: mc(θ − θf) = (2m)(4c)(θf − 0).—All 3 steps must be completed — there is no mark for a part-answer.
Step 2θ − θf = 8θf, so 9θf = θ.—
Step 3θf = θ/9.✓ 1Answer A

Answer: A  ·  3 stages of work, one mark

Every option, and why

  • ACorrect: the liquid has 8 times the heat capacity of the block, so the block falls by 8 parts and the liquid rises by 1 part.
  • BThis takes the temperature fall of the block to be the whole of θ: mcθ = 8mcθf.
  • CThis ignores the different masses: c(θ − θf) = 4cθf.
  • DThis ignores the different specific heat capacities: m(θ − θf) = 2mθf.
22B-1A-30Luminosity and apparent brightness Paper 1AMedium1 mark

Two stars X and Y have black-body spectra with the same peak wavelength. The radius of X is three times the radius of Y, and X is twice as far from the Earth as Y.

The apparent brightness of Y is b. What is the apparent brightness of X?

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Marking pointMarkNotes
Step 1Same peak wavelength, so by Wien's law the same surface temperature.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2L = σAT4 with A ∝ r2: LX = 32LY = 9LY.—
Step 3b = L/4πd2: bX = 9b/22 = 9b/4.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis takes the luminosity to be proportional to the radius instead of the surface area (×3, not ×9).
  • BCorrect: nine times the luminosity at twice the distance: 9/4.
  • CThis uses an inverse (not inverse-square) dependence on distance: 9/2.
  • DThis ignores the different distances, giving the ratio of luminosities.
23B-1A-31Molecular model and density Paper 1AMedium1 mark

At 100 °C and atmospheric pressure the density of liquid water is 958 kg m−3 and the density of steam is 0.598 kg m−3.

What is the best estimate of (mean separation of the molecules in the steam) / (mean separation of the molecules in the liquid water)?

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Step 1The same mass contains the same number of molecules, so the volume available to each molecule is inversely proportional to the density: Vsteam/Vliquid = 958/0.598 = 1.60 × 103.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Each molecule occupies a small cube of side equal to the mean separation, so the separation is proportional to (volume per molecule)1/3.—
Step 3Ratio of separations = (1.60 × 103)1/3 = 11.7 ≈ 12: in the liquid the molecules are almost touching, in the gas they are about ten molecular diameters apart.✓ 1Answer B

Answer: B  ·  3 stages of work, one mark

Every option, and why

  • AThis is the ratio inverted, 1/11.7 = 0.085: the molecules are further apart in the gas, not closer.
  • BCorrect: the separation scales as the cube root of the volume per molecule, (958/0.598)1/3 = 12.
  • CThis takes the square root, (1.60 × 103)1/2 = 40, as if the molecules were spread over an area rather than a volume.
  • DThis is the ratio of the densities, which equals the ratio of the volumes per molecule, not of the separations.
24B-1A-34Heating curves Paper 1AMedium1 mark

A solid sample is heated at constant power with negligible energy losses. The graph shows how its temperature varies with time as it warms, melts and then warms as a liquid.

What is (specific heat capacity of the liquid) / (specific heat capacity of the solid)?

012345678910t (min)020406080100120140temperature (°C)
Temperature of the sample against time (drawn to scale)
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Step 1With constant power P, the energy supplied in time t is Pt = mcΔT, so c ∝ t/ΔT for the same sample.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Solid: 60 K in 2.0 min. Liquid: 50 K in 3.0 min.—
Step 3cliquid/csolid = (3.0/50)/(2.0/60) = 1.8.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis is the ratio of the gradients, (50/3.0)/(60/2.0) = 0.56: a larger specific heat capacity gives a smaller gradient, not a larger one.
  • BThis is the ratio of the temperature rises, 60/50 = 1.2, ignoring the different heating times.
  • CThis is the ratio of the heating times, 3.0/2.0 = 1.5, ignoring the different temperature rises.
  • DCorrect: the liquid needs 0.060 min K−1 and the solid 0.033 min K−1, a ratio of 1.8.
25B-1A-36Black-body spectra Paper 1AMedium1 mark

The heating element of an electric grill behaves as a grey body of constant emissivity and surface area. Its temperature is raised from 327 °C to 627 °C.

Which row gives the ratio (peak wavelength of the emitted radiation at 627 °C) / (peak wavelength at 327 °C) and the ratio (power emitted at 627 °C) / (power emitted at 327 °C)?

Ratio of peak wavelengthsRatio of powers
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Step 1Convert to kelvin: 327 °C = 600 K and 627 °C = 900 K, so the absolute temperature rises by a factor of 1.5.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Wien's displacement law: λmaxT = constant, so the peak wavelength falls by the same factor: ratio 600/900 = 0.67.—
Step 3Stefan–Boltzmann law with constant e and A: P ∝ T⁴, so the power rises by 1.5⁴ = 5.1.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis uses the Celsius temperatures: 327/627 = 0.52 and (627/327)⁴ = 14.
  • BThis takes the power to be proportional to T instead of T⁴: 900/600 = 1.5.
  • CCorrect: λmax falls by the factor 1.5 and the power rises by 1.5⁴ = 5.1.
  • DThis inverts Wien's law: a hotter body emits its peak at a shorter wavelength, not a longer one.
26B-1A-63Stefan–Boltzmann law Paper 1AMedium1 mark

Two solid spheres P and Q are made of the same metal and have the same black surface. The radius of Q is twice the radius of P. Both spheres are at the same temperature, suspended in an evacuated enclosure whose walls are so cold that the radiation absorbed by the spheres is negligible. The temperature is uniform throughout each sphere.

What is (initial rate of fall of temperature of P)/(initial rate of fall of temperature of Q)?

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Marking pointMarkNotes
Step 1Power radiated P = σ(4πr2)T4, so at the same temperature the power is proportional to r2.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy needed per kelvin = mc = ρ(4/3)πr3c, proportional to r3.—
Step 3Rate of fall of temperature = P/(mc) ∝ r2/r3 = 1/r, so P cools twice as fast as Q: ratio = 2.✓ 1Answer C

Answer: C  ·  3 stages of work, one mark

Every option, and why

  • AThis is the ratio of the powers radiated (the area ratio), which ignores that P has only 1/8 of the heat capacity of Q.
  • BThis is the correct ratio upside down: the smaller sphere has more surface area per unit mass, so it cools faster, not slower.
  • CCorrect: the radiated power scales as r2 but the heat capacity as r3, so the rate of fall of temperature scales as 1/r.
  • DThis is the ratio of the heat capacities (masses), which ignores that P radiates only a quarter of the power radiated by Q.
27IBP2H-010-00-02Luminosity and apparent brightness Paper 1AMedium1 mark

Star X has twice the surface temperature of star Y and half the radius of star Y. Both stars may be treated as black bodies. What is \(\frac{L_X}{L_Y}\), the ratio of the luminosity of X to the luminosity of Y?

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L = σ4πR2T4, so L ∝ R2T4. The ratio is (1/2)2 × 24 = (1/4) × 16 = 4. Option 8 uses R instead of R2. Option 16 ignores the radius. Option 2 uses T2 together with R.✓ 1Answer B

Answer: B

Common error: Luminosity depends on surface area, which goes as R², not R. Square the radius ratio before multiplying by the fourth power of the temperature ratio.

28IBP2H-010-00-03Luminosity and apparent brightness Paper 1AMedium1 mark

Star P has a luminosity of 9 L☉ and is at a distance d from Earth. Star Q has a luminosity of 1 L☉ and is at a distance d/3 from Earth. What is \(\frac{b_P}{b_Q}\), the ratio of the apparent brightness of P to the apparent brightness of Q?

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Marking pointMarkNotes
b = L/(4πd2). For P, bP ∝ 9/d2. For Q, bQ ∝ 1/(d/3)2 = 9/d2. The ratio is therefore 1: the factor of 9 in luminosity is exactly cancelled by the factor of 3 in distance. Option 3 uses an inverse-distance law instead of an inverse-square law. Option 9 ignores the distances. Option 81 multiplies the two factors instead of letting them cancel.✓ 1Answer A

Answer: A

Exam tip: Write b ∝ L/d² for each star separately, then divide. Ratio questions become quick and reliable this way.

29IBP2H-010-00-04Luminosity, apparent brightness and black-body radiation Paper 1AMedium1 mark

A medical thermal-imaging camera is used to detect radiation from a patient's skin. The skin is at 34 °C. Wien's displacement constant is 2.9 × 10−3 m K. What is the approximate wavelength at which the skin's emission peaks?

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Marking pointMarkNotes
Wien's law needs an absolute temperature, so T = 34 + 273 = 307 K. Then λmax = 2.9 × 10−3 / 307 = 9.4 × 10−6 m = 9.4 μm, which is in the infrared. The distractors come from these errors:
• 85 μm comes from using 34 K instead of 307 K.
• 9.4 mm comes from converting 10−6 m as if it were 10−3 m, confusing micro with milli.
• 0.94 μm is a power-of-ten slip in the division.
✓ 1Answer B

Answer: B

Common error: Using the Celsius temperature in Wien's law gives a peak wavelength about nine times too large. Always convert to kelvin first.

30IBP2H-010-00-08Luminosity in solar units Paper 1AMedium1 mark

A main-sequence star has a luminosity of 100 L☉ and a surface temperature twice that of the Sun. What is the radius of the star in terms of the solar radius R☉?

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Marking pointMarkNotes
L ∝ R2T4, so R/R☉ = √(L/L☉) × (T☉/T)2 = √100 × (1/2)2 = 10 × 1/4 = 2.5. 6.25 R☉ gives R2 rather than R. 5.0 R☉ uses T2 instead of T4 in the luminosity relation. 25 R☉ divides 100 by 4 without taking the square root.✓ 1Answer A

Answer: A

Exam tip: Working in solar units removes σ and 4π completely. Rearrange the proportionality to make R the subject before substituting.

31IBP2H-010-01-01Conduction Paper 1AMedium1 mark

The wall of a laboratory freezer is made of two flat layers, X and Y, of equal thickness and equal area, in good thermal contact. The thermal conductivity of X is three times that of Y. Energy is conducted through the wall at a steady rate.
What is the ratio \(\frac{\text{temperature difference across X}}{\text{temperature difference across Y}}\)?

X3kYkddequal area A, steady flow
Diagram not to scale
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Marking pointMarkNotes
In the steady state no energy builds up in either layer, so the same power passes through both. From \(P = kA\frac{\Delta T}{\Delta x}\) with the same \(P\), \(A\) and \(\Delta x\), \(\Delta T \propto \frac{1}{k}\). X has three times the conductivity, so its temperature difference is one third of Y's. The ratio is 1/3. Choosing 1 assumes equal thickness means equal temperature drop. Choosing 3 inverts the relationship. Choosing 1/9 wrongly squares the ratio.✓ 1Answer B

Answer: B

Exam tip: For layers in series, the layer with the lowest conductivity (the best insulator) has the largest temperature drop across it.

32IBP2H-010-01-04Mixing and thermal equilibrium Paper 1AMedium1 mark

Equal masses of water at 20 °C and an engine coolant at 80 °C are mixed in an insulated container. The specific heat capacity of the coolant is half that of water. No energy is transferred to the surroundings or the container.
What is the final temperature of the mixture?

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Marking pointMarkNotes
The energy lost by the coolant equals the energy gained by the water: \(m\frac{c}{2}(80 - T) = mc(T - 20)\). The mass and \(c\) cancel, giving \(80 - T = 2T - 40\), so \(T = 40\) °C. The water, with the larger heat capacity, changes temperature by less. The value 50 °C assumes equal specific heat capacities. The value 60 °C reverses which substance has the larger heat capacity. The masses are equal, so they cancel and are not needed.✓ 1Answer A

Answer: A

Exam tip: The final temperature is always closer to the starting temperature of the body with the larger heat capacity, mc.

33IBP2H-010-01-05Molecular model and density Paper 1AMedium1 mark

The density of liquid water is 1000 kg m−3. The density of steam at 100 °C and atmospheric pressure is 0.60 kg m−3.
What is the best estimate of \(\frac{\text{mean molecular spacing in steam}}{\text{mean molecular spacing in liquid water}}\)?

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Marking pointMarkNotes
The number of molecules per unit volume is proportional to density. Each molecule therefore occupies a volume proportional to 1/ρ. The mean spacing is the cube root of this volume. The spacing ratio is (1000/0.60)1/3 = (1700)1/3 ≈ 12. The value 1700 is the volume ratio, not the spacing ratio. The value 41 uses a square root instead of a cube root. The value 1 confuses molecular size with molecular spacing.✓ 1Answer B

Answer: B

Common error: Density ratios compare volumes. Take the cube root to compare distances between molecules.

34IBP2H-010-01-07Luminosity and apparent brightness Paper 1AMedium1 mark

Two stars, X and Y, have the same luminosity. The peak wavelength in the spectrum of X is twice that of Y. Both stars behave as black bodies.
What is \(\frac{\text{radius of X}}{\text{radius of Y}}\)?

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Marking pointMarkNotes
By Wien's law, \(\lambda_{max} \propto \frac{1}{T}\), so \(T_X = \frac{T_Y}{2}\). Equal luminosities give \(R_X^2 T_X^4 = R_Y^2 T_Y^4\). Substituting gives \(R_X^2 = 16R_Y^2\), so \(R_X = 4R_Y\). The value 16 forgets to take the square root. The value 2 assumes radius is proportional to peak wavelength. The value 1/4 inverts the ratio by assuming the longer-wavelength star is hotter.✓ 1Answer C

Answer: C

Common error: A longer peak wavelength means a cooler surface. The cooler star must be larger to have the same luminosity.

35IBP3H-B1-20r17-03Evaporation and boiling Paper 1AMedium1 mark

A shallow dish of water in a beehive loses water by evaporation and the remaining water becomes cooler. Which explanation is correct?

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Marking pointMarkNotes
Option D is correct: the faster molecules escape, so the average kinetic energy (and hence temperature) of those that remain is lower. A is wrong because the energy to escape comes from the liquid itself, which is why the liquid cools. B is wrong because pressure above the liquid is not the cause of the temperature drop. C is wrong because losing molecules also removes their share of the internal energy; with the same energy shared among fewer molecules the temperature would rise, not fall.✓ 1Answer D

Answer: D

Common error: Do not say that evaporation 'takes heat from the surroundings' as the cause of the cooling of the liquid: the key idea is the selective loss of the most energetic molecules.

36IBP3H-B1-20r17-05Heating curves Paper 1AMedium1 mark

A solid of mass m is heated by a heater of constant power. Its temperature rises from 20 °C to 60 °C in 50 s. It then melts at 60 °C, which takes a further 100 s. After it has melted, the liquid rises from 60 °C to 100 °C in 150 s. Assume all the energy from the heater goes into the sample. What is the value of (specific heat capacity of the liquid) ÷ (specific heat capacity of the solid)?

050100150200250300020406080100120t / sθ / °C
Temperature of the solid against time (drawn to scale)
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For equal temperature rises of 40 K with the same power and mass, c is proportional to the heating time: 150 s ÷ 50 s = 3, option C. A inverts the ratio. B takes the difference of the times, 100 s ÷ 50 s. D includes the 100 s melting plateau, (150 + 100) ÷ 50, but that time is spent on latent heat and not on raising the temperature.✓ 1Answer C

Answer: C

Common error: The flat part of a heating curve is latent heat only; do not include it when comparing specific heat capacities.

37IBP3H-B1-20r17-06Evaporation and boiling Paper 1AMedium1 mark

A beekeeper dries a thin film of nectar in a warm hive. Which single change would decrease the rate at which water evaporates from the film?

