B.1 Thermal Energy Transfers Questions

1B-1A-01

Stefan–Boltzmann law·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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)?

Show mark scheme

Marking point	Mark	Notes
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 P_X/P_Y = 1/(1/4) = 4.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis is the ratio the other way up, P_Y/P_X.
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.

Syllabus understandingB.1 — quantitative analysis of energy transferred by radiation modelled by the Stefan–Boltzmann law as given by L = σAT⁴ Command term: Determine

2B-1A-02

Luminosity in solar units·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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).	✓ 1	Answer 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.

Syllabus understandingB.1 — the concept of apparent brightness b; luminosity L of a body as given by b = L/4πd² (Guidance: the luminosity of a star can be expressed in watts or in terms of the luminosity of the Sun L_⊙) Command term: Determine

3B-1A-13

Net radiation exchange·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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: σT_s⁴ = 5.67 × 10⁻⁸ × 300⁴ = 459 W m⁻².	—
Step 3Net loss = σ(T⁴ − T_s⁴) = 3544 − 459 = 3.1 × 10³ W m⁻².	✓ 1	Answer 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 − T_s)⁴, which is not how the fourth powers combine.

Syllabus understandingB.1 — quantitative analysis of energy transferred by radiation modelled by the Stefan–Boltzmann law as given by L = σAT⁴ Command term: Determine

4B-1A-17

Specific latent heat·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Mass of water in the kettle against time (drawn to scale). The m axis does not start at zero.

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific latent heat of fusion and vaporization of substances L as given by Q = mL Command term: Determine

5B-1A-18

Conduction·B.1 Thermal energy transfers

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

6B-1A-22

Mixing and thermal equilibrium·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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?

Show mark scheme

Marking point	Mark	Notes
Step 1Energy lost by the block = energy gained by the liquid: mc(θ − θ_f) = (2m)(4c)(θ_f − 0).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2θ − θ_f = 8θ_f, so 9θ_f = θ.	—
Step 3θ_f = θ/9.	✓ 1	Answer 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 = 8mcθ_f.
CThis ignores the different masses: c(θ − θ_f) = 4cθ_f.
DThis ignores the different specific heat capacities: m(θ − θ_f) = 2mθ_f.

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c as given by Q = mcΔT; that temperature difference determines the direction of the resultant thermal energy transfer between bodies Command term: Deduce

7B-1A-30

Luminosity and apparent brightness·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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?

Show mark scheme

Marking point	Mark	Notes
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: L_X = 3²L_Y = 9L_Y.	—
Step 3b = L/4πd²: b_X = 9b/2² = 9b/4.	✓ 1	Answer 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.

Syllabus understandingB.1 — the concept of apparent brightness b; luminosity L of a body as given by b = L/4πd²; the Stefan–Boltzmann law L = σAT⁴; Wien's displacement law Command term: Deduce

8B-1A-31

Molecular model and density·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksEstimate

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)?

Show mark scheme

Marking point	Mark	Notes
Step 1The same mass contains the same number of molecules, so the volume available to each molecule is inversely proportional to the density: V_steam/V_liquid = 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.	✓ 1	Answer 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.

Syllabus understandingB.1 — molecular theory in solids, liquids and gases; density as given by ρ = m/V Command term: Estimate

9B-1A-32

Kelvin and Celsius temperatures·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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

Show mark scheme

Marking point	Mark	Notes
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, E_k = (3/2)k_BT; the temperatures are 291 K and 336 K.	—
Step 3Ratio = 336/291 = 1.15.	✓ 1	Answer 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 E_k ∝ 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.

Syllabus understandingB.1 — that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales; that Kelvin temperature is a measure of the average kinetic energy of particles as given by E_k = (3/2)k_BT Command term: Determine

10B-1A-33

Internal energy and temperature·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

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

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

11B-1A-34

Heating curves·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

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)?

Temperature of the sample against time (drawn to scale)

Show mark scheme

Marking point	Mark	Notes
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 3c_liquid/c_solid = (3.0/50)/(2.0/60) = 1.8.	✓ 1	Answer 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.

12B-1A-35

Conduction·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksExplain

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

13B-1A-36

Black-body spectra·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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

Show mark scheme

Marking point	Mark	Notes
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: λ_maxT = 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.	✓ 1	Answer 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: λ_max 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.

Syllabus understandingB.1 — the emission spectrum of a black body and the determination of the temperature of the body using Wien's displacement law λ_maxT = 2.9 × 10⁻³ m K; the Stefan–Boltzmann law; conversion between the Kelvin and Celsius scales Command term: Deduce

14B-1A-61

Convection·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksExplain

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?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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).

Syllabus understandingB.1 — qualitative description of thermal energy transferred by convection due to fluid density differences; that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer; density ρ = m/V Command term: Explain

15B-1A-62

Evaporation and boiling·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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.

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingB.1 — molecular theory in solids, liquids and gases; that Kelvin temperature is a measure of the average kinetic energy of particles; that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature (Guidance: the terms melting, freezing, boiling, condensing and evaporation should be familiar) Command term: Deduce

16B-1A-63

Stefan–Boltzmann law·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

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)?

Show mark scheme

Marking point	Mark	Notes
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.	✓ 1	Answer 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.

Syllabus understandingB.1 — quantitative analysis of energy transferred by radiation modelled by the Stefan–Boltzmann law as given by L = σAT⁴; quantitative analysis of thermal energy transfers Q = mcΔT; density ρ = m/V Command term: Deduce

17B-1A-73

Mechanisms of thermal energy transfer·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A sealed double-glazed window has two panes of glass separated by a narrow gap filled with argon gas. The room on one side of the window is warmer than the air outside.

Which statements about the transfer of thermal energy through the window are correct?

I. Energy is transferred through each pane of glass by convection.
II. Energy is transferred across the argon by conduction.
III. Energy would still be transferred across the gap by thermal radiation if the argon were removed.

Show mark scheme

Marking point	Mark	Notes
Step 1I is wrong: convection needs bulk movement of a fluid; the particles of solid glass vibrate about fixed positions, so energy passes through the glass by conduction.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2II is correct: argon particles near the warmer pane have a greater mean kinetic energy and pass energy on in collisions, so conduction occurs in a gas (slowly, because the particles are far apart).	—
Step 3III is correct: thermal radiation is electromagnetic and needs no medium, so it crosses an evacuated gap.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis accepts convection in a solid and rejects both conduction in a gas and radiation through a vacuum.
BThis rejects III, as if thermal radiation needed a material medium; electromagnetic waves travel through a vacuum.
CThis accepts convection in the glass and treats a gas as unable to conduct; gases are poor conductors, not non-conductors.
DCorrect: conduction occurs in the argon and radiation crosses the gap with or without the gas; convection cannot occur in a solid.

Syllabus understandingB.1 — that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer; conduction in terms of the difference in the kinetic energy of particles; qualitative description of thermal energy transferred by convection due to fluid density differences Command term: Deduce

18B-1A-74

Black-body spectra·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

The same black body is observed at two temperatures. Each graph shows a curve for the body at 4000 K (solid line) and a curve for the body at 3000 K (dashed line), drawn to the same scales.

Which graph correctly shows the two emission spectra?

Four sketch graphs of intensity against wavelength (solid: 4000 K; dashed: 3000 K)

Show mark scheme

Marking point	Mark	Notes
Step 1Wien's displacement law: λ_max = 2.9 × 10⁻³/T, so the hotter body peaks at the shorter wavelength (0.725 μm at 4000 K, 0.97 μm at 3000 K).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2A hotter black body emits more at every wavelength, so its curve lies wholly above the cooler one; the total area under it (the power per unit area, σT⁴) is (4/3)⁴ ≈ 3.2 times larger.	—
Step 3Only graph X shows both features.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis reverses Wien's law: the peak moves to shorter, not longer, wavelengths as the temperature rises.
BCorrect: the 4000 K curve peaks at the shorter wavelength and is above the 3000 K curve at all wavelengths.
CThis draws the two curves with equal areas (equal total power), so they cross; a hotter black body emits more at every wavelength and the curves never cross.
DThis shifts the peak correctly but ignores that the emitted intensity increases strongly with temperature (the total rises as T⁴).

19B-1A-75

Mean kinetic energy and temperature·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The mean random kinetic energy of the molecules of the air in a deep cave is 5.80 × 10⁻²¹ J.

What is the temperature of the air in the cave?

Show mark scheme

Marking point	Mark	Notes
Step 1E_k = (3/2)k_BT, so T = 2E_k/(3k_B).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2T = 2 × 5.80 × 10⁻²¹/(3 × 1.38 × 10⁻²³) = 280 K.	—
Step 3In degrees Celsius: 280 − 273 = 7 °C.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the kelvin temperature is 280 K, which is 7 °C.
BThis omits the factor 3/2: E_k/k_B = 420 K, i.e. 147 °C.
CThis is the kelvin temperature, 280 K, labelled in degrees Celsius without subtracting 273.
DThis multiplies by 3/2 instead of dividing: 3E_k/(2k_B) = 630 K, i.e. 357 °C.

Syllabus understandingB.1 — that Kelvin temperature is a measure of the average kinetic energy E_k of particles as given by E_k = (3/2)k_BT; that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales Command term: Determine

20B-1A-76

Internal energy during a phase change·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

Molten solder on a circuit board solidifies. Its temperature stays at its melting point while it solidifies.