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Marking pointMarkNotes
Option C is correct: air that already holds a lot of water vapour accepts fewer extra molecules, because more water molecules return to the liquid and the net evaporation rate falls. A increases the fraction of molecules with enough energy to escape. B increases the surface area from which molecules can escape. D removes vapour near the surface so fewer molecules return.✓ 1Answer C

Answer: C

Exam tip: Four factors change the evaporation rate: temperature, surface area, humidity and air movement; check each direction carefully.

38IBP3H-B1-20r23-01Black-body spectra Paper 1AMedium1 mark

The surface of a space probe's radiator behaves as a black body. Its thermodynamic temperature is increased to three times its original value. The original peak wavelength was λ and the original total power emitted was P. Which row gives the new peak wavelength and the new total power emitted?

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Marking pointMarkNotes
Wien's law gives λmax ∝ 1/T, so tripling T divides the peak wavelength by 3. The Stefan–Boltzmann law gives P ∝ T4, so the power increases by 34 = 81. The first option uses a T² dependence for power, the second reverses the direction of the shift in peak wavelength, and the fourth makes both errors.✓ 1Answer C

Answer: C

Common error: The peak wavelength varies as 1/T, but the total power varies as T to the fourth power; do not swap the two relationships.

39IBP3H-B1-20r23-05Specific latent heat Paper 1AMedium1 mark

A sample of 0.30 kg of ice at 0 °C is taken from the deck of a bridge. A heater of power 500 W, with all of its energy going to the ice, melts the sample completely. The specific latent heat of fusion of ice is 3.3 × 105 J kg−1. What is the time taken?

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Marking pointMarkNotes
Energy needed = mL = 0.30 × 3.3 × 105 = 9.9 × 104 J, and time = energy ÷ power = 9.9 × 104 ÷ 500 = 198 s. The 660 s option omits the mass, 2200 s divides by the mass instead of multiplying, and 5.1 × 10−3 s inverts the whole calculation.✓ 1Answer D

Answer: D

Common error: Check units: time is energy divided by power, so a very small answer in seconds for melting several hundred grams of ice should prompt a recheck.

40IBP3H-B1-20r23-07Black-body spectra Paper 1AMedium1 mark

The graph of intensity against wavelength for a hot black-body surface has a single peak. The surface temperature is raised. Which statement about the new graph is correct?

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Marking pointMarkNotes
Wien's law shows the peak moves to a shorter wavelength as the temperature rises, and the area under the curve represents the total emitted intensity, which increases as T4. A longer-wavelength peak is the wrong direction, a decreasing area contradicts the Stefan–Boltzmann law, and a fixed peak wavelength ignores Wien's law.✓ 1Answer B

Answer: B

Exam tip: The area under an intensity–wavelength graph represents total intensity; the peak position tells you the temperature through Wien's law.

41IBP3H-B1-20r23-08Mixing and thermal equilibrium Paper 1AMedium1 mark

Hot coffee of mass 0.30 kg at 80 °C is mixed in an insulated cup with cold milk of mass 0.20 kg at 20 °C. Assume the coffee and the milk have the same specific heat capacity and that no energy is transferred to the surroundings. What is the final temperature of the mixture?

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Marking pointMarkNotes
Energy lost by the coffee equals energy gained by the milk: 0.30(80 − T) = 0.20(T − 20), so 24 − 0.30T = 0.20T − 4, giving 0.50T = 28 and T = 56 °C. The option 50 °C takes the simple mean of the two temperatures, 44 °C swaps the masses in the weighted average, and 36 °C is the temperature rise of the milk rather than the final temperature.✓ 1Answer C

Answer: C

Common error: The final temperature is a mass-weighted average, closer to the temperature of the larger mass; check that your answer is between the two starting values.

42B-1A-18Conduction Paper 1AHard1 mark

Two rods of the same cross-sectional area are joined end to end. Rod X has length L and thermal conductivity 2k; rod Y has length 2L and thermal conductivity k. The free end of X is at 100 °C and the free end of Y at 0 °C, and the sides are insulated.

What is the steady temperature of the junction?

XYlength Lconductivity 2klength 2Lconductivity k100 °C0 °Cjunction θ = ?sides insulated (dashed)
Diagram not to scale
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Marking pointMarkNotes
Step 1In the steady state the same rate of energy flow passes through both rods: kAΔT/L is equal for X and Y.—All 3 steps must be completed — there is no mark for a part-answer.
Step 2ΔT ∝ L/k: for X, L/2k; for Y, 2L/k — so the temperature drop across Y is four times that across X.—
Step 3Total drop 100 K shared 1 : 4, so X drops 20 K and the junction is at 80 °C.✓ 1Answer D

Answer: D  ·  3 stages of work, one mark

Every option, and why

  • AThis is the drop across X, not the junction temperature.
  • BThis would apply if the rods had equal thermal resistance.
  • CThis uses only the ratio of the thermal conductivities (or only the ratio of the lengths): the thermal resistances are then taken as 1 : 2, so X takes 1/3 of the drop and the junction is at 100 − 33 = 67 °C.
  • DCorrect: the thermal resistances (L/kA) are in the ratio 1 : 4, so X takes 1/5 of the temperature difference.
43IBP2H-010-00-06Heating curves Paper 1AHard1 mark

A solid is heated by a heater of constant power, and no energy is lost to the surroundings. While the substance is solid, its temperature rises at 2.0 K s−1. When it reaches its melting point, it takes 120 s to melt completely. What is the ratio \(\frac{L_f}{c_s}\) of the specific latent heat of fusion to the specific heat capacity of the solid?

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Marking pointMarkNotes
While the substance is solid, P = mcs(ΔT/Δt) = mcs × 2.0. While it melts, P × 120 = mLf. Dividing the second equation by the first gives Lf/cs = 120 × 2.0 = 240 K. 60 K divides the melting time by the rate instead of multiplying. 120 K ignores the heating rate. The value in K−1 is the inverse ratio.✓ 1Answer C

Answer: C

Exam tip: Check the units: L/c has units of (J kg⁻¹)/(J kg⁻¹ K⁻¹) = K. This rules out the option with units of K⁻¹ immediately.

44IBP2H-010-00-07Luminosity in solar units Paper 1AHard1 mark

A small spherical space probe behaves as a black body and has no internal power source. At distance d from the Sun its equilibrium surface temperature is T. The probe then moves to a distance 4d from the Sun. What is its new equilibrium temperature?

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Marking pointMarkNotes
The power absorbed is proportional to the intensity of sunlight, which goes as 1/d2. The power emitted goes as T4. At equilibrium these are equal, so T4 ∝ 1/d2 and therefore T ∝ d−1/2. Moving to 4d multiplies T by 4−1/2 = 1/2. T/16 assumes T ∝ 1/d2. T/4 assumes T ∝ 1/d. T/√2 uses the fourth root of 4 instead of its square root.✓ 1Answer C

Answer: C

Common error: Do not assume temperature follows the inverse-square law. It is the absorbed intensity that does. Equate absorbed and emitted power first, then take the fourth root.

45IBP2H-010-01-02Net radiation exchange Paper 1AHard1 mark

A small black sphere at a temperature of 400 K sits inside a large evacuated enclosure whose walls are at 200 K. The net power radiated by the sphere is P.
The sphere is heated to 800 K. The walls stay at 200 K. What is the new net power radiated by the sphere?

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Marking pointMarkNotes
Net power is \(\sigma A(T^4 - T_0^4)\). Working in units of 100 K, the fourth powers are 24 = 16, 44 = 256 and 84 = 4096. The ratio of net powers is (4096 − 16)/(256 − 16) = 4080/240 = 17, so the new net power is 17P. The value 16P ignores the energy absorbed from the surroundings. The value 3P treats net power as proportional to the temperature difference. The value 81P raises the temperature difference to the fourth power.✓ 1Answer C

Answer: C

Common error: Use the difference of fourth powers, T⁴ − T₀⁴. Do not use the fourth power of the difference, (T − T₀)⁴.

46IBP3H-B1-20r17-07Specific latent heat Paper 1AHard1 mark

A hospital warmer contains an 80 W heater that is switched on for 5.0 minutes. Only 60% of the energy it transfers reaches a block of ice already at 0 °C, which melts. The specific latent heat of fusion of ice is 3.34 × 105 J kg−1. What is the mass of ice that melts?

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Energy reaching the ice = 0.60 × 80 W × 300 s = 14 400 J, so m = 14 400 ÷ 3.34 × 105 = 0.0431 kg, option A. B ignores the 60% efficiency. C divides by 0.60 instead of multiplying. D uses 5.0 for the time instead of 300 s (the time was not converted to seconds).✓ 1Answer A

Answer: A

Common error: Convert minutes to seconds before using P = E/t, and multiply by the efficiency to find the useful energy.

47IBP3H-B1-20r17-08Mixing and thermal equilibrium Paper 1AHard1 mark

An insulated container holds 0.50 kg of water at 30 °C. A block of ice of mass 0.10 kg at 0 °C is added and all of it melts. The specific heat capacity of water is 4.18 × 103 J kg−1 K−1 and the specific latent heat of fusion of ice is 3.34 × 105 J kg−1. Energy transfer to the container and surroundings is negligible. What is the final temperature?

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Energy lost by the warm water = energy to melt the ice + energy to warm the melted ice: 0.50c(30 − T) = 0.10L + 0.10cT, which gives T = (0.50 × 4180 × 30 − 0.10 × 3.34 × 105) ÷ (0.60 × 4180) = 11.7 °C, option A. B forgets that the melted ice also has to be warmed from 0 °C to T. C is the mean of 0 °C and 30 °C, ignoring the different masses and the latent heat. D ignores the latent heat altogether: 0.50 × 30 ÷ 0.60 = 25.0 °C.✓ 1Answer A

Answer: A

Common error: The melted ice becomes water at 0 °C and must then be warmed to the final temperature; include both stages.

48IBP3H-B1-20r23-06Luminosity and apparent brightness Paper 1AHard1 mark

Two spherical radiators on a space probe emit as black bodies. Radiator X has radius r and thermodynamic temperature T. Radiator Y has radius 2r and thermodynamic temperature T/2. What is the value of (power emitted by Y) ÷ (power emitted by X)?

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Power = σAT4 with A = 4πr², so power ∝ r²T4. For Y the factor is 2² × (1/2)4 = 4/16 = 1/4. The option 1/8 treats area as proportional to radius, 1/16 ignores the change in radius, and 1 results from using T² instead of T4.✓ 1Answer A

Answer: A

Common error: Surface area scales with radius squared, so doubling the radius quadruples the area, not doubles it.

49B-1B-01Stefan–Boltzmann law Paper 1BMedium7 marks

A student tests the Stefan–Boltzmann law using the tungsten filament of a small lamp in an evacuated bulb. The kelvin temperature T of the filament is found from its resistance, and the electrical power P supplied is assumed to be radiated. The filament is modelled as a grey body:

P = eσAT⁴

where e is the emissivity and A = 1.0 × 10−5 m² is the surface area of the filament. The table and the graph show log₁₀ P against log₁₀ T with the line of best fit. The power is uncertain by ±2 %, the temperature by ±1.5 % and the area by ±4 %.

T / K15001800210024002700
P / W1.162.364.447.4912.1
log₁₀(T / K)3.1763.2553.3223.3803.431
log₁₀(P / W)0.0640.3730.6470.8741.083
3.153.203.253.303.353.403.45log₁₀(T / K)0.00.20.40.60.81.01.2log₁₀(P / W)
log₁₀ P against log₁₀ T with the line of best fit (drawn to scale)
(a)
(i)

Determine the gradient of the line and state whether the data support the Stefan–Boltzmann law.

(2)
(b)
(i)

Using the data point at 2700 K, determine e.

(2)
(c)
(i)

Determine the absolute uncertainty in e.

(2)
(d)
(i)

Published data give the emissivity of tungsten at 2700 K as 0.33. Comment on this value in the light of your answers to (b) and (c).

(1)
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Marking pointMarkNotes
Part (a)(i)
Gradient from the full line, e.g. (1.083 − 0.064)/(3.431 − 3.176) = 4.0✓ 1Accept 3.9–4.1. Points used must be far apart on the line (a short triangle is not accepted).
Gradient ≈ 4, so P ∝ T⁴, as the law predicts✓ 1The conclusion must refer to the value 4 (the exponent).
Part (b)(i)
e = P/(σAT⁴) = 12.1/(5.67 × 10−8 × 1.0 × 10−5 × 2700⁴)✓ 1Accept use of a point on the line.
e = 0.40✓ 1Accept 0.39–0.41. Allow ECF from (a). (a candidate who found a different exponent and uses it consistently may score the substitution mark only)
Part (c)(i)
% uncertainty = 2 + 4 × 1.5 + 4 = 12 %✓ 1The temperature uncertainty must be multiplied by 4 (power law).
Δe = 0.12 × 0.40 = ±0.05 (e = 0.40 ± 0.05)✓ 1Absolute uncertainty to 1 s.f. Allow ECF from (b). Award [1 max] for 7.5 % (temperature not multiplied by 4) giving ±0.03.
Part (d)(i)
0.33 lies outside 0.35–0.45, so the difference is not explained by the random uncertainties; a systematic error must be present (e.g. part of P is lost by conduction through the supports, so the radiated power is less than P and e is overestimated)✓ 1Allow ECF from (b) and (c). The comparison must use the candidate's own range.

Answers: (a)(i) gradient = 4.0  ·  (b)(i) 0.40  ·  (c)(i) ±0.05 (the remaining parts are explanations — see the table above)

50B-1B-08Specific heat capacity Paper 1BMedium7 marks

A cardboard tube, closed at both ends, contains 0.80 kg of lead shot. When the tube is quickly inverted, the shot falls through a mean height of 1.10 m and stops at the bottom. The student inverts the tube N times, then pours the shot into a polystyrene cup and measures the rise in temperature ΔT with a digital thermometer of resolution 0.1 K. The shot is cooled back to room temperature before each run. The table and the graph show the results with the line of best fit.

N255075100125150
ΔT / K1.93.55.57.19.010.9
020406080100120140160N024681012ΔT / K
Temperature rise ΔT against number of inversions N with the line of best fit (drawn to scale)
(a)
(i)

Determine the gradient of the line.

(1)
(b)
(i)

Show that the gradient is equal to gh/c, where c is the specific heat capacity of lead, and hence determine c.

(2)
(c)
(i)

The accepted value of c for lead is 128 J kg−1 K−1. Explain the difference between this value and your answer to (b).

(2)
(d)
(i)

Each value of ΔT is the difference of two readings of the thermometer. Determine the smallest number of inversions needed for the percentage uncertainty in ΔT to be no more than 2 %, using the same thermometer.

(2)
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Marking pointMarkNotes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. 10.8/150 = 0.072 K✓ 1Accept 0.070–0.074 K (per inversion). Unit not required.
Part (b)(i)
For each fall mgh = mcΔT1, so the rise per inversion is gh/c (the mass cancels)✓ 1All the gravitational potential energy is assumed to become internal energy of the shot.
c = 9.81 × 1.10/0.072 = 150 J kg−1 K−1✓ 1Allow ECF from (a). Accept 146–154 J kg−1 K−1.
Part (c)(i)
The experimental value is larger, because the measured temperature rise is smaller than the energy transfer mgh per fall would produce✓ 1Allow ECF from (b). The direction must agree with the candidate's own value.
Some of the energy is transferred to the tube, to the air and as sound, and the shot loses energy while it is poured into the cup (and some shot does not fall the full height)✓ 1Any one valid mechanism linked to a smaller ΔT.
Part (d)(i)
ΔT uncertainty = 2 × 0.1 = 0.2 K, so ΔT ≥ 0.2/0.02 = 10 K✓ 1The doubled reading uncertainty is required.
N ≥ 10/0.072 ≈ 139 inversions✓ 1Allow ECF from (a). Accept 135–145.

Answers: (a)(i) 0.072 K  ·  (b)(i) 150 J kg−1 K−1  ·  (d)(i) ≈ 139 (the remaining parts are explanations — see the table above)

51B-1B-11Density of a gas Paper 1BMedium7 marks

A student determines the density of air. A rigid glass flask with a tap has an internal volume of 1.25 × 10−3 m³ (±1 %). The flask is connected to a vacuum pump and a pressure gauge. The student reduces the pressure p of the air in the flask in steps; at each step the tap is closed and the mass m of the flask and the air inside it is measured on a balance. The temperature of the air in the flask is 20 °C throughout.

The table and the graph show m against p with the line of best fit.

p / kPa10080604020
m / g153.807153.506153.210152.910152.614
020406080100120p / kPa152.2152.4152.6152.8153.0153.2153.4153.6153.8154.0m / g
Mass m of the flask and the air inside it against the pressure p, with the line of best fit (drawn to scale). The m axis does not start at zero.
(a)
(i)

Determine the gradient of the line. Give a unit for your answer.

(2)
(ii)

Explain, using the ideal gas law, why the graph is a straight line that does not pass through the origin.

(1)
(b)
(i)

Determine the density of air at a pressure of 100 kPa and a temperature of 20 °C.

(2)
(c)
(i)

The uncertainty in the gradient is ±2 %. Determine the molar mass of air, with its absolute uncertainty.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (153.806 − 152.315)/(100 − 0) = 0.0149✓ 1Accept 0.0145–0.0153.
Unit g kPa−1✓ 1Accept equivalent units, e.g. 1.49 × 10−8 kg Pa−1. Unit mark is independent of the value.
Part (a)(ii)
At constant V and T the amount of air n = pV/RT is proportional to p, so the mass of air increases linearly with p; the intercept (p = 0) is the mass of the evacuated flask (≈ 152.31 g)✓ 1Both the proportionality and the meaning of the intercept are needed.
Part (b)(i)
Mass of air at 100 kPa = gradient × 100 kPa = 1.49 g✓ 1Allow ECF from (a)(i). Accept the difference 153.806 − 152.315 read from the line. Using the reading 153.807 g alone (flask included) scores [0].
ρ = 1.49 × 10−3/1.25 × 10−3 = 1.19 kg m−3✓ 1Accept 1.16–1.22 kg m−3.
Part (c)(i)
M = ρRT/p (or M = gradient × RT/V) = 1.19 × 8.31 × 293/1.00 × 105 = 0.0290 kg mol−1✓ 1Allow ECF from (b). Accept 0.0283–0.0297 kg mol−1. The temperature must be in kelvin.
Percentage uncertainty = 2 % + 1 % = 3 %, so M = (0.0290 ± 0.0009) kg mol−1✓ 1Allow ECF. Accept ±0.0008 to ±0.001 kg mol−1. Award [0] for this mark if the uncertainty in the volume is omitted (±0.0006).

Answers: (a)(i) 0.0149 g kPa−1  ·  (a)(ii) mass of air ∝ p at constant V and T; intercept = mass of the evacuated flask  ·  (b)(i) 1.19 kg m−3  ·  (c)(i) (0.0290 ± 0.0009) kg mol−1 (the remaining parts are explanations — see the table above)

52B-1B-12Cooling curve and latent heat Paper 1BMedium7 marks

A test tube containing 0.025 kg of molten lauric acid is clamped in still air in a room at 20 °C. A temperature sensor records the temperature θ of the lauric acid every minute as it cools and freezes. The graph shows the data points and a smooth curve through them. The specific heat capacity of liquid lauric acid is 2.3 × 10³ J kg−1 K−1. The rate of energy loss to the room depends only on the temperature difference between the tube and the room.

0481216202428t / min20304050607080θ / °C
Temperature of the lauric acid against time (drawn to scale)
(a)
(i)

By drawing a tangent, determine the rate of fall of temperature, in K s−1, of the liquid as it reaches the freezing point.

(2)
(b)
(i)

Show that the rate of energy loss to the room at this instant is about 5 W.

(2)
(c)
(i)

Determine the specific latent heat of fusion of lauric acid.

(2)
(d)
(i)

The start and the end of the plateau can each be judged to ±0.5 min. Calculate the percentage uncertainty in the duration you used in (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Tangent drawn to the liquid curve at 44 °C (about 4.4 min); gradient ≈ −5.0 K min−1✓ 1Accept 4.5–5.5 K min−1 in magnitude.
Rate = 5.0/60 = 0.083 K s−1✓ 1Accept 0.075–0.092 K s−1. Sign may be omitted.
Part (b)(i)
P = mc × (ΔT/Δt) = 0.025 × 2.3 × 10³ × 0.083✓ 1Allow ECF from (a).
P = 4.8 W ≈ 5 W✓ 1The answer must be given to more figures than the target.
Part (c)(i)
Freezing lasts from about 4.4 min to 20.0 min: 15.6 min = 936 s; the loss rate stays at 4.8 W because the temperature stays at 44 °C✓ 1Accept 15.0–16.5 min.
L = Pt/m = 4.8 × 936/0.025 = 1.8 × 105 J kg−1✓ 1Allow ECF from (b). Accept 1.6–2.0 × 105 J kg−1.
Part (d)(i)
Uncertainty in the duration = 0.5 + 0.5 = ±1.0 min, i.e. 1.0/15.6 = 6 %✓ 1Allow ECF from (c). The uncertainties of the two readings add; ±0.5 min (3 %) scores [0].

Answers: (a)(i) 0.083 K s−1  ·  (b)(i) 4.8 W  ·  (c)(i) 1.8 × 105 J kg−1  ·  (d)(i) 6 % (the remaining parts are explanations — see the table above)

53B-1B-13Apparent brightness and the inverse-square law Paper 1BMedium7 marks

A student investigates how the apparent brightness b of the radiation from a small filament lamp, supplied with 12 W of electrical power, varies with distance. A radiation sensor is placed at distance d from the front of the glass bulb. The filament is an unknown distance d0 behind the front of the bulb, so the student models the lamp as a point source of luminosity L:

b = L / 4π(d + d0)²

The graph shows b−1/2 against d with the line of best fit.

d / m0.100.150.200.250.300.40
b / W m−249.224.314.79.726.994.02
b−1/2 / W−1/2 m0.1430.2030.2610.3210.3780.499
0.000.050.100.150.200.250.300.350.400.45d / m0.00.10.20.30.40.5b⁻⁰·⁵ / W⁻⁰·⁵ m
b−1/2 against d with the line of best fit (drawn to scale)
(a)
(i)

Determine the gradient of the line. Give a unit for your answer.

(2)
(b)
(i)

Determine L.

(2)
(c)
(i)

Determine d0.

(2)
(d)
(i)

Suggest why your answer to (b) is less than the electrical power supplied to the lamp.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (0.499 − 0.143)/(0.40 − 0.10) = 1.18✓ 1Accept 1.15–1.21.
Unit W−1/2✓ 1Accept W−0.5.
Part (b)(i)
b−1/2 = √(4π/L)(d + d0), so gradient = √(4π/L) and L = 4π/gradient²✓ 1
L = 4π/1.18² = 9.0 W✓ 1Allow ECF from (a). Accept 8.6–9.5 W.
Part (c)(i)
Intercept on the vertical axis ≈ 0.025 W−1/2 m = gradient × d0✓ 1Accept 0.020–0.030. Alternatively the extended line meets the d axis at d = −d0.
d0 = 0.025/1.18 = 0.021 m✓ 1Allow ECF from (a). Accept 0.017–0.026 m.
Part (d)(i)
9.0 W < 12 W because part of the energy leaves the hot bulb by conduction and convection to the air (and through the leads), not as radiation✓ 1Allow ECF from (b). Accept: the glass absorbs part of the infrared. "Energy is lost" alone scores [0].

Answers: (a)(i) 1.18 W−1/2  ·  (b)(i) 9.0 W  ·  (c)(i) 0.021 m (the remaining parts are explanations — see the table above)

54B-1B-24Conduction Paper 1BMedium7 marks

A student measures the thermal conductivity k of expanded polystyrene. A small heater inside a closed polystyrene box keeps the air in the box 20.0 K above room temperature. When conditions are steady, the electrical power P supplied to the heater is recorded. The experiment is repeated with boxes that have the same inner surface area A = 0.240 m² but walls of different thickness x.

The student models the walls as flat slabs, so that P = kAΔT/x + P0, where ΔT = 20.0 K and P0 is the power lost through the gap around the lid, which does not depend on x. The uncertainty in each value of P is ±0.3 W. The graph shows P against 1/x with error bars, the line of best fit (solid) and the steepest and shallowest lines consistent with the error bars (dashed).

x / cm(1/x) / m−1P / W
1.0100.017.6
1.566.711.9
2.050.09.1
2.540.07.7
3.033.36.4
4.025.05.1
5.020.04.3
0102030405060708090100110(1/x) / m⁻¹02468101214161820P / W
Heater power against reciprocal of wall thickness, with error bars and lines of best, maximum and minimum gradient (drawn to scale)
(a)
(i)

Determine the gradient of the line of best fit. Give a unit for your answer.

(2)
(b)
(i)

Determine k.

(2)
(c)
(i)

Determine P0.

(1)
(d)
(i)

The temperature difference is uncertain by ±0.4 K. Use the dashed lines to determine the absolute uncertainty in k.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Gradient read from two well-separated points on the line, e.g. (17.5 − 0.93)/(100 − 0) = 0.166✓ 1Accept 0.160–0.176. Points from the table that are not on the line score [0] for this mark unless they happen to lie on it.
Unit: W m✓ 1Accept J s−1 m.
Part (b)(i)
Gradient = kAΔT, so k = gradient/(AΔT)✓ 1
k = 0.166/(0.240 × 20.0) = 0.035 W m−1 K−1✓ 1Allow ECF from (a). Accept 0.033–0.037 W m−1 K−1.
Part (c)(i)
P0 = intercept on the P axis (where 1/x = 0) ≈ 0.93 W✓ 1Accept 0.5–1.1 W. The intercept must be read at 1/x = 0, not at the first data point.
Part (d)(i)
Steepest ≈ 0.174 W m and shallowest ≈ 0.159 W m, so uncertainty in gradient = ±0.007 W m (4.5 %)✓ 1Accept ±0.006 to ±0.010 W m. Allow ECF from (a).
Percentage uncertainty in k = 4.5 % + 2.0 % = 6.5 %, so k = 0.035 ± 0.002 W m−1 K−1✓ 1Allow ECF from (b). Accept ±0.002 to ±0.003 W m−1 K−1. Award [1 max] if the uncertainty in ΔT is omitted.

Answers: (a)(i) 0.166 W m  ·  (b)(i) 0.035 W m−1 K−1  ·  (c)(i) 0.93 W  ·  (d)(i) ±0.002 W m−1 K−1 (the remaining parts are explanations — see the table above)

55B-1B-25Testing Wien's displacement law Paper 1BMedium7 marks

A student tests Wien's displacement law using an electrically heated ceramic element. An infrared spectrometer measures the wavelength λmax at which the emitted radiation is most intense, with an uncertainty of ±0.03 μm. The temperature θ of the element is measured with a thermocouple embedded in the middle of the element.

The table gives the results. The graph shows 1/λmax against the thermocouple temperature T in kelvin, with the line of best fit.

θ / °Cλmax / μm(1/λmax) / μm−1
4504.360.229
5503.800.263
6503.350.299
7503.010.332
8502.720.368
9502.500.400
10502.290.437
0200400600800100012001400T / K0.00.10.20.30.40.5(1/λmax) / μm⁻¹
Reciprocal of the peak wavelength against thermocouple temperature (drawn to scale)
(a)
(i)

The hypothesis is that λmaxT is constant, where T is the thermocouple temperature in kelvin. Test this hypothesis using three rows of the table.

(2)
(b)
(i)

Determine the gradient of the graph and hence the value of the Wien constant given by these data, in m K.

(2)
(c)
(i)

The line of best fit does not pass through the origin. Determine the intercept on the T axis and explain what it suggests about the thermocouple reading.

(2)
(d)
(i)

Suggest one change to the experiment that would reduce the systematic error identified in (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Temperatures converted to kelvin and at least three products calculated, e.g. 4.36 × 723 = 3152, 3.01 × 1023 = 3079, 2.29 × 1323 = 3030 (μm K)✓ 1Award [0] for this mark if Celsius temperatures are used (products 1960 to 2400).
The product falls steadily by about 4 %, more than the uncertainty of about 1 % in each product, so the hypothesis is not supported✓ 1The conclusion must refer to the size of the variation compared with the uncertainty. Allow ECF from the candidate's own products.
Part (b)(i)
Gradient = (0.436 − 0.229)/(1323 − 723) = 3.45 × 10−4 μm−1 K−1✓ 1Accept 3.3–3.6 × 10−4 μm−1 K−1.
Wien constant = 1/gradient = 2.90 × 103 μm K = 2.9 × 10−3 m K✓ 1Allow ECF from the gradient. Accept 2.8–3.0 × 10−3 m K.
Part (c)(i)
Intercept ≈ 58 K✓ 1Accept 40–80 K, from the graph or from the gradient and a point. Allow ECF from (b).
1/λmax ∝ (T − 58 K), so the radiating surface is about 58 K cooler than the thermocouple reading: the thermocouple (in the hotter middle of the element) gives a systematic error, which is why λmaxT was not constant in (a)✓ 1Reference to a systematic (constant) offset in temperature is required. Allow ECF from the intercept.
Part (d)(i)
Measure the temperature of the radiating surface itself, e.g. attach the thermocouple to the outer surface of the element (or use a much thinner element so that its temperature is uniform)✓ 1The change must reduce the difference between the measured temperature and the surface temperature. "Repeat the readings" or "use a more precise spectrometer" scores [0] (these reduce random, not systematic, error).

Answers: (b)(i) 2.9 × 10−3 m K  ·  (c)(i) 58 K (the remaining parts are explanations — see the table above)

56IBP2H-010-00-09Specific and latent heat Paper 2Easy5 marks

A physiotherapist takes an ice pack from a freezer and places it on an athlete's injured ankle. The pack contains 0.25 kg of ice at −10 °C. The ice warms to 0 °C and then melts completely.
Specific heat capacity of ice = 2100 J kg−1 K−1
Specific latent heat of fusion of ice = 3.34 × 105 J kg−1

(a)

State what is meant by the specific latent heat of fusion.

(1)
(b)

State the initial temperature of the ice in kelvin.

(1)
(c)

Calculate the total energy absorbed by the ice pack from the ankle and its surroundings as the ice warms to 0 °C and melts completely.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
The energy required to change unit mass (1 kg) of a substance from solid to liquid without any change in temperature✓ 1
Part (b)
263 K (accept 263.15 K)✓ 1
Part (c)
Energy to warm the ice: Q = mcΔT = 0.25 × 2100 × 10 = 5250 J✓ 1
Energy to melt the ice: Q = mL = 0.25 × 3.34 × 105 = 8.35 × 104 J✓ 1
Total = 8.9 × 104 J (accept 88750 J)✓ 1

Exam tip: Split a heating process into stages: temperature change, then phase change. Write a separate energy term for each stage before adding them.

57IBP3H-B1-20r17-09Specific latent heat Paper 2Easy7 marks

A hospital ward kettle contains 1.5 kg of water at 20 °C. Its heater has a power of 2.0 kW and all the energy it transfers goes into the water. Specific heat capacity of water = 4.18 × 103 J kg−1 K−1. Specific latent heat of vaporisation of water = 2.26 × 106 J kg−1.