Which row describes the changes in the solder during the solidification?

Mean random kinetic energy of the particlesInternal energy of the solder

Show mark scheme

Marking point	Mark	Notes
Step 1The temperature is constant, so the mean random kinetic energy of the particles (proportional to the kelvin temperature) does not change.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2As the particles become bound into fixed positions in the solid, the intermolecular potential energy decreases and energy is transferred to the surroundings (the latent heat).	—
Step 3Internal energy = total random kinetic energy + total intermolecular potential energy, so it decreases.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis takes the energy released to come from the kinetic energy of the particles; the temperature, and so the mean kinetic energy, is constant.
BThis combines both errors: the kinetic energy is constant, and the internal energy falls because the potential energy falls.
CThis identifies internal energy with temperature; at constant temperature the internal energy can still change through the potential energy.
DCorrect: the mean kinetic energy is unchanged and the internal energy decreases as the intermolecular potential energy falls.

21B-1A-77

Molecular size from density·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksEstimate

The density of liquid nitrogen is 808 kg m⁻³ and the molar mass of nitrogen (N₂) is 28 g mol⁻¹. In the liquid the molecules are in contact, and each molecule may be taken to occupy a cube whose side is equal to the molecular diameter d.

What is the best estimate of d?

Show mark scheme

Marking point	Mark	Notes
Step 1Number of molecules per unit volume = ρN_A/M = 808 × 6.02 × 10²³/0.028 = 1.74 × 10²⁸ m⁻³.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Volume per molecule = 1/n = 5.8 × 10⁻²⁹ m³ = d³.	—
Step 3d = (5.8 × 10⁻²⁹)^1/3 = 3.9 × 10⁻¹⁰ m.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis takes a square root instead of a cube root of the volume per molecule, which does not give a length.
BThis uses the molar mass of atomic nitrogen, 14 g mol⁻¹, doubling the number of molecules per unit volume.
CCorrect: each molecule occupies 5.8 × 10⁻²⁹ m³, a cube of side 3.9 × 10⁻¹⁰ m.
DThis uses the molar mass as 28 kg mol⁻¹ instead of 0.028 kg mol⁻¹, giving 1000 times too few molecules.

Syllabus understandingB.1 — molecular theory in solids, liquids and gases; density ρ as given by ρ = m/V; B.3 — the amount of substance n as given by n = N/N_A Command term: Estimate

22B-1A-78

Latent heat in a mixture·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

An ice cube of mass m at 0 °C is dropped into a mass M of water at a temperature θ, in degrees Celsius, in an insulated cup of negligible heat capacity. All the ice melts and the final temperature of the water is θ_f. The specific heat capacity of water is c.

What is the specific latent heat of fusion of ice?

Show mark scheme

Marking point	Mark	Notes
Step 1Energy given out by the warm water as it cools from θ to θ_f: Mc(θ − θ_f).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy taken in by the ice: mL to melt, then mcθ_f to warm the melted ice from 0 °C to θ_f.	—
Step 3So mL + mcθ_f = Mc(θ − θ_f), giving L = c[M(θ − θ_f) − mθ_f]/m. (Check: with m = 0.020 kg, M = 0.200 kg, θ = 30 °C and L = 3.34 × 10⁵ J kg⁻¹, θ_f = 20.0 °C and the expression returns 3.34 × 10⁵ J kg⁻¹.)	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis forgets that the melted ice must also be warmed from 0 °C to θ_f.
BCorrect: the energy from the cooling water melts the ice and then warms the meltwater to θ_f.
CThis has the wrong sign: the energy used to warm the meltwater must be subtracted from the energy given out by the water.
DThis warms the meltwater to the initial temperature θ of the water instead of to the final temperature θ_f.

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; that temperature difference determines the direction of the resultant thermal energy transfer between bodies Command term: Deduce

23B-1A-79

Heating graph against energy·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A 0.050 kg sample of a solid is heated in an insulated container. The graph shows how its temperature θ varies with the energy Q supplied to it.

Which row gives the specific latent heat of fusion of the substance and the specific heat capacity of the liquid?

Temperature of the sample against energy supplied (drawn to scale)

Specific latent heat of fusionSpecific heat capacity of the liquid

Show mark scheme

Marking point	Mark	Notes
Step 1The horizontal section (melting) runs from 4.0 kJ to 12.0 kJ: L = 8.0 × 10³/0.050 = 1.6 × 10⁵ J kg⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The liquid warms by 120 − 70 = 50 K while 17.0 − 12.0 = 5.0 kJ is supplied.	—
Step 3c = 5.0 × 10³/(0.050 × 50) = 2.0 × 10³ J kg⁻¹ K⁻¹.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThe latent heat is right, but c uses the final temperature 120 °C instead of the temperature change of 50 K: 5.0 × 10³/(0.050 × 120) = 8.3 × 10².
BCorrect: 8.0 kJ melts 0.050 kg, and 5.0 kJ warms the liquid by 50 K.
CThis reads the energy at the end of the horizontal section (12.0 kJ) instead of the length of the section: 12.0 × 10³/0.050 = 2.4 × 10⁵.
DThis makes both errors: the total energy at the end of melting, and the final temperature instead of the temperature change.

24B-1A-80

Conduction through a two-layer wall·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A wall consists of a layer of insulating board fixed to the inside of a layer of brick. The graph shows the variation of temperature θ with distance x through the wall, measured from the inside surface, when the conditions are steady. The rate of thermal energy transfer is the same at every point across the wall.

What is (thermal conductivity of the board)/(thermal conductivity of the brick)?

Temperature against distance through the wall (drawn to scale)

Show mark scheme

Marking point	Mark	Notes
Step 1In the steady state the same rate ΔQ/Δt passes through each layer, and the area A is the same, so k × (temperature gradient) is the same in both layers.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradients from the graph: board (20 − 5)/0.050 = 300 K m⁻¹; brick (5 − 2)/0.200 = 15 K m⁻¹.	—
Step 3k_board/k_brick = 15/300 = 0.050.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the board needs a temperature gradient 20 times steeper to pass the same rate, so its conductivity is 1/20 of the brick's.
BThis uses only the temperature drops, 3 K/15 K, ignoring the different thicknesses.
CThis uses only the thicknesses, 0.050 m/0.200 m, ignoring the different temperature drops.
DThis is the ratio inverted: a steeper temperature gradient means a smaller, not a larger, thermal conductivity.

25B-1A-81

Wien and Stefan–Boltzmann for stars·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two stars X and Y radiate as black bodies. The luminosity of Y is four times the luminosity of X and the radius of Y is four times the radius of X.

What is (peak wavelength of the radiation from Y)/(peak wavelength of the radiation from X)?

Show mark scheme

Marking point	Mark	Notes
Step 1L = σ4πR²T⁴, so T⁴ ∝ L/R²: (T_Y/T_X)⁴ = 4/4² = 1/4.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2T_Y/T_X = (1/4)^1/4 = 1/√2 = 0.71.	—
Step 3Wien: λ_max ∝ 1/T, so the ratio of peak wavelengths = √2 = 1.4.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis is the ratio of the temperatures; the peak wavelength is inversely proportional to the temperature.
BThis takes the surface area to be proportional to R instead of R², so the luminosity and radius effects seem to cancel.
CCorrect: Y is cooler by the factor √2, so its peak is at a wavelength √2 times longer.
DThis takes the square root instead of the fourth root: T ratio = (1/4)^1/2 = 1/2, giving 2.0.

Syllabus understandingB.1 — quantitative analysis of energy transferred by radiation as a result of the emission of electromagnetic waves from the surface of a body, which in the case of a black body can be modelled by the Stefan–Boltzmann law as given by L = σAT⁴; the emission spectrum of a black body and the determination of the temperature of the body using Wien's displacement law as given by λ_maxT = 2.9 × 10⁻³ m K (Guidance: the luminosity of a star may be expressed in terms of the luminosity of the Sun) Command term: Deduce

26B-1A-82

Equilibrium temperature of a sunlit plate·B.1 Thermal energy transfers

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The Sun radiates as a black body of radius R and surface temperature T. A space probe at distance d from the centre of the Sun carries a thin flat plate that faces the Sun. The front of the plate is black; the back is perfectly insulated, so the plate emits radiation only from its front surface. Radiation from all other sources is negligible.

What is the temperature of the plate when it is in thermal equilibrium?

Show mark scheme

Marking point	Mark	Notes
Step 1Luminosity of the Sun L = 4πR²σT⁴; apparent brightness at the probe b = L/4πd² = σT⁴R²/d².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2A plate of area A absorbs bA and emits σAT_p⁴ from its front only. In equilibrium σT_p⁴ = σT⁴R²/d².	—
Step 3T_p = T(R/d)^1/2. (For the Sun at 1.50 × 10¹¹ m this is 395 K.)	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis treats the temperature itself as following an inverse-square law; it is the power per unit area that falls as 1/d², and the temperature is its fourth root.
BThis takes the emitting area of the Sun as its cross-section πR² instead of its surface area 4πR², giving one quarter of the true brightness.
CThis lets the plate emit from both faces, ignoring the insulation on the back: 2σT_p⁴ = b.
DCorrect: the power absorbed per unit area, σT⁴(R/d)², equals the power emitted per unit area, σT_p⁴.