(a)

Calculate the energy required to raise the temperature of the water to 100 °C.

(2)
(b)

Calculate the time taken for the water to reach 100 °C. Use the unrounded answer to (a).

(2)
(c)

The kettle is not switched off and the water keeps boiling for a further 3.0 minutes. Calculate the mass of water that is turned into steam in this time.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 use of Q = mcΔT with ΔT = 80 K✓ 1
A1 Q = 1.5 × 4180 × 80 = 5.02 × 105 J✓ 1
Part (b)
M1 t = E/P = 501 600 ÷ 2000 (ECF from (a))✓ 1
A1 t = 251 s✓ 1
Part (c)
M1 energy supplied = 2000 × 180 = 3.6 × 105 J✓ 1
M1 m = E/L = 3.6 × 105 ÷ 2.26 × 106✓ 1
A1 m = 0.159 kg✓ 1

Common error: While the water boils, the temperature stays at 100 °C and all the energy supplied goes into the latent heat, not into mcΔT.

58IBP3H-B1-20r17-10Specific and latent heat Paper 2Easy7 marks

A 0.50 kg block of a solid material for storing heat in a Formula 1 team garage is heated by an immersed 250 W heater. Its temperature is 20 °C at t = 0 and 60 °C at t = 300 s. The temperature then stays at 60 °C until t = 800 s, when all the solid has melted. Assume all the energy from the heater goes into the material.

010020030040050060070080090001020304050607080t / sθ / °C
Temperature of the block against time (drawn to scale)
(a)

Describe, in terms of the molecules, the changes in their kinetic energy and potential energy between t = 300 s and t = 800 s.

(2)
(b)

Calculate the specific heat capacity of the solid.

(3)
(c)

Calculate the specific latent heat of fusion of the material.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the total (mean random) kinetic energy of the molecules is constant (because the temperature is constant)✓ 1
B1 the potential energy of the molecules increases (as the bonds between them are loosened)✓ 1
Part (b)
M1 Pt = mcΔT with t = 300 s and ΔT = 40 K✓ 1
M1 c = (250 × 300) ÷ (0.50 × 40)✓ 1
A1 c = 3750 J kg−1 K−1✓ 1
Part (c)
M1 L = Pt/m with t = 500 s✓ 1
A1 L = (250 × 500) ÷ 0.50 = 2.5 × 105 J kg−1✓ 1

Exam tip: On a heating curve the horizontal section lasts for t = 800 − 300 = 500 s; use that duration, not the 800 s, for the latent heat.

59IBP3H-B1-20r23-11Density of a gas Paper 2Easy6 marks

Air in a coffee roaster's chimney is at a thermodynamic temperature of 353 K, while the air outside is at 293 K. Both are at a pressure of 1.01 × 105 Pa. The molar mass of air is 0.0290 kg mol−1. Use the molar gas constant R = 8.31 J mol−1 K−1.

(a)

Show that the density of the outside air is about 1.20 kg m−3.

(2)
(b)

Explain, in terms of particles, why the air in the chimney rises and cooler air is drawn in at the bottom.

(3)
(c)

Calculate the density of the air in the chimney.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 pV = nRT with n/V = ρ/M, so ρ = pM/(RT)✓ 1
A1 ρ = (1.01 × 105 × 0.0290)/(8.31 × 293) = 1.20 kg m−3✓ 1
Part (b)
A1 the particles in the hot air move faster and collide more strongly with each other and push apart, so the air expands✓ 1
A1 the same mass occupies a larger volume, so the density of the hot air is lower than that of the surrounding air✓ 1
A1 the denser cold air pushes the hot air upwards and flows in to replace it, creating a convection current✓ 1
Part (c)
A1 ρ = (1.01 × 105 × 0.0290)/(8.31 × 353) = 0.998 kg m−3✓ 1

Exam tip: In an 'explain' question about convection, link three ideas in order: expansion, lower density, then the surrounding fluid pushing the warm fluid upwards.

60IBP3H-B1-20r23-14Mixing and thermal equilibrium Paper 2Easy7 marks

A student determines the specific heat capacity of an aluminium block of mass 0.80 kg that will be used in a probe's thermal test. A heater of power 60 W is switched on for 5.0 minutes and the temperature of the block rises by 21.0 K.

(a)

Calculate the specific heat capacity of the block that the student obtains.

(3)
(b)

The accepted value for aluminium is 900 J kg−1 K−1. Suggest why the student's value is larger.

(2)
(c)

Using the accepted value of 900 J kg−1 K−1, determine the percentage of the energy supplied that raised the temperature of the block.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 energy supplied = 60 × 300 = 1.8 × 104 J✓ 1
M1 c = E/(mΔθ) = 18 000/(0.80 × 21.0)✓ 1
A1 c = 1.07 × 103 J kg−1 K−1✓ 1
Part (b)
A1 some of the energy from the heater is transferred to the surroundings✓ 1
A1 so the energy that raised the temperature of the block is less than 18 000 J, but the calculation assumed all of it was used✓ 1
Part (c)
M1 energy gained by block = 0.80 × 900 × 21.0 = 15 120 J✓ 1
A1 percentage = (15 120/18 000) × 100 = 84.0%✓ 1

Common error: If a heat loss occurs, the measured specific heat capacity comes out too large because the temperature rise is smaller than expected for the energy supplied.

61B-2-01Specific and latent heat Paper 2Medium10 marks

A mountaineer melts snow in a pan on a small gas stove. The pan contains 0.80 kg of snow, initially at −24 °C, and the stove transfers thermal energy to the contents of the pan at a constant rate P. Neglect the energy absorbed by the pan and the energy lost to the surroundings.

The graph shows how the temperature θ of the contents varies with time t until the water reaches 25 °C. Specific heat capacity of ice = 2.1 × 103 J kg−1 K−1; specific latent heat of fusion of ice = 3.34 × 105 J kg−1.

01002003004005006007008009001000t / s-30-20-100102030θ / °C
Temperature θ of the contents of the pan against time t (drawn to scale)
(a)
(i)

Outline why the temperature of the contents does not change between t ≈ 100 s and t ≈ 770 s.

(1)
(ii)

Show, using the graph, that P is about 400 W.

(2)
(b)
(i)

Show that (gradient of the graph while the ice warms)/(gradient of the graph while the water warms) is equal to (specific heat capacity of water)/(specific heat capacity of ice), and use the graph to determine the specific heat capacity of water.

(2)
(c)
(i)

The stove burns butane, of energy density 4.6 × 107 J kg−1. Only 35 % of the energy released by the burning butane is transferred to the contents of the pan. Determine the mass of butane burned between t = 0 and the end of the graph.

(3)
(d)
(i)

In fact the pan has a heat capacity of 270 J K−1. State and explain the effect of this on your answers to (a)(ii) and (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
The energy supplied increases the intermolecular potential energy (bonds are broken as the ice melts), not the random kinetic energy of the molecules, so the temperature (a measure of the mean kinetic energy) stays at 0 °C✓ 1Reference to potential energy or to breaking bonds is required; "the energy is used for melting" alone is not enough.
Part (a)(ii)
Duration of melting read from the graph: 770 − 100 = 670 s✓ 1Accept 660–680 s.
P = mL/t = 0.80 × 3.34 × 105/670 = 399 W✓ 1Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
Each gradient is Δθ/Δt = P/(mc), so the ratio is (P/mcice)/(P/mcwater) = cwater/cice (P and m cancel)✓ 1The cancellation of both P and m must be seen.
Gradients ≈ 24/101 = 0.24 K s−1 and 25/210 = 0.12 K s−1; ratio = 2.0, so cwater = 2.0 × 2.1 × 103 = 4.2 × 103 J kg−1 K−1✓ 1Accept 3.9–4.5 × 103 J kg−1 K−1 (graph reading).
Part (c)(i)
Energy transferred to the contents = 399 × 979 = 3.9 × 105 J✓ 1Or the sum mciceΔθ + mL + mcwaterΔθ = 3.9 × 105 J. Allow ECF from (a)(ii).
Energy released by the butane = 3.9 × 105/0.35 = 1.1 × 106 J✓ 1
Mass = 1.1 × 106/4.6 × 107 = 0.024 kg (about 24 g)✓ 1Accept 0.023–0.025 kg. Award [2 max] for 3 × 10−3 kg (efficiency multiplied instead of divided).
Part (d)(i)
(a)(ii) is unaffected: during melting the pan stays at 0 °C, so it absorbs no energy and all of P still goes into melting the ice✓ 1Allow ECF from (a)(ii).
(b): both gradients become P/(mc + C) and the ice gradient is reduced relatively more; the measured ratio becomes (mcwater + C)/(mcice + C) ≈ 1.86, not cwater/cice, so the value found for the specific heat capacity of water is too small (if the true value were 4.2 × 103 J kg−1 K−1, the method would give 1.86 × 2.1 × 103 ≈ 3.9 × 103 J kg−1 K−1)✓ 1Allow ECF from (b). A correct qualitative argument (the pan takes a larger fraction of the energy while the ice warms) scores the mark without the number.

Answers: (a)(ii) 399 W  ·  (b)(i) 4.2 × 103 J kg−1 K−1  ·  (c)(i) 0.024 kg (the remaining parts are explanations — see the table above)

62B-2-06Luminosity, apparent brightness and black-body radiation Paper 2Medium12 marks

A thermal imaging camera detects the infrared radiation emitted by people at night. A person has a surface area of 1.8 m² at a mean surface temperature of 33 °C. The emissivity of the surface is 0.98.

(a)
(i)

Calculate, in μm, the wavelength at which the radiation emitted by the person is most intense.

(1)
(b)
(i)

Calculate the total power radiated by the person.

(1)
(c)
(i)

Estimate the number of photons emitted per second, assuming that all the photons have the wavelength found in (a).

(2)
(ii)

State why the actual number of photons emitted per second is different from your estimate in (c)(i).

(1)
(d)
(i)

The camera is 50 m from the person and has a circular lens of diameter 3.0 cm. Treating the person as a point source radiating equally in all directions, determine the power of the radiation from the person that enters the lens.

(3)
(e)
(i)

The person does not cool down rapidly, although radiating this power. Suggest two reasons why.

(2)
(f)
(i)

Ordinary window glass transmits visible light but absorbs strongly at wavelengths longer than about 5 μm. Explain why the camera cannot detect a person standing behind a closed window, although the person can be seen in visible light.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
λmax = 2.9 × 10−3/306 = 9.5 × 10−6 m = 9.5 μm✓ 1Kelvin temperature and the unit μm are required.
Part (b)(i)
P = eσAT⁴ = 0.98 × 5.67 × 10−8 × 1.8 × 306⁴ = 8.8 × 102 W✓ 1Accept 870–880 W. Using 33 instead of 306 K (about 0.1 W) scores 0.
Part (c)(i)
Photon energy = hc/λ = 2.1 × 10−20 J✓ 1Allow ECF from (a).
Number per second = 877/2.1 × 10−20 = 4.2 × 1022 s−1✓ 1Allow ECF from (b).
Part (c)(ii)
The radiation has a continuous spectrum: photons of many wavelengths, and so many energies, are emitted (most of the energy is at wavelengths longer than λmax, so the actual number is larger)✓ 1Either idea.
Part (d)(i)
b = L/4πd² = 877/(4π × 50²) = 2.8 × 10−2 W m−2✓ 1Allow ECF from (b).
Area of the lens = π × (0.015)² = 7.07 × 10−4 m²✓ 1Using the diameter as the radius scores 0 for this mark.
Power = 2.8 × 10−2 × 7.07 × 10−4 = 2.0 × 10−5 W✓ 1Allow ECF from the intensity.
Part (e)(i)
The person also absorbs radiation from the surroundings, so the net loss by radiation is much smaller than P✓ 1
Respiration (metabolism) in the body continuously releases thermal energy that replaces the energy lost✓ 1Accept clothing, which lowers the temperature of the radiating surface.
Part (f)(i)
The person's radiation is emitted mainly at wavelengths around 10 μm, longer than 5 μm, so it is absorbed by the glass✓ 1Allow ECF from (a).
The camera receives instead the infrared emitted by the glass itself, which is at the temperature of the glass, not of the person✓ 1

Answers: (a)(i) 9.5 μm  ·  (b)(i) 8.8 × 102 W  ·  (c)(i) 4.2 × 1022 s−1  ·  (d)(i) 2.0 × 10−5 W (the remaining parts are explanations — see the table above)

63B-2-09Specific and latent heat Paper 2Medium10 marks

A runner of mass 65 kg runs at a constant speed of 4.0 m s−1 on level ground against a total resistive force of 15 N. Her muscles convert chemical energy into mechanical work with an efficiency of 20 %; the rest of this energy is transferred as thermal energy to her body. Specific latent heat of vaporization of sweat at skin temperature = 2.4 × 106 J kg−1; mean specific heat capacity of the body = 3.5 × 103 J kg−1 K−1.

(a)
(i)

Show that the rate at which her muscles use chemical energy is about 300 W.

(2)
(ii)

Calculate the rate at which thermal energy is transferred to her body.

(1)
(b)
(i)

75 % of this thermal energy is removed by the evaporation of sweat. Determine the mass of sweat evaporated in one hour.

(2)
(ii)

Explain, in terms of molecules, why the evaporation of sweat cools the skin.

(2)
(c)
(i)

On a humid day the sweat cannot evaporate, so this 75 % is not removed. Estimate the rise in her body temperature in 10 minutes of running.

(2)
(ii)

State one assumption you made in (c)(i).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Mechanical power = Fv = 15 × 4.0 = 60 W✓ 1The constant speed means the driving force equals the resistive force.
Rate of use of chemical energy = 60/0.20 = 300 W✓ 1
Part (a)(ii)
300 − 60 = 240 W✓ 1Allow ECF from (a)(i). Award [0] for 0.80 × 60 = 48 W.
Part (b)(i)
Rate of evaporation = 0.75 × 240/2.4 × 106 = 7.5 × 10−5 kg s−1✓ 1Allow ECF from (a)(ii).
Mass in one hour = 7.5 × 10−5 × 3600 = 0.27 kg✓ 1
Part (b)(ii)
Only the molecules with the largest kinetic energies can overcome the intermolecular attractions and escape from the surface of the liquid✓ 1
So the mean kinetic energy of the molecules left behind falls, and the temperature of the sweat (and of the skin in contact with it) falls✓ 1"Evaporation takes in heat" alone is not enough.
Part (c)(i)
Extra energy retained = 0.75 × 240 × 600 = 1.08 × 105 J✓ 1Allow ECF from (a)(ii).
ΔT = 1.08 × 105/(65 × 3.5 × 103) = 0.47 K✓ 1Accept 0.4–0.5 K.
Part (c)(ii)
Any one: the thermal energy is shared uniformly through the whole body; the other energy losses (radiation, convection, breathing) stay the same; the specific heat capacity is constant✓ 1

Answers: (a)(i) 300 W  ·  (a)(ii) 240 W  ·  (b)(i) 0.27 kg  ·  (c)(i) 0.47 K (the remaining parts are explanations — see the table above)

64B-2-18Specific and latent heat Paper 2Medium11 marks

In laser spot welding, a single pulse of infrared light melts a small cylinder of steel of diameter 0.40 mm and depth 0.20 mm. The steel is initially at 20 °C. Each pulse has an energy of 0.50 J and the wavelength of the light is 1.06 μm.

Data for steel: density = 7.9 × 103 kg m−3; specific heat capacity = 490 J kg−1 K−1; melting point = 1800 K; specific latent heat of fusion = 2.7 × 105 J kg−1.