27B-1A-83

Thermal equilibrium·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A metal block P of mass 0.20 kg is placed in a liquid Q of mass 0.30 kg in an insulated container. The graph shows how the temperatures of P and Q vary with time t.

What is (specific heat capacity of P)/(specific heat capacity of Q)?

Temperatures of P and Q against time (drawn to scale)

Show mark scheme

Marking point	Mark	Notes
Step 1The net transfer of energy is from P (hotter) to Q until both reach the same temperature, 40 °C; no energy is lost from the container.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Energy lost by P = energy gained by Q: 0.20 × c_P × (100 − 40) = 0.30 × c_Q × (40 − 20).	—
Step 3c_P/c_Q = (0.30 × 20)/(0.20 × 60) = 0.50.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis inverts the mass ratio: (0.20 × 20)/(0.30 × 60) = 0.22.
BThis ignores the different masses and uses only the temperature changes, 20/60.
CCorrect: 0.20 × 60 × c_P = 0.30 × 20 × c_Q, so 12c_P = 6c_Q and the ratio is 0.50.
DThis is the ratio inverted: the block with the larger temperature change has the smaller heat capacity, not the larger.

Syllabus understandingB.1 — that temperature difference determines the direction of the resultant thermal energy transfer between bodies; quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL Command term: Determine

28B-1A-84

Cooling by evaporation·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksEstimate

Drinking water is kept in a porous clay pot. Water seeps through the clay and evaporates from the outer surface, and this evaporation cools the water in the pot from 35 °C to 20 °C. Assume that no energy enters the pot from the surroundings.

Specific heat capacity of water = 4.2 × 10³ J kg⁻¹ K⁻¹
Specific latent heat of vaporization of water at these temperatures = 2.4 × 10⁶ J kg⁻¹
Specific latent heat of fusion of ice = 3.3 × 10⁵ J kg⁻¹

What is the best estimate of the percentage of the water that evaporates?

Show mark scheme

Marking point	Mark	Notes
Step 1The energy to vaporize the water that escapes comes from the water that remains: m_evapL_v ≈ mcΔT.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2m_evap/m = cΔT/L_v = 4.2 × 10³ × 15/(2.4 × 10⁶) = 0.026.	—
Step 3About 2.6 % of the water evaporates (the temperature change is the same, 15 K, in kelvin or in degrees Celsius).	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the latent heat of vaporization is large, so the evaporation of about 2.6 % of the water cools the rest by 15 K.
BThis uses the final temperature, 20 °C, instead of the temperature change of 15 K.
CThis uses the initial temperature, 35 °C, instead of the temperature change of 15 K.
DThis uses the specific latent heat of fusion; the water changes from liquid to vapour, not from solid to liquid.

Syllabus understandingB.1 — that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales Command term: Estimate

29B-1A-85

Radiation from an electrically heated wire·B.1 Thermal energy transfers

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A wire of constant resistance is enclosed in an evacuated glass bulb. When the current in the wire is I, the steady temperature of the wire is 1200 °C. The wire loses energy only by radiation, and the radiation it absorbs from the bulb is negligible.

The current is increased to 2I. What is the new steady temperature of the wire?

Show mark scheme

Marking point	Mark	Notes
Step 1The electrical power P = I²R is four times larger; in the steady state it equals the power radiated, eσAT⁴.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2So T⁴ increases by a factor 4 and T by 4^1/4 = √2. The kelvin temperature must be used: 1200 °C = 1473 K.	—
Step 3New temperature = 1473 × √2 = 2083 K = 1810 °C.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis takes the power to be proportional to I instead of I²: 1473 × 2^1/4 = 1752 K = 1480 °C.
BThis applies the factor √2 to the Celsius temperature: 1200 × √2 = 1700 °C. The Stefan–Boltzmann law needs the kelvin temperature.
CCorrect: T⁴ ∝ I², so the kelvin temperature rises by √2 to 2083 K.
DThis is the new kelvin temperature, 2083 K, not converted back to degrees Celsius.

Syllabus understandingB.1 — quantitative analysis of energy transferred by radiation as a result of the emission of electromagnetic waves from the surface of a body, which in the case of a black body can be modelled by the Stefan–Boltzmann law as given by L = σAT⁴; that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales; B.5 — electrical power dissipated P = I²R Command term: Determine

30B-1B-01

Stefan–Boltzmann law·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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 / K	1500	1800	2100	2400	2700
P / W	1.16	2.36	4.44	7.49	12.1
log₁₀(T / K)	3.176	3.255	3.322	3.380	3.431
log₁₀(P / W)	0.064	0.373	0.647	0.874	1.083

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)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient from the full line, e.g. (1.083 − 0.064)/(3.431 − 3.176) = 4.0	✓ 1	Accept 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	✓ 1	The 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)	✓ 1	Accept use of a point on the line.
e = 0.40	✓ 1	Accept 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 %	✓ 1	The temperature uncertainty must be multiplied by 4 (power law).
Δe = 0.12 × 0.40 = ±0.05 (e = 0.40 ± 0.05)	✓ 1	Absolute 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)	✓ 1	Allow 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)

Syllabus understandingB.1 — quantitative analysis of energy transferred by radiation … modelled by the Stefan–Boltzmann law as given by L = σAT⁴; B.2 — emissivity as the ratio of the power radiated per unit area by a surface compared to that of an ideal black surface at the same temperature; Tools — linearising a power law with logarithms, propagation of uncertainties Command term: Determine

31B-1B-08

Specific heat capacity·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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.

N	25	50	75	100	125	150
ΔT / K	1.9	3.5	5.5	7.1	9.0	10.9

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)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. 10.8/150 = 0.072 K	✓ 1	Accept 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)	✓ 1	All 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	✓ 1	Allow 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	✓ 1	Allow 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)	✓ 1	Any 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	✓ 1	The doubled reading uncertainty is required.
N ≥ 10/0.072 ≈ 139 inversions	✓ 1	Allow 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)

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c as given by Q = mcΔT; A.3 — the principle of the conservation of energy; gravitational potential energy; Tools — uncertainty of a difference of two readings, improving an experiment Command term: Determine

32B-1B-11

Density of a gas·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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 / kPa	100	80	60	40	20
m / g	153.807	153.506	153.210	152.910	152.614

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 point	Mark	Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (153.806 − 152.315)/(100 − 0) = 0.0149	✓ 1	Accept 0.0145–0.0153.
Unit g kPa⁻¹	✓ 1	Accept 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)	✓ 1	Both 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	✓ 1	Allow 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	✓ 1	Accept 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	✓ 1	Allow 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	✓ 1	Allow 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)

Syllabus understandingB.1 — density ρ as given by ρ = m/V; B.3 — the equations governing the behaviour of ideal gases as given by PV = nRT; the amount of substance n; Tools — gradient and intercept with units, propagation of uncertainties Command term: Determine

33B-1B-12

Cooling curve and latent heat·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)(i)
Tangent drawn to the liquid curve at 44 °C (about 4.4 min); gradient ≈ −5.0 K min⁻¹	✓ 1	Accept 4.5–5.5 K min⁻¹ in magnitude.
Rate = 5.0/60 = 0.083 K s⁻¹	✓ 1	Accept 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	✓ 1	Allow ECF from (a).
P = 4.8 W ≈ 5 W	✓ 1	The 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	✓ 1	Accept 15.0–16.5 min.
L = Pt/m = 4.8 × 936/0.025 = 1.8 × 10⁵ J kg⁻¹	✓ 1	Allow 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 %	✓ 1	Allow 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)

Syllabus understandingB.1 — that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; quantitative analysis of thermal energy transfers with Q = mcΔT and Q = mL; Tools — gradient of a tangent, uncertainty of a difference of two readings Command term: Determine

34B-1B-13

Apparent brightness and the inverse-square law·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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 / m	0.10	0.15	0.20	0.25	0.30	0.40
b / W m⁻²	49.2	24.3	14.7	9.72	6.99	4.02
b^−1/2 / W^−1/2 m	0.143	0.203	0.261	0.321	0.378	0.499

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 point	Mark	Notes
Part (a)(i)
Gradient from two well-separated points on the line, e.g. (0.499 − 0.143)/(0.40 − 0.10) = 1.18	✓ 1	Accept 1.15–1.21.
Unit W^−1/2	✓ 1	Accept 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	✓ 1	Allow 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	✓ 1	Accept 0.020–0.030. Alternatively the extended line meets the d axis at d = −d₀.
d₀ = 0.025/1.18 = 0.021 m	✓ 1	Allow 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	✓ 1	Allow 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)

Syllabus understandingB.1 — the concept of apparent brightness b; luminosity L of a body as given by b = L/4πd²; that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer; Tools — linearising a relationship, gradient with unit, intercept as a systematic offset Command term: Determine

35B-1B-24

Conduction·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

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⁻¹	P / W
1.0	100.0	17.6
1.5	66.7	11.9
2.0	50.0	9.1
2.5	40.0	7.7
3.0	33.3	6.4
4.0	25.0	5.1
5.0	20.0	4.3

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 point	Mark	Notes
Part (a)(i)
Gradient read from two well-separated points on the line, e.g. (17.5 − 0.93)/(100 − 0) = 0.166	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	Allow 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	✓ 1	Accept 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 %)	✓ 1	Accept ±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	✓ 1	Allow 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)

Syllabus understandingB.1 — quantitative analysis of rate of thermal energy transfer by conduction in terms of the type of material and cross-sectional area of the material and the temperature gradient as given by ΔQ/Δt = kAΔT/Δx; Tools — linearizing a relationship, gradient and intercept with units, lines of maximum and minimum gradient, propagation of uncertainties Command term: Determine

36B-1B-25

Testing Wien's displacement law·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A student tests Wien's displacement law using an electrically heated ceramic element. An infrared spectrometer measures the wavelength λ_max 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/λ_max against the thermocouple temperature T in kelvin, with the line of best fit.