(a)
(i)

Show that the mass of steel melted is about 2.0 × 10−7 kg.

(2)
(b)
(i)

Determine the energy needed to melt this steel.

(3)
(c)
(i)

Determine the number of photons in one pulse.

(2)
(d)
(i)

Determine the fraction of the pulse energy that is used to melt the steel, and suggest one reason why the rest of the energy is needed.

(2)
(e)
(i)

The diameter of the spot is halved and its depth is unchanged. Assuming the same fraction of the pulse energy is used, deduce the new pulse energy, and the ratio (new pulse energy per unit area of the spot)/(original pulse energy per unit area of the spot).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
V = π × (0.20 × 10−3)² × 0.20 × 10−3 = 2.51 × 10−11 m³✓ 1Award [0] if the diameter is used as the radius.
m = ρV = 7.9 × 103 × 2.51 × 10−11 = 1.99 × 10−7 kg✓ 1An answer to at least 3 s.f. is required.
Part (b)(i)
Temperature rise = 1800 − 293 = 1507 K✓ 1Award [2 max] for 0.23 J (1780 K used).
mcΔT = 1.99 × 10−7 × 490 × 1507 = 0.147 J✓ 1Allow ECF from (a).
mL = 0.054 J, total = 0.20 J✓ 1
Part (c)(i)
Photon energy = hc/λ = 6.63 × 10−34 × 3.00 × 108/1.06 × 10−6 = 1.88 × 10−19 J✓ 1
Number = 0.50/1.88 × 10−19 = 2.7 × 1018✓ 1
Part (d)(i)
Fraction = 0.20/0.50 = 0.40✓ 1Allow ECF from (b).
Any one: part of the light is reflected by the steel surface; thermal energy is conducted into the surrounding metal during the pulse; some steel around the spot is heated but not melted✓ 1
Part (e)(i)
The melted mass is proportional to d², so the energy needed is ¼ of the original: 0.50/4 = 0.125 J✓ 1Allow ECF from (d).
The area of the spot is also ¼ of the original, so the energy per unit area is unchanged: ratio = 1✓ 1

Answers: (a)(i) 1.99 × 10−7 kg  ·  (b)(i) 0.20 J  ·  (c)(i) 2.7 × 1018  ·  (d)(i) 0.40  ·  (e)(i) 0.125 J; 1 (the remaining parts are explanations — see the table above)

65B-2-19Specific and latent heat Paper 2Medium11 marks

A reusable hand warmer is a sealed plastic pouch containing 0.090 kg of liquid sodium acetate trihydrate at room temperature, 20 °C. The liquid is below its normal freezing temperature. When a small metal disc in the pouch is clicked, the liquid starts to crystallise (solidify). Its temperature rises rapidly to the melting point, 58 °C, and then stays at 58 °C for a long time while crystallisation continues.

Specific heat capacity of the liquid = 3.0 × 103 J kg−1 K−1; specific heat capacity of the solid = 2.0 × 103 J kg−1 K−1; specific latent heat of fusion = 2.64 × 105 J kg−1.

(a)

Temperatures.

(i)

State the melting point in kelvin and the rise in temperature in kelvin.

(1)
(b)

Energy released. The initial rise in temperature is so rapid that no energy is lost to the surroundings during it; take the specific heat capacity of the contents during the rise to be that of the liquid.

(i)

Show that about 0.04 kg of the liquid crystallises during the initial rise from 20 °C to 58 °C.

(2)
(ii)

Determine the energy transferred to the surroundings while the rest of the liquid crystallises at 58 °C.

(2)
(iii)

The pouch transfers energy to a hand at a mean rate of 4.0 W. Estimate, in minutes, how long the hand warmer stays at 58 °C.

(1)
(c)

Resetting.

(i)

After use the pouch is solid at 20 °C. It is reset by heating it in boiling water until all of it has melted. Calculate the minimum energy that must be supplied to melt all of the contents, starting from 20 °C.

(2)
(ii)

The melted pouch is then allowed to cool to 20 °C and remains liquid. Compare the internal energy of the pouch as a liquid at 20 °C with its internal energy as a solid at 20 °C, and determine the difference between them.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
58 °C = 331 K; the rise is 38 K✓ 1Both needed.
Part (b)(i)
Latent heat released = energy to warm the contents: mx × 2.64 × 105 = 0.090 × 3.0 × 103 × 38✓ 1
mx = 1.03 × 104/2.64 × 105 = 0.039 kg ≈ 0.04 kg✓ 1Answer to at least 2 s.f. needed.
Part (b)(ii)
Mass still liquid = 0.090 − 0.039 = 0.051 kg✓ 1
Q = 0.051 × 2.64 × 105 = 1.35 × 104 J✓ 1Accept 1.3–1.4 × 104 J (1.32 × 104 J if 0.040 kg is used).
Part (b)(iii)
t = 1.35 × 104/4.0 = 3.4 × 103 s ≈ 56 min✓ 1Accept 55–57 min; ECF from (b)(ii).
Part (c)(i)
Warming the solid: 0.090 × 2.0 × 103 × 38 = 6.84 × 103 J; melting: 0.090 × 2.64 × 105 = 2.38 × 104 J✓ 1
Total = 3.1 × 104 J✓ 1Accept 3.06 × 104 J.
Part (c)(ii)
The liquid has the greater internal energy✓ 1
Because its intermolecular potential energy is greater; the total random kinetic energy of the particles is the same in both states, since they are at the same temperature✓ 1
Difference ≈ 2.0 × 104 J (= 1.35 × 104 J from (b)(ii) + 0.090 × 2.0 × 103 × 38 J released as the solid cools from 58 °C to 20 °C)✓ 1Accept 2.0–2.1 × 104 J; ECF from (b)(ii). Do not award for 2.4 × 104 J (mL), which ignores the different specific heat capacities.

Answers: (a)(i) 331 K; 38 K  ·  (b)(i) 0.039 kg  ·  (b)(ii) 1.35 × 104 J  ·  (b)(iii) ≈ 56 min  ·  (c)(i) 3.1 × 104 J  ·  (c)(ii) liquid greater by 2.0 × 104 J (the remaining parts are explanations — see the table above)

66IBP2H-010-00-10Specific and latent heat Paper 2Medium6 marks

During a hot-weather marathon, a runner's body must lose excess thermal energy at a rate of 600 W to keep a constant core temperature. Assume that all of this energy is removed by the evaporation of sweat.
Specific latent heat of vaporization of sweat at skin temperature = 2.4 × 106 J kg−1
Mass of runner = 65 kg
Average specific heat capacity of the human body = 3500 J kg−1 K−1

(a)

Explain, with reference to particles, why the evaporation of sweat cools the skin.

(2)
(b)

Calculate the mass of sweat that must evaporate each hour to remove the excess energy.

(2)
(c)

Suppose instead that no energy is lost from the body. Calculate the rise in the runner's body temperature in 10 minutes.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The fastest-moving (highest kinetic energy) molecules have enough energy to overcome the intermolecular attractions and escape from the surface of the sweat✓ 1
so the mean kinetic energy of the remaining liquid falls and its temperature falls; energy then flows from the skin to the liquid✓ 1
Part (b)
Energy per hour = 600 × 3600 = 2.16 × 106 J✓ 1
m = Q/L = 2.16 × 106/2.4 × 106 = 0.90 kg✓ 1
Part (c)
Q = 600 × 600 = 3.6 × 105 J✓ 1
ΔT = Q/(mc) = 3.6 × 105/(65 × 3500) = 1.6 K✓ 1

Common error: Convert the time to seconds before multiplying by power. Using minutes or hours gives an energy that is wrong by a factor of 60 or 3600.

67IBP2H-010-00-11Specific heat capacity Paper 2Medium6 marks

An engineer tests an aluminium alloy block of mass 0.80 kg for use as a heat sink. A 50 W electrical heater is embedded in the block. The temperature of the block is recorded every 60 s.

Time / s: 0, 60, 120, 180, 240
Temperature / °C: 20.0, 24.0, 28.1, 32.0, 36.0

060120180240152025303540t / stemperature / °C
Readings from the table
(a)

Determine the mean rate of increase of temperature of the block.

(2)
(b)

Determine the specific heat capacity of the alloy.

(2)
(c)

The accepted value for the alloy is 900 J kg−1 K−1. Explain why the experimental value is larger than the accepted value.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Uses the full range of the data or a best-fit gradient: ΔT/Δt = (36.0 − 20.0)/240✓ 1
= 0.067 K s−1 (accept °C s−1)✓ 1
Part (b)
P = mc(ΔT/Δt), so c = P/(m × ΔT/Δt)✓ 1
c = 50/(0.80 × 0.0667) ≈ 940 J kg−1 K−1 (accept 930–940; allow error carried forward from the rate in the first part)✓ 1
Part (c)
Some energy from the heater is lost to the surroundings, so less than 50 W heats the block✓ 1
The temperature rises more slowly than it would with no losses, so using P = 50 W with this smaller rate in c = P/(m × ΔT/Δt) gives a larger value of c✓ 1

Exam tip: In data questions, use the first and last readings or a line of best fit rather than a single interval. This reduces the effect of random error in individual readings.

68IBP2H-010-00-13Specific latent heat Paper 2Medium6 marks

A 2.2 kW electric kettle contains 1.2 kg of water at 15 °C. It takes 215 s to bring the water to its boiling point of 100 °C. The kettle's switch is faulty, so the kettle stays on after the water boils.
Specific heat capacity of water = 4180 J kg−1 K−1
Specific latent heat of vaporization of water = 2.26 × 106 J kg−1

(a)

Calculate the energy needed to raise the temperature of the water to 100 °C.

(2)
(b)

Determine the efficiency of the kettle.

(2)
(c)

The kettle stays on for a further 60 s after the water boils. Assuming the efficiency is unchanged, calculate the mass of water converted to steam in this time.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Q = mcΔT = 1.2 × 4180 × 85✓ 1
= 4.26 × 105 J (≈ 4.3 × 105 J)✓ 1
Part (b)
Electrical energy supplied = 2200 × 215 = 4.73 × 105 J✓ 1
Efficiency = 4.26 × 105 / 4.73 × 105 = 0.90 (90%); accept 0.90–0.91 if the rounded value 4.3 × 105 J is used✓ 1
Allow error carried forward from the previous part.—
Part (c)
Useful energy = 0.90 × 2200 × 60 = 1.19 × 105 J✓ 1
m = Q/L = 1.19 × 105 / 2.26 × 106 = 0.053 kg✓ 1
Allow error carried forward from the efficiency found in the previous part.—

Common error: Once the water is boiling, its temperature no longer changes. Use Q = mL, not Q = mcΔT, for the energy that turns water into steam. Carry unrounded values forward between parts to avoid rounding drift.

69IBP2H-010-00-16Specific and latent heat Paper 2Medium6 marks

A cyclist adds 0.050 kg of crushed ice at 0 °C to a bottle containing 0.40 kg of sports drink at 28 °C. The bottle is well insulated and has negligible heat capacity. All the ice melts.
Specific heat capacity of the drink and of water = 4180 J kg−1 K−1
Specific latent heat of fusion of ice = 3.34 × 105 J kg−1

(a)

Show that the energy required to melt the ice is about 1.7 × 104 J.

(1)
(b)

Determine the final temperature of the drink.

(3)
(c)

Explain why 0.050 kg of ice at 0 °C cools the drink more than 0.050 kg of water at 0 °C.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Q = mL = 0.050 × 3.34 × 105 = 1.67 × 104 J✓ 1
Part (b)
Energy lost by drink = energy to melt ice + energy to warm melted ice: 0.40 × 4180 × (28 − T) = 1.67 × 104 + 0.050 × 4180 × T✓ 1
46 816 − 1672T = 16 700 + 209T✓ 1
T = 16 °C✓ 1
ECF from (a)—
Part (c)
The ice must first absorb latent heat to melt, with no change in its temperature✓ 1
so more energy is taken from the drink before the melted ice starts to warm, giving a lower final temperature✓ 1

Common error: Remember the melted ice must also warm from 0 °C to the final temperature. Leaving out this mcΔT term gives a final temperature that is too low.

70IBP2H-010-01-11Specific and latent heat Paper 2Medium7 marks

A student finds the specific heat capacity of steel. Steel ball bearings are heated in boiling water and then quickly transferred into water in an insulated polystyrene cup. The data are shown below.
Quantity — Value
mass of steel balls — 0.250 kg
initial temperature of balls — 100.0 °C
mass of water in cup — 0.180 kg
initial temperature of water — 18.0 °C
final temperature of mixture — 28.5 °C
Specific heat capacity of water = 4180 J kg−1 K−1. Assume the cup absorbs no energy.

(a)

Determine the specific heat capacity of steel from these data.

(3)
(b)

The accepted value is 490 J kg−1 K−1. Explain whether the cooling of the balls during transfer could account for the difference.

(2)
(c)

Using the accepted value, determine the temperature of the balls as they entered the water.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Energy gained by water = 0.180 × 4180 × 10.5 = 7.90×103 J .✓ 1
Energy lost by steel = 0.250 × c × 71.5 .✓ 1
c = 442 J kg−1 K−1 .✓ 1
Part (b)
The balls cool during transfer, so their temperature on entering the water is below 100 °C and their real temperature drop is less than 71.5 K .✓ 1
Using 71.5 K overestimates ΔT, which makes the calculated c too small. This is consistent with 442 being less than 490 .✓ 1
Part (c)
ΔT = 7.90×103 / (0.250 × 490) = 64.5 K (ECF from (a)) .✓ 1
T = 28.5 + 64.5 = 93 °C .✓ 1

Common error: The temperature change of the hot object is its initial temperature minus the final temperature of the mixture, not its initial temperature minus the water's initial temperature.

71IBP2H-010-01-13Molecular model and density Paper 2Medium8 marks

Liquid nitrogen is used in medicine to freeze and destroy skin lesions.
Density of liquid nitrogen = 808 kg m−3.
Density of nitrogen gas at 20 °C and atmospheric pressure = 1.16 kg m−3.
Molar mass of nitrogen = 0.028 kg mol−1. NA = 6.02×1023 mol−1.

(a)

Calculate the number of nitrogen molecules per cubic metre of liquid nitrogen.

(2)
(b)

Estimate the mean spacing of molecules in the liquid.

(2)
(c)

Determine the mean spacing of molecules in the gas at 20 °C.

(2)
(d)

Outline, using the molecular model, why liquid nitrogen must never be stored in a tightly sealed container.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
Amount per m3 = 808 / 0.028 = 2.89×104 mol .✓ 1
n = 2.89×104 × 6.02×1023 = 1.74×1028 m−3 .✓ 1
Part (b)
Volume per molecule = 1/n = 5.76×10−29 m3 .✓ 1
Spacing ≈ (5.76×10−29)1/3 = 3.9×10−10 m .✓ 1
Part (c)
Either: spacing ratio = (808/1.16)1/3 = 8.9, or: number of molecules per m3 of gas = (1.16/0.028) × 6.02×1023 = 2.49×1025 m−3, so volume per molecule = 4.0×10−26 m3 .✓ 1
Spacing = 3.4×10−9 m. Accept 3.4–3.5×10−9 m; ECF from (b) applies to the ratio method .✓ 1
Part (d)
As the liquid boils, the molecules move from being close together to being far apart, so the gas would need a volume about 700 times that of the liquid .✓ 1
In a sealed container the gas cannot expand. The molecules therefore collide with the walls far more often, the pressure rises enormously and the container may burst .✓ 1

Exam tip: Molecules in a liquid are spaced at about the size of one molecule (~10⁻¹⁰ m). Use this to check that your answer for the liquid is sensible.