θ / °C	λ_max / μm	(1/λ_max) / μm⁻¹
450	4.36	0.229
550	3.80	0.263
650	3.35	0.299
750	3.01	0.332
850	2.72	0.368
950	2.50	0.400
1050	2.29	0.437

Reciprocal of the peak wavelength against thermocouple temperature (drawn to scale)

(a)

(i)

The hypothesis is that λ_maxT 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 point	Mark	Notes
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)	✓ 1	Award [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	✓ 1	The 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	✓ 1	Accept 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	✓ 1	Allow ECF from the gradient. Accept 2.8–3.0 × 10⁻³ m K.
Part (c)(i)
Intercept ≈ 58 K	✓ 1	Accept 40–80 K, from the graph or from the gradient and a point. Allow ECF from (b).
1/λ_max ∝ (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 λ_maxT was not constant in (a)	✓ 1	Reference 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)	✓ 1	The 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)

Syllabus understandingB.1 — the emission spectrum of a black body and the determination of the temperature of the body using Wien's displacement law as given by λ_maxT = 2.9 × 10⁻³ m K; the conversion between the Kelvin and Celsius scales; Tools — testing a hypothesis, linearizing a relationship, gradient with a unit, intercept, identifying and reducing systematic error Command term: Determine

37B-1B-26

Density from buoyancy·B.1 Thermal energy transfers

Paper 1BEasy7 marks

Data-based question7 steps to full marksDetermine

A student determines the density of a small stone. The stone hangs from a force sensor by a thin thread and is fully immersed in a liquid of known density ρ_L. The reading F of the sensor is recorded. The stone is dried and the measurement is repeated with ethanol–water mixtures and sugar solutions of different densities. The uncertainty in each reading of F is ±0.005 N.

The buoyancy force on the stone is ρ_LVg, where V is the volume of the stone, so F = mg − ρ_LVg, where m is the mass of the stone. The table and the graph show the results with the line of best fit, extended to ρ_L = 0.

ρ_L / kg m⁻³	790	900	1000	1100	1200	1300
F / N	1.260	1.195	1.131	1.034	0.999	0.935

Reading of the force sensor against density of the liquid, with the line of best fit (drawn to scale). The F axis does not start at zero.

(a)

(i)

One reading is anomalous. Identify it and suggest a cause that is consistent with the position of the point.

(2)

(b)

(i)

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

(2)

(c)

(i)

Determine the volume V of the stone.

(1)

(d)

(i)

Determine the density of the stone.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The reading at ρ_L = 1100 kg m⁻³ (1.034 N), which is about 0.03 N below the line, far outside ±0.005 N	✓ 1
Air bubbles clinging to the stone increase the volume of liquid displaced, so the buoyancy force is larger and F is smaller	✓ 1	The cause must make F too small. Accept: the stone touched the side or bottom of the beaker (part of the weight supported). "Human error" or "misreading" scores [0].
Part (b)(i)
Gradient from two well-separated points on the line, e.g. (0.872 − 1.77)/(1400 − 0) = −6.4 × 10⁻⁴	✓ 1	Accept −6.1 × 10⁻⁴ to −6.7 × 10⁻⁴; the sign may be omitted.
Unit: N m³ kg⁻¹	✓ 1	Accept m⁴ s⁻².
Part (c)(i)
Gradient = −Vg, so V = 6.4 × 10⁻⁴/9.81 = 6.5 × 10⁻⁵ m³	✓ 1	Allow ECF from (b). Accept 6.2–6.8 × 10⁻⁵ m³.
Part (d)(i)
Intercept on the F axis = mg ≈ 1.77 N	✓ 1	Accept 1.74–1.79 N.
ρ = m/V = 1.77/(9.81 × 6.5 × 10⁻⁵) = 2.8 × 10³ kg m⁻³	✓ 1	Allow ECF from (c). ALTERNATIVE: ρ = intercept/\|gradient\|. Accept 2.6–2.9 × 10³ kg m⁻³. A reading from the table alone (e.g. 1.26 N) used as mg scores [0] for the first mark.

Answers: (b)(i) −6.4 × 10⁻⁴ N m³ kg⁻¹ · (c)(i) 6.5 × 10⁻⁵ m³ · (d)(i) 2.8 × 10³ kg m⁻³ (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — density ρ as given by ρ = m/V; A.2 — buoyancy F_b = ρVg; Tools — identifying an anomalous result, gradient and intercept with units, extrapolation Command term: Determine

38B-1B-27

Specific latent heat by the two-power method·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A student determines the specific latent heat of vaporization L of water. An electric heater keeps water boiling in a beaker that stands on a balance. When the water is boiling steadily, the student records the potential difference V across the heater, the current I in it, and the loss of mass Δm of the water in 600 s. The experiment is repeated for different heater powers P = VI. The rate of loss of mass is r = Δm/600 s.

The student models the results by P = Lr + h, where h is the rate at which energy is transferred from the beaker to the surroundings. The table and the graph show the results, with the line of best fit extended to r = 0.

V / V	I / A	P / W	Δm / g	r / 10⁻⁵ kg s⁻¹
6.0	2.40	14.4	1.70	0.283
8.0	3.20	25.6	4.62	0.770
10.0	4.00	40.0	8.55	1.425
12.0	4.80	57.6	13.09	2.182
14.0	5.60	78.4	18.61	3.102

Heater power against rate of loss of mass of the water, with the line of best fit (drawn to scale)

(a)

(i)

The potential difference is measured to ±0.1 V and the current to ±0.05 A. Determine the absolute uncertainty in P for the reading V = 10.0 V.

(2)

(b)

(i)

Determine L from the graph.

(2)

(c)

(i)

Determine h and explain why it is reasonable to treat h as the same for every heater power.

(2)

(d)

(i)

Another student calculates L from the single reading P = 40.0 W as P/r and obtains 2.8 × 10⁶ J kg⁻¹. Explain, with reference to your answer to (c), why this value is larger than your answer to (b).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Fractional uncertainty = 0.1/10.0 + 0.05/4.00 = 0.010 + 0.0125 = 0.0225 (2.25 %)	✓ 1	Fractional (percentage) uncertainties must be added.
Absolute uncertainty = 0.0225 × 40.0 = ±0.9 W	✓ 1	Accept ±0.9 W or ±1 W. Award [1 max] for ±0.15 (absolute uncertainties added).
Part (b)(i)
L is the gradient of the graph of P against r	✓ 1
Gradient e.g. (87.4 − 8.0)/(3.5 × 10⁻⁵) = 2.3 × 10⁶ J kg⁻¹	✓ 1	Accept 2.1–2.4 × 10⁶ J kg⁻¹. The power of ten of the r axis must be used.
Part (c)(i)
h = intercept on the P axis ≈ 8.0 W	✓ 1	Accept 6–10 W.
The water is at its boiling point (100 °C) in every run, so the temperature difference between the beaker and the surroundings, and hence the rate of energy loss, is the same	✓ 1	Reference to the constant temperature of the boiling water is required.
Part (d)(i)
P/r includes the power h lost to the surroundings, which does not vaporize water; only (P − h) vaporizes water, so P/r overestimates L (the graph method removes h as the intercept)	✓ 1	Allow ECF from (b) and (c). A statement that energy is lost is not enough: the loss must be linked to h.

Answers: (a)(i) ±0.9 W · (b)(i) 2.3 × 10⁶ J kg⁻¹ · (c)(i) 8.0 W (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; B.5 — electrical power P = VI; Tools — propagation of uncertainties, gradient and intercept of a linear graph, eliminating a systematic loss Command term: Determine

39B-1B-28

Method of mixtures with a linear graph·B.1 Thermal energy transfers

Paper 1BMedium7 marks

Data-based question7 steps to full marksDetermine

A student determines the specific heat capacity c_b of a metal. A block of the metal of mass M = 0.200 kg is heated in boiling water to θ₀ = 100.0 °C and then transferred quickly into a mass m of water at θ_i = 20.0 °C in a polystyrene cup. The highest temperature θ_f reached by the water is recorded. The experiment is repeated with different masses of water, always starting at 20.0 °C. The cup and thermometer have a total heat capacity C, and the specific heat capacity of water is c_w = 4.18 × 10³ J kg⁻¹ K⁻¹. The thermometer reads to ±0.1 °C.

The student calculates y = (θ₀ − θ_f)/(θ_f − θ_i). The table and the graph show the results, with the line of best fit extended to m = 0.

m / kg	θ_f / °C	y
0.050	40.2	2.96
0.100	32.0	5.67
0.150	28.6	8.30
0.200	26.6	11.12
0.250	25.4	13.81

y = (θ₀ − θ_f)/(θ_f − θ_i) against mass of water, with the line of best fit (drawn to scale)

(a)

(i)

Show that y = (c_w/Mc_b)m + C/Mc_b, assuming no energy is transferred to the surroundings.

(2)

(b)

(i)

Determine c_b.

(2)

(c)

(i)

Determine C.

(1)

(d)

(i)

For m = 0.250 kg, determine the percentage uncertainty in (θ_f − θ_i), and suggest one change to the procedure, using the same thermometer, that would reduce it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Energy lost by the block = energy gained by the water and the cup: Mc_b(θ₀ − θ_f) = (mc_w + C)(θ_f − θ_i)	✓ 1	The heat capacity C must be included.
Dividing by Mc_b(θ_f − θ_i) gives y = (c_w/Mc_b)m + C/Mc_b	✓ 1	The division step must be seen.
Part (b)(i)
Gradient of the line, e.g. (16.5 − 0.23)/0.30 = 54.3 kg⁻¹	✓ 1	Accept 52–57 kg⁻¹.
c_b = c_w/(M × gradient) = 4.18 × 10³/(0.200 × 54.3) = 385 J kg⁻¹ K⁻¹	✓ 1	Allow ECF from (a). Accept 370–400 J kg⁻¹ K⁻¹.
Part (c)(i)
Intercept ≈ 0.23, so C = intercept × Mc_b = 0.23 × 0.200 × 385 = 17 J K⁻¹	✓ 1	Allow ECF from (b). Accept an intercept of 0.1–0.4, giving 8–31 J K⁻¹.
Part (d)(i)
Uncertainty in the difference = 2 × 0.1 = 0.2 K; θ_f − θ_i = 5.4 K, so the uncertainty is 3.7 %	✓ 1	Accept 3.5–4 %. The uncertainties of the two readings must be added; 0.1/5.4 (1.9 %) scores [0].
Use a block of larger mass (or smaller masses of water), so that the temperature rise of the water is larger	✓ 1	The change must increase the temperature rise. "Use a more precise thermometer" or "repeat the readings" scores [0].

Answers: (b)(i) 385 J kg⁻¹ K⁻¹ · (c)(i) 17 J K⁻¹ · (d)(i) 3.7 % (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; that temperature difference determines the direction of the resultant thermal energy transfer between bodies; Tools — deriving and linearising a relationship, gradient and intercept, uncertainty of a difference of two readings, improving an experiment Command term: Determine

40B-2-01

Specific and latent heat·B.1 Thermal energy transfers

Paper 2Medium10 marks

Short answer & extended response10 steps to full marksDetermine

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.

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 point	Mark	Notes
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	✓ 1	Reference 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	✓ 1	Accept 660–680 s.
P = mL/t = 0.80 × 3.34 × 10⁵/670 = 399 W	✓ 1	Full 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/mc_ice)/(P/mc_water) = c_water/c_ice (P and m cancel)	✓ 1	The 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 c_water = 2.0 × 2.1 × 10³ = 4.2 × 10³ J kg⁻¹ K⁻¹	✓ 1	Accept 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	✓ 1	Or the sum mc_iceΔθ + mL + mc_waterΔθ = 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)	✓ 1	Accept 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	✓ 1	Allow ECF from (a)(ii).
(b): both gradients become P/(mc + C) and the ice gradient is reduced relatively more; the measured ratio becomes (mc_water + C)/(mc_ice + C) ≈ 1.86, not c_water/c_ice, 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)	✓ 1	Allow 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)

Syllabus understandingB.1 — that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; quantitative analysis of thermal energy transfers with specific heat capacity and specific latent heat as given by Q = mcΔT and Q = mL; A.3 — energy density of the fuel sources; efficiency η = E_output/E_input Command term: Determine

41B-2-06

Luminosity, apparent brightness and black-body radiation·B.1 Thermal energy transfers

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine

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 point	Mark	Notes
Part (a)(i)
λ_max = 2.9 × 10⁻³/306 = 9.5 × 10⁻⁶ m = 9.5 μm	✓ 1	Kelvin 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	✓ 1	Accept 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	✓ 1	Allow ECF from (a).
Number per second = 877/2.1 × 10⁻²⁰ = 4.2 × 10²² s⁻¹	✓ 1	Allow 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 λ_max, so the actual number is larger)	✓ 1	Either idea.
Part (d)(i)
b = L/4πd² = 877/(4π × 50²) = 2.8 × 10⁻² W m⁻²	✓ 1	Allow ECF from (b).
Area of the lens = π × (0.015)² = 7.07 × 10⁻⁴ m²	✓ 1	Using the diameter as the radius scores 0 for this mark.
Power = 2.8 × 10⁻² × 7.07 × 10⁻⁴ = 2.0 × 10⁻⁵ W	✓ 1	Allow 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	✓ 1	Accept 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	✓ 1	Allow 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)

Syllabus understandingB.1 — the emission spectrum of a black body and Wien's displacement law λ_maxT = 2.9 × 10⁻³ m K; the Stefan–Boltzmann law L = σAT⁴; apparent brightness b = L/4πd²; B.2 — emissivity; E.1 — photons of energy E = hf Command term: Determine

42B-2-09

Specific and latent heat·B.1 Thermal energy transfers

Paper 2Medium10 marks

Short answer & extended response10 steps to full marksDetermine

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 point	Mark	Notes
Part (a)(i)
Mechanical power = Fv = 15 × 4.0 = 60 W	✓ 1	The 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	✓ 1	Allow 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	✓ 1	Allow 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	✓ 1	Allow ECF from (a)(ii).
ΔT = 1.08 × 10⁵/(65 × 3.5 × 10³) = 0.47 K	✓ 1	Accept 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)

Syllabus understandingB.1 — molecular theory in solids, liquids and gases; that a phase change represents a change in particle behaviour; Q = mcΔT and Q = mL; A.3 — power P = Fv; efficiency η = P_output/P_input Command term: Determine

43B-2-11

Conduction·B.1 Thermal energy transfers

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine

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.

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 point	Mark	Notes
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	✓ 1	Reference 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)	✓ 1	Using 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	✓ 1	Accept 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)	✓ 1	Allow 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	✓ 1	Using πR² (9.6 × 10¹² W in the answer) scores [0] for this mark.
Total rate = 5.10 × 10¹⁴ × 0.0754 = 3.8 × 10¹³ W	✓ 1	Allow 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	✓ 1	Award [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)	✓ 1	Allow 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	✓ 1	Accept 88–92 °C and 11–13 °C.
η_Carnot = 1 − 285/363 = 0.21	✓ 1	Accept 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)	✓ 1	Allow 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	✓ 1	Equivalent: 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 (T_h) falls, and the maximum (and actual) efficiency of the station decreases	✓ 1	Allow 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)

Syllabus understandingB.1 — conduction in terms of the difference in the kinetic energy of particles; quantitative analysis of rate of thermal energy transfer by conduction in terms of the type of material and cross-sectional area of the material and the temperature gradient as given by ΔQ/Δt = kAΔT/Δx; that Kelvin and Celsius scales are used to express temperature; B.2 — the solar constant S; that the incoming radiative power is dependent on the projected surface of a planet along the direction of the path of the rays, resulting in a mean value of the incoming intensity being S/4; albedo; B.4 — that the Carnot cycle sets a limit for the efficiency of a heat engine at the temperatures of its heat reservoirs as given by η_Carnot = 1 − T_c/T_h; Tools — interpret features of graphs including gradient Command term: Determine

44B-2-18

Specific and latent heat·B.1 Thermal energy transfers

Paper 2Medium11 marks

Short answer & extended response11 steps to full marksDetermine

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 point	Mark	Notes
Part (a)(i)
V = π × (0.20 × 10⁻³)² × 0.20 × 10⁻³ = 2.51 × 10⁻¹¹ m³	✓ 1	Award [0] if the diameter is used as the radius.
m = ρV = 7.9 × 10³ × 2.51 × 10⁻¹¹ = 1.99 × 10⁻⁷ kg	✓ 1	An answer to at least 3 s.f. is required.
Part (b)(i)
Temperature rise = 1800 − 293 = 1507 K	✓ 1	Award [2 max] for 0.23 J (1780 K used).
mcΔT = 1.99 × 10⁻⁷ × 490 × 1507 = 0.147 J	✓ 1	Allow 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	✓ 1	Allow 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	✓ 1	Allow 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)

Syllabus understandingB.1 — density ρ = m/V; quantitative analysis of thermal energy transfers Q = mcΔT and Q = mL; that a phase change occurs at constant temperature; the conversion between Kelvin and Celsius scales; E.1 — photon energy E = hf Command term: Determine

45B-2-19

Specific and latent heat·B.1 Thermal energy transfers

Paper 2Medium11 marks

Short answer & extended response8 steps to full marksDetermine

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 point	Mark	Notes
Part (a)(i)
58 °C = 331 K; the rise is 38 K	✓ 1	Both needed.
Part (b)(i)
Latent heat released = energy to warm the contents: m_x × 2.64 × 10⁵ = 0.090 × 3.0 × 10³ × 38	✓ 1
m_x = 1.03 × 10⁴/2.64 × 10⁵ = 0.039 kg ≈ 0.04 kg	✓ 1	Answer 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	✓ 1	Accept 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	✓ 1	Accept 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	✓ 1	Accept 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)	✓ 1	Accept 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)

Syllabus understandingB.1 — that Kelvin and Celsius scales are used to express temperature; that the internal energy of a system is the total intermolecular potential energy plus the total random kinetic energy of the molecules; that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; quantitative analysis of thermal energy transfers Q = mcΔT and Q = mL Command term: Determine

46B-2-20

Specific and latent heat·B.1 Thermal energy transfers

Paper 2Hard11 marks

Short answer & extended response11 steps to full marksDetermine

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.