72IBP2H-010-01-14Specific and latent heat Paper 2Medium7 marks

A home energy store uses 12 kg of paraffin wax as a phase-change material. The wax is heated from a solid at 20 °C to a liquid at 70 °C. The melting point of the wax is 58 °C.
Specific heat capacity of solid wax = 2100 J kg−1 K−1.
Specific heat capacity of liquid wax = 2400 J kg−1 K−1.
Specific latent heat of fusion of wax = 2.0×105 J kg−1.
Specific heat capacity of water = 4180 J kg−1 K−1.

(a)

Describe what happens to the mean kinetic energy and the potential energy of the wax molecules while the wax melts.

(2)
(b)

Show that the total energy stored is about 3.7 MJ.

(3)
(c)

Determine the mass of water that would store the same energy when heated from 20 °C to 70 °C.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
The mean random kinetic energy does not change because the temperature stays constant .✓ 1
The intermolecular potential energy increases as bonds are broken and the molecules separate, so the internal energy increases .✓ 1
Part (b)
Heating the solid: 12 × 2100 × 38 = 9.58×105 J .✓ 1
Melting: 12 × 2.0×105 = 2.40×106 J .✓ 1
Heating the liquid: 12 × 2400 × 12 = 3.46×105 J, giving a total of 3.70×106 J .✓ 1
Part (c)
m = 3.70×106 / (4180 × 50) (ECF from (b)) .✓ 1
m = 18 kg .✓ 1

Common error: Use a separate specific heat capacity for each phase, and include the latent heat term only once, at the melting point.

73IBP2H-010-01-16Convection Paper 2Medium9 marks

The envelope of a hot-air balloon used for aerial filming has a volume of 2800 m3. The outside air is at 15 °C and has a density of 1.22 kg m−3. The air inside the envelope is heated to 100 °C. The air inside and outside is at the same pressure. At constant pressure, the density of air is inversely proportional to its kelvin temperature.

air at 100 °CV = 2800 m³burneroutside air15 °C, 1.22 kg m⁻³same pressure
Diagram not to scale
(a)

Explain, using the particle model, why the heated air is less dense.

(2)
(b)

Show that the density of the air inside the envelope is about 0.94 kg m−3.

(2)
(c)

Determine the maximum total mass of envelope fabric, basket, crew and camera equipment for which the balloon can just hover.

(3)
(d)

Outline how convection distributes the energy from the burner, which is at the bottom of the envelope.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
At a higher temperature the molecules have a greater mean kinetic energy and move faster .✓ 1
For the pressure to stay the same there must be fewer molecules per unit volume (they are further apart), so there is less mass per unit volume .✓ 1
Part (b)
Uses kelvin temperatures: ρ = 1.22 × 288/373 .✓ 1
ρ = 0.942 kg m−3 .✓ 1
Part (c)
Mass of displaced outside air = 1.22 × 2800 = 3416 kg, and mass of hot air inside = 0.942 × 2800 = 2638 kg (ECF from (b)) .✓ 1
For hovering, upthrust = weight of hot air + weight of load, so the load mass equals the difference in air masses (g cancels). Accept the equivalent force route: net upward force = (3416 − 2638) × 9.8 = 7.6×103 N, then divide by g .✓ 1
Maximum mass = 3416 − 2638 ≈ 780 kg (7.8×102 kg) .✓ 1
Part (d)
Air heated by the burner expands, becomes less dense and rises to the top of the envelope .✓ 1
Cooler, denser air sinks to replace it and is heated in turn, setting up a convection current that spreads energy through the envelope .✓ 1

Common error: The upthrust is the weight of the displaced outside air. The weight of the hot air inside must still be subtracted before you find the load the balloon can support.

74IBP3H-B1-20r17-11Density of a gas Paper 2Medium8 marks

Air in a beehive is at a temperature of 35 °C (308 K) and a pressure of 1.01 × 105 Pa. The molar mass of air is 0.0290 kg mol−1 and the molar gas constant is R = 8.31 J mol−1 K−1. Treat air as an ideal gas.

(a)

Show that the density of an ideal gas is given by ρ = pM/(RT), where M is the molar mass.

(3)
(b)

Calculate the density of the air in the hive.

(2)
(c)

The air outside the hive is at the same pressure but at 15 °C (288 K). Calculate its density and explain why warm air tends to leave the hive through an opening near the top.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 start from pV = nRT with n = m/M✓ 1
M1 pV = (m/M)RT so p = (m/V)RT/M✓ 1
A1 with ρ = m/V: ρ = pM/(RT) (AG)✓ 1
Part (b)
M1 ρ = (1.01 × 105 × 0.0290) ÷ (8.31 × 308)✓ 1
A1 ρ = 1.14 kg m−3✓ 1
Part (c)
M1 ρ = (1.01 × 105 × 0.0290) ÷ (8.31 × 288)✓ 1
A1 ρ = 1.22 kg m−3✓ 1
B1 the air in the hive has a lower density than the outside air, so it is pushed upwards (buoyancy) and out through the top while denser outside air enters lower down✓ 1

Exam tip: In ρ = pM/(RT), M must be in kg mol⁻¹ (not g mol⁻¹) and T in kelvin, or the density is wrong by a factor of 1000.

75IBP3H-B1-20r17-12Specific and latent heat Paper 2Medium8 marks

During a 90-minute race a Formula 1 driver loses 2.1 kg of sweat, of which 70% evaporates from the skin. Specific latent heat of vaporisation of sweat = 2.4 × 106 J kg−1. The driver's mass is 72 kg and the specific heat capacity of the body is 3500 J kg−1 K−1. The driver's body generates thermal energy at an average rate of 920 W.

(a)

Calculate the energy removed from the driver's body by evaporation during the race.

(2)
(b)

Calculate the average rate at which energy is removed by evaporation.

(2)
(c)

Estimate the rise in the driver's body temperature if evaporation were the only way in which thermal energy leaves the body. Use the unrounded answer to (a).

(3)
(d)

Suggest why the actual temperature rise of a driver is much smaller than your answer to (c).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 mass evaporated = 0.70 × 2.1 = 1.47 kg✓ 1
A1 E = 1.47 × 2.4 × 106 = 3.5 × 106 J (3.528 × 106 J)✓ 1
Part (b)
M1 P = E/t with t = 90 × 60 = 5400 s✓ 1
A1 P = 653 W✓ 1
Part (c)
M1 energy generated = 920 × 5400 = 4.968 × 106 J✓ 1
M1 net energy retained = 4.968 × 106 − 3.528 × 106 = 1.44 × 106 J and ΔT = E/(mc)✓ 1
A1 ΔT = 1.44 × 106 ÷ (72 × 3500) = 5.7 K✓ 1
Part (d)
B1 energy is also lost by conduction, convection and radiation from the body (or in the airflow through the cockpit, cooling vest or drinks)✓ 1

Common error: Only the evaporated fraction (70%) removes energy; using the full 2.1 kg of sweat in the latent heat calculation overestimates the cooling.

76IBP3H-B1-20r17-13Heating curves Paper 2Medium9 marks

A 0.40 kg block of beeswax is initially a solid at 25 °C. It is heated in a water bath by a 600 W heater, of which 80% of the energy transferred goes into the wax. The wax melts at a constant temperature of 62 °C and is then heated to 80 °C. Specific heat capacity of solid wax = 2500 J kg−1 K−1. Specific heat capacity of liquid wax = 2900 J kg−1 K−1. Specific latent heat of fusion of wax = 1.8 × 105 J kg−1.

(a)

Sketch, in words, the graph of the temperature of the wax against time during the heating.

(3)
(b)

Calculate the total energy that must be transferred to the wax.

(4)
(c)

Calculate the time for which the heater must be on. Use the unrounded answer to (b).

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 a straight rising line from 25 °C to 62 °C✓ 1
B1 a horizontal section at 62 °C while the wax melts✓ 1
B1 a second rising line from 62 °C to 80 °C with a smaller gradient than the first line (because the specific heat capacity of the liquid is greater)✓ 1
Part (b)
M1 solid stage: 0.40 × 2500 × (62 − 25) = 3.70 × 104 J✓ 1
M1 melting: 0.40 × 1.8 × 105 = 7.2 × 104 J✓ 1
M1 liquid stage: 0.40 × 2900 × (80 − 62) = 2.088 × 104 J✓ 1
A1 total = 1.30 × 105 J (129 880 J)✓ 1
Part (c)
M1 useful power = 0.80 × 600 = 480 W and t = 129 880 ÷ 480✓ 1
A1 t = 271 s✓ 1

Model sketch

256280tθ25 °C62 °C (melting)80 °C

Exam tip: For multi-stage heating problems, calculate each stage separately (solid, melting, liquid) and add; only the stages that change temperature use mcΔT.

77IBP3H-B1-20r17-14Specific and latent heat Paper 2Medium8 marks

A hospital sterilises 0.80 kg of steel instruments, initially at 20 °C, in a steam chamber. Steam at 100 °C condenses on the instruments until they reach 100 °C. Assume that all the energy released by the condensing steam goes to the instruments and that the condensed water stays at 100 °C. Specific heat capacity of steel = 480 J kg−1 K−1. Specific latent heat of vaporisation of water = 2.26 × 106 J kg−1. Specific heat capacity of water = 4.18 × 103 J kg−1 K−1.

(a)

Calculate the energy gained by the instruments.

(2)
(b)

Calculate the mass of steam that condenses. Use the unrounded answer to (a).

(2)
(c)

Explain, in terms of molecules, why condensing steam at 100 °C transfers much more energy than the same mass of liquid water cooling from 100 °C to 80 °C.

(2)
(d)

Calculate the mass of boiling water at 100 °C that would give the same energy to the instruments if it cooled from 100 °C to 20 °C. Use the unrounded value of Q.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 Q = mcΔT = 0.80 × 480 × 80✓ 1
A1 Q = 3.07 × 104 J (30 720 J)✓ 1
Part (b)
M1 m = Q/L = 30 720 ÷ 2.26 × 106✓ 1
A1 m = 0.0136 kg✓ 1
Part (c)
B1 when steam condenses the molecules lose potential energy as they come together (intermolecular forces do work), which is released as thermal energy at constant temperature✓ 1
B1 the energy released per kilogram on condensing (latent heat) is much greater than that released by cooling the liquid, which only reduces the kinetic energy of molecules a little✓ 1
Part (d)
M1 m = Q/(cΔT) = 30 720 ÷ (4180 × 80)✓ 1
A1 m = 0.0919 kg✓ 1

Exam tip: Per kilogram, condensing steam releases about 27 times as much energy as water cooling by 20 K, so keep the latent-heat and specific-heat stages separate.

78IBP3H-B1-20r17-16Evaporation and boiling Paper 2Medium8 marks

Honeybees keep a hive cool in hot weather by fanning their wings over water droplets that they have placed on the comb.

(a)

Outline, in terms of molecules, why the evaporation of the droplets cools the water that remains.

(3)
(b)

State two reasons why the fanning increases the rate of evaporation of the droplets.

(2)
(c)

Explain why water in an open pan boils at a lower temperature at a mountain apiary than at sea level.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
B1 the molecules with the highest kinetic energies are the ones that can escape through the surface✓ 1
B1 these molecules take their energy with them so the average kinetic energy of the remaining molecules decreases✓ 1
B1 the temperature (of the remaining water) therefore falls✓ 1
Part (b)
B1 first valid reason, e.g. it carries away water vapour from near the surface so fewer molecules return to the liquid✓ 1
B1 second valid reason, different from the first, e.g. it brings in drier air (lower humidity) so the net rate of escape is greater✓ 1
Part (c)
B1 boiling begins when the saturated vapour pressure of the water equals the external pressure✓ 1
B1 atmospheric pressure is lower on the mountain✓ 1
B1 so the saturated vapour pressure reaches the external pressure at a lower temperature✓ 1

Common error: Boiling does not occur when the water 'gets hot enough to turn to steam': it occurs when saturated vapour pressure equals external pressure.

79IBP3H-B1-20r23-09Black-body spectra Paper 2Medium8 marks

A re-entry probe has a nose cone that can be modelled as a black body with a total surface area of 8.0 m². Its surface is at a thermodynamic temperature of 1400 K. Use the Stefan–Boltzmann constant σ = 5.67 × 10−8 W m−2 K−4 and Wien's constant 2.90 × 10−3 m K.

(a)

Calculate the peak wavelength of the radiation emitted by the nose cone.

(2)
(b)

Calculate the total power radiated by the nose cone.

(2)
(c)

The surface cools until the power radiated is half of its original value. Determine the new thermodynamic temperature of the surface. Assume the area is unchanged.

(3)
(d)

State how the peak wavelength after cooling compares with the value in (a).

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses λmax = (2.90 × 10−3)/T✓ 1
A1 λmax = 2.07 × 10−6 m✓ 1
Part (b)
M1 uses P = σAT4✓ 1
A1 P = 5.67 × 10−8 × 8.0 × 14004 = 1.74 × 106 W✓ 1
Part (c)
M1 uses P ∝ T4, so (T/1400)4 = 1/2✓ 1
M1 T = 1400 × (1/2)1/4✓ 1
A1 T = 1.18 × 103 K✓ 1
Part (d)
A1 it is longer, because the peak wavelength is inversely proportional to the temperature✓ 1

Exam tip: When a quantity scales as a power of T, write the ratio of the new and old values equal to the required factor, then take the root.

80IBP3H-B1-20r23-10Specific latent heat Paper 2Medium8 marks

A coffee roaster heats 12.0 kg of green beans, of which 1.2 kg is water. Assume the following.
• The 10.8 kg of dry bean material, specific heat capacity 1600 J kg−1 K−1, is heated from 22 °C to 200 °C.
• The 1.2 kg of water, specific heat capacity 4200 J kg−1 K−1, is heated from 22 °C to 100 °C and then completely vaporised. The specific latent heat of vaporisation of water is 2.26 × 106 J kg−1.
• Energy needed to heat the drum is ignored.

(a)

Calculate the energy needed to heat the dry bean material.

(2)
(b)

Calculate the energy needed to heat and then vaporise the water.

(3)
(c)

The roaster's heater has a power of 12.0 kW and 85% of its output energy is transferred to the beans and the water. Determine the time needed to complete the process. Use the unrounded values from (a) and (b).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 uses E = mcΔθ with Δθ = 178 K✓ 1
A1 E = 10.8 × 1600 × 178 = 3.08 × 106 J✓ 1
Part (b)
A1 heating the water: 1.2 × 4200 × 78 = 3.93 × 105 J✓ 1
M1 uses E = mL for vaporisation✓ 1
A1 vaporising: 1.2 × 2.26 × 106 = 2.71 × 106 J, so the total for the water is 3.11 × 106 J✓ 1
Part (c)
M1 total energy = 3 075 840 + 3 105 120 = 6.18 × 106 J✓ 1
M1 useful power = 0.85 × 12 000 = 10 200 W and time = energy ÷ power✓ 1
A1 t = 606 s✓ 1

Common error: Latent heat transfers energy with no temperature change, so do not multiply it by a temperature change; add it as a separate stage.

81IBP3H-B1-20r23-12Luminosity, apparent brightness and black-body radiation Paper 2Medium8 marks

A probe's thermal camera records the radiation emitted by two hot components, X and Y, which can be modelled as black bodies. The peak wavelength of the radiation from X is 1.45 µm and that from Y is 4.35 µm. Wien's constant is 2.90 × 10−3 m K.

(a)

Determine the thermodynamic temperature of each component.

(2)
(b)

Calculate the ratio (power emitted per unit area by X) ÷ (power emitted per unit area by Y).

(2)
(c)

Describe two differences between the intensity–wavelength graphs for X and Y.

(2)
(d)

Component X appears dull red to a person looking at it, but component Y emits no visible light. Explain this observation.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)
A1 TX = (2.90 × 10−3)/(1.45 × 10−6) = 2.00 × 103 K✓ 1
A1 TY = (2.90 × 10−3)/(4.35 × 10−6) = 667 K✓ 1
Part (b)
M1 power per unit area ∝ T4, so ratio = (TX/TY)4✓ 1
A1 ratio = 81 (ECF from (a); equivalently (4.35/1.45)4)✓ 1
Part (c)
A1 the peak of the graph for X is at a shorter wavelength than the peak for Y✓ 1
A1 the graph for X is higher at all wavelengths and has a much greater area under it✓ 1
Part (d)
A1 the graph for X has a tail that extends into the visible region, including the red end, with non-zero intensity✓ 1
A1 for Y the intensity at visible wavelengths (about 0.4 µm to 0.7 µm) is negligible, as its peak is far into the infrared✓ 1

Exam tip: When describing graph differences, name the feature (peak position, area, height) and say which curve is greater or shorter for each.

82IBP3H-B1-20r23-13Specific and latent heat Paper 2Medium9 marks

A layer of ice of uniform thickness 3.0 mm covers a bridge deck of area 120 m². The ice is at −6.0 °C. Electric heating cables raise the ice to 0 °C and then melt it. Use the following data: density of ice 920 kg m−3; specific heat capacity of ice 2100 J kg−1 K−1; specific latent heat of fusion of ice 3.34 × 105 J kg−1.

(a)

Show that the mass of the ice is about 331 kg.

(2)
(b)

Calculate the energy needed to raise the temperature of the ice to 0 °C.

(2)
(c)

Calculate the energy needed to melt all of the ice at 0 °C.

(2)
(d)

The cables deliver 25.0 kW and all of this energy is transferred to the ice. Determine the time needed to melt all of the ice, starting from −6.0 °C. Use the unrounded values from (b) and (c).

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
M1 volume = 120 × 3.0 × 10−3 = 0.36 m³✓ 1
A1 mass = 920 × 0.36 = 331 kg✓ 1
Part (b)
M1 uses E = mcΔθ with Δθ = 6.0 K✓ 1
A1 E = 331.2 × 2100 × 6.0 = 4.17 × 106 J✓ 1
Part (c)
M1 uses E = mL✓ 1
A1 E = 331.2 × 3.34 × 105 = 1.11 × 108 J✓ 1
Part (d)
M1 total energy = 4 173 120 + 110 620 800 = 1.148 × 108 J✓ 1
M1 time = energy ÷ power✓ 1
A1 t = 4.59 × 103 s (about 76.5 minutes)✓ 1

Exam tip: Split a heating-then-melting problem into separate stages, calculate each energy, then add them before dividing by power.

83B-2-11Conduction Paper 2Hard13 marks

Thermal energy flows from the hot interior of the Earth to its surface. In a region of solid granite, the temperature T of the rock was measured at different depths z in a deep borehole. The graph shows the results.

Thermal conductivity of granite = 2.9 W m−1 K−1; radius of the Earth = 6.37 × 106 m; solar constant = 1.36 × 103 W m−2; mean albedo of the Earth = 0.30.

0.00.51.01.52.02.53.0z / km020406080100T / °C
Temperature T of the granite against depth z below the surface (drawn to scale)
(a)
(i)

Outline how thermal energy is transferred upwards through the solid granite by conduction.

(1)
(b)
(i)

Use the graph to determine the temperature gradient in the granite, in K m−1.

(2)
(c)
(i)

Show that the rate at which thermal energy is conducted upwards through each square metre of the granite is about 0.08 W.

(1)
(d)
(i)

Assume that the value in (c) is typical of the whole surface of the Earth. Determine the total rate at which thermal energy is conducted out of the Earth's interior.

(2)
(e)
(i)

By comparing your answer to (c) with the mean intensity of the sunlight absorbed by the Earth, deduce whether the energy from the Earth's interior has a significant effect on the temperature of the Earth's surface.

(2)
(f)
(i)

A geothermal power station extracts thermal energy from hot rock at a depth of 3.0 km and rejects thermal energy at the surface temperature. Use the graph to determine the maximum possible efficiency of the station.

(2)
(g)
(i)

The station extracts thermal energy from the rock at a rate of 5.0 MW. Explain, with the help of a calculation, why the temperature of the rock around the station falls during its years of operation, and state the consequence for the efficiency of the station.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Particles in the hotter (deeper) rock vibrate with a greater mean kinetic energy; through their interactions (collisions) with neighbouring particles they pass kinetic energy to the particles in the cooler rock above, so there is a net transfer of energy from hot to cold✓ 1Reference to the transfer of kinetic energy between neighbouring particles is required; "heat rises" scores [0].
Part (b)(i)
Line of best fit drawn through the points and its gradient found from two well-separated points on the line, e.g. (90 − 12)/(3.0 − 0)✓ 1Using two data points that are not on the line scores [0] for this mark unless they lie on it.
Gradient = 26 K km−1 = 0.026 K m−1✓ 1Accept 0.025–0.027 K m−1. The conversion to K m−1 is required for this mark.
Part (c)(i)
ΔQ/Δt per unit area = k × gradient = 2.9 × 0.026 = 0.075 W (per m²)✓ 1Allow ECF from (b). An answer to at least 2 s.f. or the full substitution is required.
Part (d)(i)
Surface area of the Earth = 4π(6.37 × 106)² = 5.10 × 1014 m²✓ 1Using πR² (9.6 × 1012 W in the answer) scores [0] for this mark.
Total rate = 5.10 × 1014 × 0.0754 = 3.8 × 1013 W✓ 1Allow ECF from (c). Accept 3.7–4.1 × 1013 W (4.1 × 1013 W if 0.08 W m−2 is used).
Part (e)(i)
Mean absorbed solar intensity = (1 − 0.30) × 1360/4 = 238 W m−2✓ 1Award [0] for this mark if the factor ¼ is omitted (952 W m−2), but allow the comparison mark.
The geothermal flux is only about 3 × 10−4 of this (0.075 W m−2 against 238 W m−2), so its effect on the surface temperature is negligible: the surface temperature is set by the absorbed sunlight (and the greenhouse effect)✓ 1Allow ECF from (c) and from the solar intensity. The conclusion must be supported by the comparison.
Part (f)(i)
Temperature at 3.0 km read from the graph ≈ 90 °C = 363 K; surface temperature 12 °C = 285 K✓ 1Accept 88–92 °C and 11–13 °C.
ηCarnot = 1 − 285/363 = 0.21✓ 1Accept 0.20–0.22. Award [0] for 1 − 12/90 = 0.87 (Celsius temperatures).
Part (g)(i)
Area of surface through which 5.0 MW would be conducted naturally = 5.0 × 106/0.075 = 6.6 × 107 m² (about 66 km²)✓ 1Allow ECF from (c). Accept 6.2–6.9 × 107 m².
Conduction from below replaces the energy only at this very small rate per unit area, so the energy removed from the limited volume of rock around the wells comes mainly from the internal energy stored in that rock✓ 1Equivalent: the rate of extraction is far greater than the rate at which conduction resupplies the rock near the station.
So the temperature of this rock (Th) falls, and the maximum (and actual) efficiency of the station decreases✓ 1Allow ECF from (f). Dependent on a correct reason for the cooling.

Answers: (b)(i) 0.026 K m−1  ·  (c)(i) 0.075 W  ·  (d)(i) 3.8 × 1013 W  ·  (e)(i) no: about 3 × 10−4 of the absorbed solar intensity  ·  (f)(i) 0.21  ·  (g)(i) 6.6 × 107 m²; the rock cools and the efficiency falls (the remaining parts are explanations — see the table above)

84B-2-20Specific and latent heat Paper 2Hard11 marks

A discarded aluminium fuel tank of mass 40 kg re-enters the Earth's atmosphere. The graph shows how its speed v varies with time t as it is slowed by air resistance. During this time the gravitational force on the tank is negligible compared with the air resistance.

Aluminium: specific heat capacity = 900 J kg−1 K−1; melting point = 933 K; specific latent heat of fusion = 3.97 × 105 J kg−1. The initial temperature of the tank is 250 K.

020406080100120140160t / s012345678v / km s⁻¹
Speed v of the tank against time t (drawn to scale)
(a)
(i)

Determine the maximum force exerted on the tank by the air.

(3)
(b)
(i)

Determine the kinetic energy lost per kilogram of the tank between t = 0 and t = 150 s.

(2)
(c)
(i)

Calculate the energy needed to raise the temperature of 1.0 kg of aluminium from 250 K to its melting point and then to melt it.

(2)
(d)
(i)

Only 3.0 % of the kinetic energy lost is transferred to the internal energy of the tank. Deduce the fraction of the tank that melts.

(3)
(e)
(i)

Explain why the answer to (d) does not depend on the mass of the tank.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Tangent drawn at the steepest part of the curve (t ≈ 60 s)✓ 1
Deceleration = gradient ≈ 1.5 × 102 m s−2✓ 1Accept 130–165 m s−2.
F = ma = 40 × 146 = 5.8 × 103 N✓ 1Allow ECF from the gradient. Accept 5.2–6.6 × 103 N.
Part (b)(i)
Speeds read: 7.45 km s−1 and 0.50 km s−1✓ 1Accept 7.4–7.5 and 0.5 km s−1.
ΔEk/m = ½(7450² − 500²) = 2.8 × 107 J kg−1✓ 1
Part (c)(i)
900 × (933 − 250) = 6.15 × 105 J✓ 1
6.15 × 105 + 3.97 × 105 = 1.01 × 106 J✓ 1
Part (d)(i)
0.030 × 2.8 × 107 = 8.3 × 105 J kg−1✓ 1Allow ECF from (b).
This exceeds 6.15 × 105 J kg−1, so the tank reaches its melting point with 2.1 × 105 J kg−1 left over✓ 1Allow ECF from (c).
Fraction melted = 2.1 × 105/3.97 × 105 = 0.54 (about half)✓ 1Accept 0.5–0.6. Award [2 max] for 0.82 (energy available divided by the total 1.01 × 106 J kg−1 instead of first subtracting the energy needed to reach the melting point).
Part (e)(i)
Both the kinetic energy transferred and the energy needed to heat and melt the aluminium are proportional to the mass, so the mass cancels✓ 1Allow ECF from (d).

Answers: (a)(i) 5.8 × 103 N  ·  (b)(i) 2.8 × 107 J kg−1  ·  (c)(i) 1.01 × 106 J  ·  (d)(i) 0.54 (the remaining parts are explanations — see the table above)

85B-2-38Convection Paper 2Hard13 marks

In a thermosiphon solar water heater no pump is used. Water warmed in a flat collector on a roof rises up a pipe into a storage tank placed above the collector, while cooler water from the bottom of the tank flows down a second pipe into the bottom of the collector, as shown.

The collector has an area of 2.0 m² and receives sunlight of intensity 850 W m−2 at right angles to its surface. The absorbing surface has an albedo of 0.06, an emissivity of 0.12 and a mean temperature of 345 K; it absorbs thermal radiation from its surroundings at the rate that a surface of the same emissivity at 300 K would emit. The back of the collector is covered by insulation of thermal conductivity 0.040 W m−1 K−1 and thickness 50 mm, across which the temperature difference is 45 K. Neglect all other energy losses.

collectorsunlightstorage tankwarmcoolh
Thermosiphon solar water heater (not to scale)
(a)
(i)

Explain why the water circulates without a pump.

(2)
(b)
(i)

Calculate the power absorbed by the collector from the sunlight.

(1)
(ii)

Determine the net rate at which the absorbing surface loses energy by radiation.

(2)
(c)
(i)

Water enters the collector at 25 °C and leaves it at 40 °C. Determine the mass of water flowing through the collector per second. Specific heat capacity of water = 4.18 × 103 J kg−1 K−1.

(3)
(d)
(i)

The pipes are vertical over a height h. The water in the warm pipe has density ρh and that in the cool pipe has density ρc. Show that, at the bottom of the pipes, the difference between the pressures due to the two columns of water is (ρc − ρh)gh.

(2)
(ii)

h = 1.6 m. The density of water is 997.0 kg m−3 at 25 °C and 992.2 kg m−3 at 40 °C. Calculate this pressure difference.

(1)
(e)
(i)

At night the collector cools below the temperature of the water in the tank. Explain, with reference to (d), why the water does not circulate in the reverse direction, and why the tank must be placed above the collector.

(2)
Show mark scheme
Marking pointMarkNotes
Part (a)(i)
Water warmed in the collector expands, so its density decreases✓ 1
The less dense warm water rises (the upward buoyancy force exceeds its weight) and the denser cool water in the other pipe sinks to replace it, so a convection current circulates✓ 1Reference to the density difference driving the flow is required; "heat rises" scores [0].
Part (b)(i)
(1 − 0.06) × 850 × 2.0 = 1.60 × 103 W✓ 1Accept 1.6 × 103 W.
Part (b)(ii)
Net rate = eσA(T4 − Ts4) = 0.12 × 5.67 × 10−8 × 2.0 × (3454 − 3004)✓ 1Award [1 max] for 193 W (absorption from the surroundings ignored).
= 83 W✓ 1Accept 82–83 W.
Part (c)(i)
Rate of conduction through the insulation = 0.040 × 2.0 × 45/0.050 = 72 W✓ 1
Power transferred to the water = 1598 − 82.6 − 72 = 1443 W✓ 1Allow ECF from (b)(i) and (b)(ii).
Mass per second = 1443/(4.18 × 103 × 15) = 0.0230 kg s−1✓ 1Accept 0.022–0.024 kg s−1. A temperature rise of 15 K (not 40 K) must be used.
Part (d)(i)
A column of cross-sectional area A has weight ρAhg, which acts on the area A✓ 1
Pressure = weight/area = ρgh for each column, so the difference is ρcgh − ρhgh = (ρc − ρh)gh✓ 1Use of p = F/A with the weight is required.
Part (d)(ii)
Δp = (997.0 − 992.2) × 9.81 × 1.6 = 75 Pa✓ 1Allow ECF from (d)(i).
Part (e)(i)
The cooled, denser water collects in the collector, which is the lowest part of the circuit, while the warm water stays in the tank above: there is no column of denser water beside a column of less dense water at the same height, so no pressure difference of the form (ρc − ρh)gh drives a flow✓ 1Allow ECF from (d)(i). OWTTE.
If the tank were below the collector, the cold dense water from the collector would sink into the tank and warm tank water would rise into the collector, so the stored energy would be lost at night✓ 1

Answers: (b)(i) 1.60 × 103 W  ·  (b)(ii) 83 W  ·  (c)(i) 0.0230 kg s−1  ·  (d)(ii) 75 Pa (the remaining parts are explanations — see the table above)

86IBP2H-010-00-12Luminosity in solar units Paper 2Hard8 marks

A space telescope observes a hot star. The spectrum of the star peaks at a wavelength of 290 nm. Other observations show that the luminosity of the star is 40 L☉. The apparent brightness of the star measured at the telescope is 4.0 × 10−9 W m−2.
Wien's displacement constant = 2.9 × 10−3 m K
L☉ = 3.83 × 1026 W
Surface temperature of the Sun = 5800 K

(a)

Calculate the surface temperature of the star.

(2)
(b)

Determine the distance from the telescope to the star.