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 point	Mark	Notes
Part (a)(i)
Tangent drawn at the steepest part of the curve (t ≈ 60 s)	✓ 1
Deceleration = gradient ≈ 1.5 × 10² m s⁻²	✓ 1	Accept 130–165 m s⁻².
F = ma = 40 × 146 = 5.8 × 10³ N	✓ 1	Allow 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	✓ 1	Accept 7.4–7.5 and 0.5 km s⁻¹.
ΔE_k/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	✓ 1	Allow 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	✓ 1	Allow ECF from (c).
Fraction melted = 2.1 × 10⁵/3.97 × 10⁵ = 0.54 (about half)	✓ 1	Accept 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	✓ 1	Allow 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)

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q = mcΔT and Q = mL; that a phase change represents a change in particle behaviour at constant temperature; A.1 — acceleration as the gradient of a velocity–time graph; A.2 — Newton's second law; A.3 — kinetic energy E_k = ½mv² and the conservation of energy Command term: Determine

47B-2-38

Convection·B.1 Thermal energy transfers

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine

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.

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 ρ_h and that in the cool pipe has density ρ_c. Show that, at the bottom of the pipes, the difference between the pressures due to the two columns of water is (ρ_c − ρ_h)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 point	Mark	Notes
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	✓ 1	Reference 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	✓ 1	Accept 1.6 × 10³ W.
Part (b)(ii)
Net rate = eσA(T⁴ − T_s⁴) = 0.12 × 5.67 × 10⁻⁸ × 2.0 × (345⁴ − 300⁴)	✓ 1	Award [1 max] for 193 W (absorption from the surroundings ignored).
= 83 W	✓ 1	Accept 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	✓ 1	Allow ECF from (b)(i) and (b)(ii).
Mass per second = 1443/(4.18 × 10³ × 15) = 0.0230 kg s⁻¹	✓ 1	Accept 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 ρ_cgh − ρ_hgh = (ρ_c − ρ_h)gh	✓ 1	Use of p = F/A with the weight is required.
Part (d)(ii)
Δp = (997.0 − 992.2) × 9.81 × 1.6 = 75 Pa	✓ 1	Allow 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 (ρ_c − ρ_h)gh drives a flow	✓ 1	Allow 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)

Syllabus understandingB.1 — qualitative description of thermal energy transferred by convection due to fluid density differences; quantitative analysis of rate of thermal energy transfer by conduction as given by ΔQ/Δt = kAΔT/Δx; the Stefan–Boltzmann law; Q = mcΔT; B.2 — albedo and emissivity; A.2 — weight and the definition of pressure as force per unit area (B.3 — p = F/A) Command term: Determine

48B-2-40

Latent heat in a paraffin-wax bath·B.1 Thermal energy transfers

Paper 2Easy13 marks

Short answer & extended response13 steps to full marksDetermine

In paraffin-wax therapy a patient dips a hand several times into molten wax kept at 52 °C. A layer of wax forms on the skin, solidifies at the melting point of the wax, 50 °C, and then cools to the skin temperature of 34 °C. The energy released keeps the hand warm.

Data for the wax: density of solid wax = 900 kg m⁻³; specific heat capacity of liquid wax = 2.9 × 10³ J kg⁻¹ K⁻¹; specific heat capacity of solid wax = 2.5 × 10³ J kg⁻¹ K⁻¹; specific latent heat of fusion = 2.1 × 10⁵ J kg⁻¹.

(a)

(i)

Outline, in terms of the particles of the wax, why energy is released as the wax solidifies although its temperature does not change.

(2)

(b)

(i)

The layer of solid wax covers an area of 0.040 m² of skin with a uniform thickness of 2.0 mm. Show that the mass of the layer is about 0.07 kg.

(2)

(ii)

Determine the total energy transferred from this wax as it cools from 52 °C to 34 °C.

(3)

(iii)

Calculate the percentage of the energy in (b)(ii) that is released as the wax solidifies.

(1)

(c)

(i)

Calculate (mean random kinetic energy of the wax particles at 52 °C)/(mean random kinetic energy of the wax particles at 34 °C).

(1)

(d)

(i)

The bath holds 4.0 kg of wax. Initially the wax is solid at 20 °C. Calculate the minimum energy needed to raise it to the working condition of the bath.

(2)

(ii)

The heater of the bath has a power of 400 W. Determine the minimum time, in minutes, needed to prepare the bath.

(1)

(iii)

The heater operates from a 230 V mains supply. Calculate the resistance of the heater.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
As the wax solidifies the particles become bound in fixed positions (forces of attraction pull them together), so the total intermolecular potential energy decreases	✓ 1	"Bonds form" is accepted for this mark.
The temperature is constant, so the mean random kinetic energy of the particles is unchanged; the decrease in internal energy is transferred to the hand as thermal energy	✓ 1	Both the constant kinetic energy and the link to the energy transferred are required.
Part (b)(i)
V = 0.040 × 2.0 × 10⁻³ = 8.0 × 10⁻⁵ m³	✓ 1	The thickness must be converted to metres.
m = ρV = 900 × 8.0 × 10⁻⁵ = 0.072 kg	✓ 1	An answer to at least 2 s.f. or the full substitution is required.
Part (b)(ii)
Liquid 52 °C → 50 °C: 0.072 × 2.9 × 10³ × 2 = 418 J	✓ 1	Allow ECF from (b)(i). Award [2 max] if the cooling of the liquid wax is omitted (1.80 × 10⁴ J).
Solidifying: 0.072 × 2.1 × 10⁵ = 1.51 × 10⁴ J; solid 50 °C → 34 °C: 0.072 × 2.5 × 10³ × 16 = 2.88 × 10³ J	✓ 1	The temperature changes are the same in kelvin and in degrees Celsius.
Total = 1.84 × 10⁴ J	✓ 1	Accept 1.8–1.9 × 10⁴ J (1.79 × 10⁴ J if 0.07 kg is used).
Part (b)(iii)
1.51 × 10⁴/1.84 × 10⁴ = 82 %	✓ 1	Allow ECF from (b)(ii). Accept 80–83 %.
Part (c)(i)
E_k ∝ T in kelvin: (52 + 273)/(34 + 273) = 325/307 = 1.06	✓ 1	Award [0] for 52/34 = 1.53 (Celsius temperatures).
Part (d)(i)
Solid 20 °C → 50 °C: 4.0 × 2.5 × 10³ × 30 = 3.0 × 10⁵ J; melting: 4.0 × 2.1 × 10⁵ = 8.4 × 10⁵ J	✓ 1	The latent heat term is required.
Liquid 50 °C → 52 °C: 2.3 × 10⁴ J; total = 1.16 × 10⁶ J	✓ 1	Accept 1.16–1.17 × 10⁶ J. Award [1 max] for 1.14 × 10⁶ J (warming of the liquid to 52 °C omitted) only if the melting term is present.
Part (d)(ii)
t = 1.16 × 10⁶/400 = 2.91 × 10³ s = 48 min	✓ 1	Allow ECF from (d)(i). Accept 48–49 min.
Part (d)(iii)
R = V²/P = 230²/400 = 132 Ω	✓ 1

Answers: (b)(i) 0.072 kg · (b)(ii) 1.84 × 10⁴ J · (b)(iii) 82 % · (c)(i) 1.06 · (d)(i) 1.16 × 10⁶ J · (d)(ii) 48 min · (d)(iii) 132 Ω (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — that the internal energy of a system is the total intermolecular potential energy arising from the forces between the molecules plus the total random kinetic energy of the molecules arising from their random motion; that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; density ρ as given by ρ = m/V; that Kelvin temperature is a measure of the average kinetic energy E_k of particles as given by E_k = (3/2)k_BT; that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales; A.3 — power as the rate of energy transfer; B.5 — electrical power P = V²/R Command term: Determine

49B-2-41

Boiling of liquid nitrogen in a storage flask·B.1 Thermal energy transfers

Paper 2Easy14 marks

Short answer & extended response14 steps to full marksDetermine

Liquid nitrogen is stored in a double-walled flask. The space between the walls is evacuated and the facing surfaces of the walls are silvered. The liquid is at its boiling point, 77 K, and the gas formed escapes through a loose cap at the top of a narrow neck. The flask stands in a room at 20 °C.

Data for nitrogen: specific latent heat of vaporization = 2.0 × 10⁵ J kg⁻¹; density of the liquid = 808 kg m⁻³; molar mass = 0.028 kg mol⁻¹.

(a)

(i)

Calculate the mean random kinetic energy of a molecule of the nitrogen gas just above the liquid.

(1)

(b)

(i)

Energy enters the liquid continuously. Explain, in terms of the molecules, why the temperature of the liquid does not rise.

(2)

(c)

(i)

The flask loses 0.20 kg of nitrogen per day by boiling. Show that the rate at which thermal energy enters the liquid is about 0.5 W.

(2)

(ii)

80 % of this energy enters by conduction down the glass wall of the neck. The wall has a cross-sectional area of 3.1 × 10⁻⁴ m² and a length of 0.20 m. The top of the neck is at room temperature and the bottom is at the temperature of the liquid. Determine the thermal conductivity of the glass.

(3)

(d)

(i)

The nitrogen that boils off in one day warms to room temperature at a pressure of 1.01 × 10⁵ Pa. Calculate its volume.

(2)

(ii)

Calculate (volume of this gas)/(volume of the liquid from which it formed).

(1)

(iii)

Suggest, with reference to (d)(ii), why the flask must never be sealed.

(1)

(e)

(i)

Explain how the evacuated space and the silvered surfaces reduce the rate at which energy reaches the liquid.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E_k = (3/2) × 1.38 × 10⁻²³ × 77 = 1.59 × 10⁻²¹ J	✓ 1
Part (b)(i)
The energy supplied is used to separate molecules, overcoming the intermolecular forces, so it increases the intermolecular potential energy as liquid changes to gas	✓ 1	"The energy is used to boil the liquid" alone scores [0].
The mean random kinetic energy of the molecules, of which the temperature is a measure, does not change	✓ 1	Accept: the most energetic molecules leave, removing the energy supplied.
Part (c)(i)
Energy per day = 0.20 × 2.0 × 10⁵ = 4.0 × 10⁴ J	✓ 1
Rate = 4.0 × 10⁴/(24 × 3600) = 0.463 W	✓ 1	An answer to at least 2 s.f. is required; the conversion of a day to seconds must be seen.
Part (c)(ii)
Temperature difference = (20 + 273) − 77 = 216 K	✓ 1	Award [0] for this mark for 20 − 77 or 97 K.
Rate by conduction = 0.80 × 0.463 = 0.37 W	✓ 1	Allow ECF from (c)(i).
k = (ΔQ/Δt)Δx/(AΔT) = 0.37 × 0.20/(3.1 × 10⁻⁴ × 216) = 1.1 W m⁻¹ K⁻¹	✓ 1	Accept 1.1–1.2 W m⁻¹ K⁻¹ (1.2 if 0.5 W is used).
Part (d)(i)
n = 0.20/0.028 = 7.14 mol	✓ 1
V = nRT/p = 7.14 × 8.31 × 293/1.01 × 10⁵ = 0.17 m³	✓ 1	The temperature must be in kelvin; 0.012 m³ (20 used) scores [1 max].
Part (d)(ii)
Volume of liquid = 0.20/808 = 2.5 × 10⁻⁴ m³; ratio = 0.17/2.5 × 10⁻⁴ ≈ 7 × 10²	✓ 1	Allow ECF from (d)(i). Accept 680–710.
Part (d)(iii)
The liquid keeps boiling, and the gas would occupy several hundred times the volume of the liquid; in a closed fixed volume the amount of gas, and so the pressure, would rise until the flask burst	✓ 1	Allow ECF from (d)(ii). Reference to the rise in pressure is required.
Part (e)(i)
The vacuum contains (almost) no particles, so energy cannot cross the gap by conduction or convection, which both need a material medium	✓ 1	[2 max]. Both mechanisms must be named.
Radiation can cross a vacuum; silvered surfaces are poor emitters and good reflectors of thermal radiation, so they reduce the energy transferred by radiation	✓ 1	Accept: low emissivity / low absorption of the silvered surfaces.

Answers: (a)(i) 1.59 × 10⁻²¹ J · (c)(i) 0.463 W · (c)(ii) 1.1 W m⁻¹ K⁻¹ · (d)(i) 0.17 m³ · (d)(ii) ≈ 7 × 10² (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — that Kelvin temperature is a measure of the average kinetic energy E_k of particles as given by E_k = (3/2)k_BT; that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; quantitative analysis of rate of thermal energy transfer by conduction in terms of the type of material and cross-sectional area of the material and the temperature gradient as given by ΔQ/Δt = kAΔT/Δx; that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer; that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales; B.3 — the amount of substance n; the ideal gas equation pV = nRT Command term: Determine

50B-2-42

Conduction into a cool box·B.1 Thermal energy transfers

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

A closed cool box is a rectangular box with walls, base and lid of expanded polystyrene, each 3.0 cm thick. The inside of the box measures 0.40 m × 0.30 m × 0.30 m. The box contains 2.40 kg of ice and stands in a room at 25 °C. At intervals the remaining ice is removed briefly and weighed. The graph shows how the mass m of ice varies with time t.

Assume that the inner surfaces of the box are at 0 °C and the outer surfaces at 25 °C, and that energy enters only by conduction through the walls, base and lid, using the inside area. Specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹.

Mass of ice in the box against time, with the line of best fit (drawn to scale)

(a)

(i)

Explain why the graph is a straight line.

(2)

(b)

(i)

Use the graph to determine the rate at which thermal energy enters the box.

(2)

(c)

(i)

Determine the thermal conductivity of the polystyrene.

(3)

(d)

(i)

A different cool box is a cube of inside side a, with walls of thickness x and thermal conductivity k, completely filled with ice of density ρ at 0 °C. Show that the time for all the ice to melt is t = ρaLx/(6kΔT), where ΔT is the temperature difference across the walls.

(2)

(ii)

Deduce how the time in (d)(i) changes when a is doubled, everything else being unchanged.

(1)

(e)

(i)

The lid of the original box makes an airtight seal. The lid was closed when the air inside was at 25 °C and atmospheric pressure, 1.01 × 10⁵ Pa. The air inside is then cooled to 0 °C at constant volume. Determine the resultant force due to the air pressure on the lid, which has an area of 0.40 m × 0.30 m.

(2)

(f)

(i)

After all the ice has melted, the 2.40 kg of water at 0 °C warms. Using your answer to (b), estimate the initial rate of rise of the temperature of the water, and explain how this rate changes as the water warms. Specific heat capacity of water = 4.18 × 10³ J kg⁻¹ K⁻¹.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
While ice remains, the ice and meltwater stay at the melting point, 0 °C (a phase change at constant temperature)	✓ 1
So the temperature difference across the walls is constant, the rate of conduction is constant, and ice melts at a constant rate	✓ 1	The chain constant ΔT → constant rate of energy transfer → constant rate of melting is required.
Part (b)(i)
Gradient = 0.209 kg h⁻¹ = 5.8 × 10⁻⁵ kg s⁻¹	✓ 1	Accept 0.20–0.22 kg h⁻¹. The conversion to seconds must be seen.
Rate = 5.8 × 10⁻⁵ × 3.34 × 10⁵ = 19 W	✓ 1	Accept 18.5–20.5 W.
Part (c)(i)
Total inside area = 2(0.40 × 0.30 + 0.40 × 0.30 + 0.30 × 0.30) = 0.66 m²	✓ 1	All six faces must be included; award [2 max] for one face or for 5 faces (no lid).
k = (ΔQ/Δt)Δx/(AΔT)	✓ 1	ΔT = 25 K and Δx = 0.030 m.
k = 19 × 0.030/(0.66 × 25) = 0.035 W m⁻¹ K⁻¹	✓ 1	Allow ECF from (b). Accept 0.033–0.037 W m⁻¹ K⁻¹.
Part (d)(i)
Energy needed = ρa³L (mass of ice = ρa³)	✓ 1
Rate of conduction = k(6a²)ΔT/x; dividing: t = ρa³Lx/(6a²kΔT) = ρaLx/(6kΔT)	✓ 1	The total area 6a² must be seen.
Part (d)(ii)
t ∝ a, so the time doubles: the mass of ice increases 8 times but the area of the walls only 4 times	✓ 1	Allow ECF from (d)(i). The reason in terms of volume and area is required.
Part (e)(i)
p_in = 1.01 × 10⁵ × 273/298 = 9.25 × 10⁴ Pa	✓ 1	Kelvin temperatures are required; award [0] for this mark for a ratio 0/25.
Force = (1.01 × 10⁵ − 9.25 × 10⁴) × 0.12 = 1.0 × 10³ N (downwards, into the box)	✓ 1	Accept 1.0 × 10³ N. The direction is not required.
Part (f)(i)
ΔT/Δt = 19/(2.40 × 4.18 × 10³) = 1.9 × 10⁻³ K s⁻¹ (about 7.0 K per hour)	✓ 1	Allow ECF from (b). Accept 1.8–2.0 × 10⁻³ K s⁻¹.
As the water warms, the temperature difference across the walls decreases, so the rate of conduction and hence the rate of rise of temperature decrease (towards zero as the water approaches 25 °C)	✓ 1	The link to the decreasing temperature difference is required.

Answers: (b)(i) 19 W · (c)(i) 0.035 W m⁻¹ K⁻¹ · (d)(ii) it doubles · (e)(i) 1.0 × 10³ N · (f)(i) 1.9 × 10⁻³ K s⁻¹; it decreases (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — quantitative analysis of rate of thermal energy transfer by conduction in terms of the type of material and cross-sectional area of the material and the temperature gradient as given by ΔQ/Δt = kAΔT/Δx; that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature; quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; that temperature difference determines the direction of the resultant thermal energy transfer between bodies; density ρ as given by ρ = m/V; B.3 — the equations governing the behaviour of ideal gases (pressure law at constant volume); pressure p = F/A Command term: Determine

51B-2-43

Heating and cooling of brake discs·B.1 Thermal energy transfers

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine

A car of mass 1400 kg descends a long straight road inclined at 7.0° to the horizontal. The driver uses the brakes to keep the speed constant at 14 m s⁻¹. Air resistance and rolling resistance together provide a constant force of 500 N opposing the motion. Assume that all the work done by the braking force is transferred to the four identical brake discs, which have a total mass of 30 kg and a specific heat capacity of 490 J kg⁻¹ K⁻¹.