(3)
(c)

Determine the radius of the star in terms of the solar radius R☉.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 2.9 × 10−3 / (290 × 10−9)✓ 1
= 1.0 × 104 K✓ 1
Part (b)
L = 40 × 3.83 × 1026 = 1.53 × 1028 W✓ 1
d = √(L / (4πb))✓ 1
= √(1.53 × 1028 / (4π × 4.0 × 10−9)) = 5.5 × 1017 m✓ 1
Part (c)
L ∝ R2T4, so R/R☉ = √(L/L☉) × (T☉/T)2✓ 1
= √40 × (5800 / (1.0 × 104))2✓ 1
R = 2.1 R☉✓ 1
Allow error carried forward from the temperature calculated in the first part.—

Exam tip: Find the radius by comparing with the Sun in solar units. This avoids carrying σ, 4π and very large powers of ten through the calculation.

87IBP2H-010-00-14Black-body spectra Paper 2Hard7 marks

The electronics on a satellite dissipate 1.5 kW of thermal power. This power is removed by a flat radiator panel that emits from one face only. The panel has an emissivity of 0.85 and operates at 20 °C. Ignore any radiation absorbed by the panel.
Stefan–Boltzmann constant σ = 5.67 × 10−8 W m−2 K−4

(a)

Outline why the temperature must be expressed in kelvin in the Stefan–Boltzmann law.

(1)
(b)

Show that the area of the panel is about 4.2 m2.

(3)
(c)

A redesign allows the panel to run at 40 °C. Determine the percentage reduction in the panel area needed to emit the same power.

(2)
(d)

Suggest why the real panel must be larger than the area calculated in the 'show that' part.

(1)
Show mark scheme
Marking pointMarkNotes
Part (a)
The emitted power is proportional to the fourth power of the absolute temperature (zero emission only at 0 K), so a scale whose zero is not absolute zero, such as Celsius, gives wrong values✓ 1
Part (b)
T = 293 K✓ 1
A = P/(eσT4) = 1500/(0.85 × 5.67 × 10−8 × 2934)✓ 1
A = 4.22 m2✓ 1
Part (c)
A ∝ 1/T4, so Anew/Aold = (293/313)4 = 0.768✓ 1
Percentage reduction = 23%✓ 1
(A method that calculates the new area of 3.24 m2 and then the reduction is also acceptable.)—
Part (d)
The real panel also absorbs radiation, for example sunlight or infrared from Earth. To remove a net 1.5 kW it must therefore emit more than 1.5 kW in total, which needs a larger area✓ 1

Exam tip: For 'show that' questions, give your final answer to one more significant figure than the value stated in the question. This shows the answer came from your working.

88IBP2H-010-00-15Stefan–Boltzmann law Paper 2Hard8 marks

A patient lies in a hospital room whose walls are at 20 °C. The patient's skin has a surface area of 1.8 m2, an emissivity of 0.97 and a temperature of 33 °C. The patient exchanges thermal radiation with the walls.
Stefan–Boltzmann constant σ = 5.67 × 10−8 W m−2 K−4
Wien's displacement constant = 2.9 × 10−3 m K

(a)

Calculate the peak wavelength of the radiation emitted by the skin, and state the region of the electromagnetic spectrum in which it lies.

(2)
(b)

Show that the net power radiated by the patient is about 140 W.

(3)
(c)

The patient develops a fever and the skin temperature rises by 2.0 K. The walls stay at 20 °C. Determine the percentage increase in the net radiated power, and comment on your answer.

(3)
Show mark scheme
Marking pointMarkNotes
Part (a)
T = 33 + 273 = 306 K; λmax = 2.9 × 10−3/306 = 9.5 × 10−6 m✓ 1
Infrared✓ 1
Part (b)
Pnet = eσA(Tskin4 − Twalls4), with Twalls = 293 K✓ 1
= 0.97 × 5.67 × 10−8 × 1.8 × (3064 − 2934)✓ 1
= 138 W, which is about 140 W✓ 1
Part (c)
New skin temperature = 308 K, and 3084 − 2934 = 1.63 × 109 K4 compared with 1.40 × 109 K4. Alternatively, the new Pnet = 161 W compared with 138 W✓ 1
Percentage increase = 16.6% ≈ 17% (accept 16–17%)✓ 1
This is far larger than the rise of about 0.7% in absolute skin temperature, because the net power depends on a difference of fourth powers✓ 1

Common error: Do not use (Tskin − Twalls)⁴. The net power depends on Tskin⁴ − Twalls⁴, with each temperature in kelvin.

89IBP2H-010-01-09Net radiation exchange Paper 2Hard9 marks

A marathon runner has a skin surface area of 1.8 m2 and the skin has an emissivity of 0.97. During a race the mean skin temperature is 33 °C and the surroundings are at 25 °C.
Stefan–Boltzmann constant σ = 5.67×10−8 W m−2 K−4.

(a)

Explain why the net rate of energy loss by radiation from the runner depends on the temperature of the surroundings.

(2)
(b)

Show that the net rate of energy loss by radiation is about 90 W.

(2)
(c)

The runner's body must lose thermal energy at a total rate of 750 W. Part of this is the net radiation loss from (b) and 150 W is removed by convection. All the remaining power is removed by evaporation of sweat. The specific latent heat of vaporization of sweat is 2.4×106 J kg−1. Determine the mass of sweat that evaporates in 1.0 hour.

(3)
(d)

On a different day, the air and the surroundings are both at 36 °C, and the skin temperature is still 33 °C. Suggest why running is more dangerous on this day.

(2)
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Marking pointMarkNotes
Part (a)
The runner emits radiation at a rate eσAT4 and also absorbs radiation from the surroundings at a rate eσAT04 .✓ 1
The net loss is the difference between these two rates, so warmer surroundings give a smaller net loss .✓ 1
Part (b)
Uses P = eσA(T4 − T04) with T = 306 K and T0 = 298 K .✓ 1
P = 0.97 × 5.67×10−8 × 1.8 × (3064 − 2984) = 87 W .✓ 1
Part (c)
Power removed by evaporation = 750 − 87 − 150 = 513 W (allow ECF from (b)) .✓ 1
Energy removed in 1.0 h = 513 × 3600 = 1.85×106 J .✓ 1
m = 1.85×106 / 2.4×106 = 0.77 kg .✓ 1
Part (d)
The air and surroundings are now hotter than the skin, so there is a net transfer of energy into the runner by radiation and by convection .✓ 1
Evaporation must now remove more than 750 W, so the sweat rate rises, which risks dehydration or overheating .✓ 1

Exam tip: Convert both temperatures to kelvin before raising them to the fourth power. Never subtract the Celsius temperatures first and then raise the result to the fourth power.

90IBP2H-010-01-10Conduction Paper 2Hard9 marks

A double-glazed window has an area of 1.5 m2. It consists of two glass panes, each 4.0 mm thick, separated by a 12 mm layer of still air. The inner surface of the window is at 18 °C and the outer surface is at 2 °C. The window is in a steady state. Assume energy flows only perpendicular to the window.
Thermal conductivity of glass = 0.80 W m−1 K−1.
Thermal conductivity of air = 0.025 W m−1 K−1.

glassstill airglass4.0 mm12 mm4.0 mminner surface18 °Couter surface2 °Csteady energy flowwindow area 1.5 m²
Diagram not to scale
(a)

Explain why the rate of thermal energy transfer is the same through each of the three layers.

(2)
(b)

Show that the temperature difference across the air gap is about 16 K.

(3)
(c)

Calculate the rate of thermal energy transfer through the window.

(2)
(d)

Outline why making the air gap much wider would not greatly reduce the energy loss.

(2)
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Part (a)
In the steady state the temperatures are constant, so no energy accumulates in any layer .✓ 1
The energy entering each layer per second therefore equals the energy leaving it, so the same power passes through every layer .✓ 1
Part (b)
With the same P and A, ΔT is proportional to Δx/k .✓ 1
Δx/k = 0.0050 m2 K W−1 for each pane and 0.48 m2 K W−1 for the air, giving a total of 0.49 m2 K W−1 .✓ 1
ΔTair = (0.48/0.49) × 16 = 15.7 K .✓ 1
Part (c)
Uses P = kAΔT/Δx for the air: P = 0.025 × 1.5 × 15.7 / 0.012 (ECF from (b)), OR uses the total: P = 1.5 × 16 / 0.49 .✓ 1
P = 49 W .✓ 1
Part (d)
In a wide gap the air is no longer still: convection currents can form in it .✓ 1
Convection (or thermal radiation across the gap, which does not depend on the gap width) transfers energy much faster than conduction through still air, so the rate does not fall in proportion to the gap width as the conduction formula predicts .✓ 1

Exam tip: For layers in series, add the Δx/k values just as you would add resistances in series.

91IBP2H-010-01-12Net radiation exchange Paper 2Hard9 marks

The electronics on a satellite dissipate 1.20 kW. All of this energy is transferred to a flat radiator panel that emits from one face only. The face has an emissivity of 0.90 and points towards deep space. Radiation absorbed from deep space is negligible.
σ = 5.67×10−8 W m−2 K−4. Wien constant = 2.9×10−3 m K.

(a)

Explain why the panel can lose energy to its surroundings only by radiation.

(2)
(b)

The panel operates at 300 K. Show that its area is about 2.9 m2.

(2)
(c)

Calculate the peak wavelength emitted by the panel and state the region of the electromagnetic spectrum in which it lies.

(2)
(d)

At one point in its orbit the panel also absorbs 140 W of radiation from the Earth. Determine the new equilibrium temperature of the panel.

(3)
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Marking pointMarkNotes
Part (a)
Space is a vacuum, so there is no medium in contact with the panel .✓ 1
Conduction and convection both need a medium, so they cannot transfer energy away from the panel .✓ 1
Part (b)
A = P / (eσT4) .✓ 1
A = 1200 / (0.90 × 5.67×10−8 × 3004) = 2.90 m2 .✓ 1
Part (c)
λ = 2.9×10−3 / 300 = 9.7×10−6 m .✓ 1
The peak lies in the infrared .✓ 1
Part (d)
At equilibrium the emitted power = 1200 + 140 = 1340 W .✓ 1
T4 = 1340 / (0.90 × 5.67×10−8 × 2.90) (ECF from (b)) .✓ 1
T = 308 K .✓ 1

Exam tip: At equilibrium, the power emitted equals the total power supplied (internal dissipation plus power absorbed). Because P ∝ T⁴, scaling the original temperature by (1340/1200)¹/⁴ is a quick check.

92IBP3H-B1-20r17-15Density of a gas Paper 2Hard11 marks

A sealed Formula 1 tyre of fixed volume 0.020 m3 contains nitrogen at an absolute pressure of 1.40 × 105 Pa and a temperature of 20 °C (293 K). The molar mass of nitrogen is 0.028 kg mol−1 and the molar gas constant is R = 8.31 J mol−1 K−1. Treat nitrogen as an ideal gas.

(a)

Calculate the mass of nitrogen in the tyre.

(3)
(b)

During the race the gas temperature rises to 78 °C (351 K). State and explain what happens to the density of the gas in the sealed tyre.

(2)
(c)

Calculate the pressure of the gas at 351 K.

(2)
(d)

A pressure-release valve then lets gas escape until the pressure returns to 1.40 × 105 Pa at 351 K, with the volume unchanged. Calculate the mass of nitrogen that escapes. Use unrounded values.

(4)
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Part (a)
M1 n = pV/(RT) = (1.40 × 105 × 0.020) ÷ (8.31 × 293) = 1.15 mol✓ 1
M1 m = nM✓ 1
A1 m = 0.0322 kg✓ 1
Part (b)
B1 the density stays constant (1.61 kg m−3)✓ 1
B1 the mass of gas and the volume of the tyre are both constant✓ 1
Part (c)
M1 p/T constant (or p = nRT/V): p = 1.40 × 105 × 351/293✓ 1
A1 p = 1.68 × 105 Pa✓ 1
Part (d)
M1 mass remaining m' = pVM/(RT) with T = 351 K✓ 1
A1 m' = (1.40 × 105 × 0.020 × 0.028) ÷ (8.31 × 351) = 0.02688 kg (density 1.34 kg m−3)✓ 1
M1 mass escaping = initial mass − remaining mass = 0.03220 − 0.02688✓ 1
A1 mass escaping = 5.32 × 10−3 kg✓ 1

Common error: At constant volume the pressure rise comes from faster molecules hitting the walls harder, not from more molecules: the density stays constant until gas is released.

93IBP3H-B1-20r23-15Density of a gas Paper 2Hard9 marks

A space probe has a thruster tank of volume 0.050 m³ that contains nitrogen gas at a pressure of 2.0 × 107 Pa and a thermodynamic temperature of 290 K. Treat the nitrogen as an ideal gas with molar mass 0.0280 kg mol−1. Use R = 8.31 J mol−1 K−1.

(a)

Calculate the amount of nitrogen in the tank.

(2)
(b)

Determine the density of the nitrogen in the tank.

(2)
(c)

The thruster uses 40% of the mass of the nitrogen. The temperature of the gas in the tank falls to 250 K. Calculate the new pressure in the tank.

(3)
(d)

Explain, in terms of the motion of the particles, why the pressure falls when the number of particles and the temperature both decrease.

(2)
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Marking pointMarkNotes
Part (a)
M1 uses n = pV/(RT)✓ 1
A1 n = (2.0 × 107 × 0.050)/(8.31 × 290) = 415 mol✓ 1
Part (b)
M1 mass = nM = 414.95 × 0.0280 = 11.6 kg and density = mass ÷ volume✓ 1
A1 ρ = 232 kg m−3✓ 1
Part (c)
M1 amount remaining = 0.60 × 414.95 = 249 mol✓ 1
M1 uses p = nRT/V with T = 250 K✓ 1
A1 p = 1.03 × 107 Pa✓ 1
Part (d)
A1 fewer particles means fewer collisions per second with the walls✓ 1
A1 a lower temperature means a lower mean speed, so each collision exerts a smaller force on the wall✓ 1

Exam tip: Keep unrounded intermediate values when a later part uses them, or note that the new pressure scales as (amount remaining) × (temperature ratio).

94IBP3H-B1-20r23-16Net radiation exchange Paper 2Hard9 marks

A spherical probe of radius r orbits Mars, where the intensity of sunlight is 590 W m−2. Treat the probe as a black body that absorbs all incident sunlight and radiates uniformly from its whole surface. Use σ = 5.67 × 10−8 W m−2 K−4 and Wien's constant 2.90 × 10−3 m K.

(a)

Explain why the power absorbed by the probe is 590πr² while the power it radiates is 4πr²σT4.

(2)
(b)

Determine the equilibrium thermodynamic temperature of the probe.

(3)
(c)

Calculate the peak wavelength of the radiation from the probe and state the region of the electromagnetic spectrum in which it lies.

(2)
(d)

The probe is now covered by a coating that absorbs only 40% of the incident sunlight but still radiates as a black body. Determine the new equilibrium thermodynamic temperature.

(2)
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Marking pointMarkNotes
Part (a)
A1 the sunlight is intercepted over the probe's circular cross-sectional area πr²✓ 1
A1 the probe radiates from its entire surface area 4πr²✓ 1
Part (b)
M1 equates 590πr² = 4πr²σT4, with r cancelling✓ 1
M1 T = (590/(4 × 5.67 × 10−8))1/4✓ 1
A1 T = 226 K✓ 1
Part (c)
A1 λmax = (2.90 × 10−3)/225.8 = 1.28 × 10−5 m✓ 1
A1 infrared✓ 1
Part (d)
M1 0.40 × 590πr² = 4πr²σT4, or T = 226 × 0.401/4✓ 1
A1 T = 180 K✓ 1

Common error: Absorption depends on the cross-sectional area (a circle) but emission depends on the whole surface area; mixing these up changes the factor of 4.