The graph shows how the temperature θ of the discs varies with time t from the start of the descent.

Temperature of the brake discs against time (drawn to scale)

(a)

(i)

Show that the rate at which energy is transferred to the discs is about 16 kW.

(2)

(b)

(i)

Calculate the initial rate of rise of the temperature of the discs, assuming that the energy lost by the discs at 30 °C is negligible.

(2)

(c)

(i)

By drawing a tangent to the graph where θ = 300 °C, determine the total rate at which the four discs lose energy to the surroundings at this temperature.

(3)

(d)

(i)

Use the graph to estimate the steady temperature of the discs, and calculate the wavelength at which the intensity of the radiation they emit is greatest at this temperature.

(1)

(ii)

In the dark the hot discs are seen to glow dull red. Explain this observation, given your answer to (d)(i).

(1)

(e)

(i)

Each disc has an emissivity of 0.55 and a total radiating surface area of 0.12 m². The surroundings are at 20 °C. Calculate the net rate at which one disc loses energy by radiation at the steady temperature.

(2)

(ii)

Deduce the percentage of the energy transferred to each disc that is removed by convection (and conduction) at the steady temperature.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Constant speed, so braking force = mg sin 7.0° − 500 = 1674 − 500 = 1174 N	✓ 1	The resistive force must be subtracted.
P = Fv = 1174 × 14 = 1.64 × 10⁴ W	✓ 1	An answer to at least 3 s.f. is required. Award [1 max] for 23 kW (resistance ignored).
Part (b)(i)
ΔT/Δt = P/(mc) = 1.64 × 10⁴/(30 × 490)	✓ 1	Allow ECF from (a).
= 1.12 K s⁻¹	✓ 1	Accept 1.1 K s⁻¹ (1.09 K s⁻¹ if 16 kW is used).
Part (c)(i)
Tangent drawn at 300 °C (about 300 s); gradient ≈ 0.69 K s⁻¹	✓ 1	Accept 0.60–0.78 K s⁻¹.
Rate of increase of internal energy of the discs = 30 × 490 × 0.69 = 1.0 × 10⁴ W	✓ 1
Rate of loss = 1.64 × 10⁴ − 1.0 × 10⁴ = 6.3 × 10³ W	✓ 1	Allow ECF from (a). Accept 5.0–7.6 kW.
Part (d)(i)
Steady temperature ≈ 560 °C = 833 K; λ_max = 2.9 × 10⁻³/833 = 3.5 × 10⁻⁶ m	✓ 1	Accept 550–570 °C, giving 3.4–3.6 × 10⁻⁶ m. The kelvin temperature is required.
Part (d)(ii)
The peak is in the infrared, but the spectrum is continuous: a small fraction of the radiation is emitted at the long-wavelength (red) end of the visible spectrum, and none of significance at shorter visible wavelengths	✓ 1	Allow ECF from (d)(i). Reference to the spread of wavelengths of the black-body spectrum is required.
Part (e)(i)
P = eσA(T⁴ − T_s⁴) = 0.55 × 5.67 × 10⁻⁸ × 0.12 × (833⁴ − 293⁴)	✓ 1	Allow ECF from (d)(i). Award [0] for this mark if Celsius temperatures are used.
= 1.77 × 10³ W	✓ 1	Accept 1.7–1.9 × 10³ W.
Part (e)(ii)
Input to one disc = 1.64 × 10⁴/4 = 4.11 × 10³ W	✓ 1	Allow ECF from (a).
Convection = (4.11 × 10³ − 1.77 × 10³)/4.11 × 10³ = 57 %	✓ 1	Allow ECF from (e)(i). Accept 52–60 %.

Answers: (a)(i) 1.64 × 10⁴ W · (b)(i) 1.12 K s⁻¹ · (c)(i) 6.3 × 10³ W · (d)(i) 3.5 × 10⁻⁶ m · (e)(i) 1.77 × 10³ W · (e)(ii) 57 % (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; quantitative analysis of energy transferred by radiation as a result of the emission of electromagnetic waves from the surface of a body, which in the case of a black body can be modelled by the Stefan–Boltzmann law as given by L = σAT⁴; the emission spectrum of a black body and the determination of the temperature of the body using Wien's displacement law as given by λ_maxT = 2.9 × 10⁻³ m K; that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer; that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales; A.2 — Newton's first law and the components of the weight on a slope; A.3 — power P = Fv; B.2 — emissivity Command term: Determine

52B-2-44

Radiation from molten glass·B.1 Thermal energy transfers

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine

A glassblower collects a sphere of molten glass, called a gather, on the end of a long steel blowpipe. The gather has a radius of 4.0 cm and a surface temperature of 1100 °C. Its surface behaves as a grey body of emissivity 0.90.

Data for the glass: density = 2.5 × 10³ kg m⁻³; specific heat capacity = 1.2 × 10³ J kg⁻¹ K⁻¹.

(a)

(i)

Calculate the wavelength at which the radiation from the gather is most intense.

(1)

(b)

(i)

Show that the power radiated by the gather is about 3.6 kW.

(2)

(ii)

Ignoring the radiation absorbed from the surroundings, determine the initial rate at which the temperature of the gather falls. Assume the temperature is uniform throughout the gather.

(3)

(c)

(i)

The face of the glassblower is 0.50 m from the centre of the gather. Determine the intensity of the radiation from the gather at the face, and compare it with the intensity of bright sunlight at the Earth's surface, about 1.0 × 10³ W m⁻².

(2)

(d)

(i)

The gather is reheated in a furnace whose walls are at 1250 °C. Explain why there is a net transfer of thermal energy to the gather, and determine the net rate of transfer by radiation when the gather is at 1100 °C. Assume that the gather absorbs radiation from the walls as a grey body of the same emissivity.

(2)

(e)

(i)

The glassblower holds the blowpipe horizontally. The centre of the gather is 1.40 m from the glassblower's front hand. Calculate the torque of the weight of the gather about this hand.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
λ_max = 2.9 × 10⁻³/(1100 + 273) = 2.1 × 10⁻⁶ m	✓ 1	Award [0] for 2.6 × 10⁻⁶ m (1100 K used).
Part (b)(i)
A = 4π(0.040)² = 2.01 × 10⁻² m²	✓ 1
P = 0.90 × 5.67 × 10⁻⁸ × 2.01 × 10⁻² × 1373⁴ = 3.65 × 10³ W	✓ 1	An answer to at least 3 s.f. or the full substitution is required.
Part (b)(ii)
m = 2.5 × 10³ × (4/3)π(0.040)³ = 0.670 kg	✓ 1	The mass must be found from the density (hidden step).
ΔT/Δt = P/(mc)	✓ 1
= 3.65 × 10³/(0.670 × 1.2 × 10³) = 4.5 K s⁻¹	✓ 1	Allow ECF from (b)(i). Accept 4.4–4.6 K s⁻¹ (4.5 K s⁻¹ if 3.6 kW is used).
Part (c)(i)
b = L/4πd² = 3.65 × 10³/(4π × 0.50²) = 1.2 × 10³ W m⁻²	✓ 1	Allow ECF from (b)(i).
This is similar to (slightly greater than) the intensity of bright sunlight	✓ 1	The comparison must agree with the candidate's value.
Part (d)(i)
The walls are at a higher temperature than the gather, and the direction of the net transfer is set by the temperature difference: the gather absorbs more radiation than it emits	✓ 1
Net rate = 0.90 × 5.67 × 10⁻⁸ × 2.01 × 10⁻² × (1523⁴ − 1373⁴) = 1.9 × 10³ W	✓ 1	Allow ECF from (b)(i). Accept 1.8–1.9 × 10³ W. Award [0] for this mark for (1523 − 1373)⁴.
Part (e)(i)
Weight = 0.670 × 9.81 = 6.57 N	✓ 1	Allow ECF from (b)(ii).
Torque = 6.57 × 1.40 = 9.2 N m	✓ 1	The weight of the blowpipe is not included.

Answers: (a)(i) 2.1 × 10⁻⁶ m · (b)(i) 3.65 × 10³ W · (b)(ii) 4.5 K s⁻¹ · (c)(i) 1.2 × 10³ W m⁻² · (d)(i) 1.9 × 10³ W · (e)(i) 9.2 N m (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — the emission spectrum of a black body and the determination of the temperature of the body using Wien's displacement law as given by λ_maxT = 2.9 × 10⁻³ m K; quantitative analysis of energy transferred by radiation as a result of the emission of electromagnetic waves from the surface of a body, which in the case of a black body can be modelled by the Stefan–Boltzmann law as given by L = σAT⁴; the concept of apparent brightness b; luminosity L of a body as given by b = L/4πd²; that temperature difference determines the direction of the resultant thermal energy transfer between bodies; quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c and specific latent heat of fusion and vaporization of substances L as given by Q = mcΔT and Q = mL; density ρ as given by ρ = m/V; that Kelvin and Celsius scales are used to express temperature; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales; B.2 — emissivity; A.4 — torque τ = Fr sin θ Command term: Determine

B.1 Thermal energy transfers: IB Physics HL exam-style questions

25 practice questions on B.1 Thermal energy transfers

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