B.1 Thermal Energy Transfers Questions

1B-1A-01

Density & the molecular model·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksCalculate

A block of metal has a mass of 0.500 kg and a volume of 2.5 × 10⁻⁴ m³.

What is its density?

Show mark scheme

Marking point	Mark	Notes
Step 1ρ = m/V = 0.500 / 2.5 × 10⁻⁴ = 2.0 × 10³ kg m⁻³.	✓ 1	Answer C

Answer: C · 1 stage of work, one mark

Every option, and why

AThis multiplies the two quantities. Density is a ratio, not a product.
BThis is V/m. Density is mass per unit volume, so the mass goes on top.
CCorrect: ρ = m/V = 0.500 / 2.5 × 10⁻⁴ = 2000 kg m⁻³.
DThis uses the mass in grams. The unit kg m⁻³ requires the mass in kilograms.

Syllabus understandingB.1 — density ρ as given by ρ = m/V Command term: Calculate

2B-1A-02

Temperature scales·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksState

A sample of water is at 300 K. Its temperature then rises by 20 °C.

What is its initial temperature in degrees Celsius, and what is the rise in kelvin?

Initial temperature / °CRise / K

Show mark scheme

Marking point	Mark	Notes
Step 1Subtract 273 to convert a Kelvin temperature to Celsius: 300 − 273 = 27 °C. A kelvin and a degree Celsius are the same size, so a change of 20 °C is a change of 20 K.	✓ 1	Answer D

Answer: D · 1 stage of work, one mark

Every option, and why

A273 has been added to the change as well. Only a temperature itself is converted; a change is the same number on both scales.
B273 has been added instead of subtracted. Kelvin values are larger than Celsius ones, so converting K to °C means subtracting.
CThis leaves the temperature unconverted. 300 °C and 300 K are very different temperatures.
DCorrect: 300 − 273 = 27 °C, and because the two scales have the same size of degree a rise of 20 °C is a rise of 20 K.

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 Command term: State

3B-1A-03

Specific heat capacity·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

An electric heater of power P warms a liquid of mass m for a time t. The temperature of the liquid rises by Δθ. Only 80 % of the energy supplied by the heater is transferred to the liquid.

What is the specific heat capacity of the liquid?

Show mark scheme

Marking point	Mark	Notes
Step 1Energy transferred to the liquid = 80 % of the energy supplied = 0.8Pt.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Q = mcΔθ, so c = Q / (mΔθ) = 0.8Pt / (mΔθ).	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThis assumes that all of the heater's energy reaches the liquid; the 20 % lost to the surroundings has been ignored.
BCorrect: the liquid receives 0.8Pt, and c = Q / (mΔθ) = 0.8Pt / (mΔθ).
CThe 0.8 has been divided into the energy supplied instead of multiplying it, crediting the liquid with 1.25Pt — more than the heater supplied.
DThe equation has been inverted. More energy for the same rise means a larger c, so the energy belongs on top.

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c as given by Q = mcΔT Command term: Determine

4B-1A-04

Latent heat & phase change·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

A block of ice at 0 °C is heated at a constant rate until all of the water formed from it has boiled away. No energy is lost to the surroundings.

The specific latent heat of vaporization of water is about seven times its specific latent heat of fusion.

Which sketch graph shows the variation with time t of the temperature θ of the sample?

Sketch graphs

Show mark scheme

Marking point	Mark	Notes
Step 1During each phase change the energy supplied increases the intermolecular potential energy while the mean kinetic energy, and so the temperature, stays constant: the graph must be horizontal at 0 °C and at 100 °C.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2At a constant heating rate the time for a phase change is proportional to mL. The same mass melts and boils, so the section at 100 °C lasts about seven times as long as the section at 0 °C.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

AThe two horizontal sections are equal. The time for each phase change is proportional to mL, and L for vaporization is about seven times L for fusion.
BThe temperature keeps rising during melting and boiling. During a phase change the energy increases the intermolecular potential energy, not the mean kinetic energy, so the temperature is constant.
CCorrect: horizontal sections at 0 °C and 100 °C, with the boiling section about seven times as long as the melting section.
DThe lengths of the two horizontal sections are reversed. That would need the latent heat of fusion to be the larger of the two.

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 latent heat of fusion and vaporization of substances L as given by Q = mL Command term: Identify

5B-1A-05

Method of mixtures·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark1 step to full marksIdentify

A hot metal block is placed in cold water in an insulated container. Neither the block nor the water changes phase. After some time thermal equilibrium is reached.

Which statement about the block and the water at thermal equilibrium is correct?

Show mark scheme

Marking point	Mark	Notes
Step 1The direction of the resultant energy transfer is set by the temperature difference. Energy passes from block to water until their temperatures are equal; then there is no resultant transfer.	✓ 1	Answer A

Answer: A · 1 stage of work, one mark

Every option, and why

ACorrect: temperature difference determines the direction of the resultant transfer, so once the temperatures are equal there is no resultant transfer.
BInternal energy depends on the mass, the material and the phase as well as on the temperature. Equilibrium means equal temperatures, not equal internal energies.
CEqual energies are transferred, not equal temperature changes. The temperature changes are in the inverse ratio of the thermal capacities mc, which are not equal in general.
DThe water gains energy and does not change phase, so its temperature must rise. The common final temperature lies between the two starting temperatures.

Syllabus understandingB.1 — that temperature difference determines the direction of the resultant thermal energy transfer between bodies Command term: Identify

6B-1A-06

Temperature & molecular energy·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

The temperature of a sample of an ideal gas is increased from 27 °C to 327 °C.

What is the ratio (average kinetic energy of the molecules at 327 °C) / (average kinetic energy of the molecules at 27 °C)?

Show mark scheme

Marking point	Mark	Notes
Step 1Convert to absolute temperatures: 27 °C = 300 K and 327 °C = 600 K.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2E_k = ³⁄₂k_BT, so E_k ∝ T and the ratio is 600/300 = 2.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThis is the ratio of the molecular speeds, which are proportional to √T, not the ratio of the kinetic energies.
BCorrect: in kelvin the temperature doubles from 300 K to 600 K, and E_k = ³⁄₂k_BT doubles with it.
CThis takes the speed to be proportional to T, so that the kinetic energy goes as T². In fact E_k itself is proportional to T.
DThis is 327/27, a ratio of Celsius temperatures. E_k is proportional to the absolute temperature.

Syllabus understandingB.1 — that Kelvin temperature is a measure of the average kinetic energy of particles as given by E_k = ³⁄₂k_BT; that Kelvin and Celsius scales are used to express temperature Command term: Determine

7B-1A-07

Conduction·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark1 step to full marksIdentify

One end of a long glass rod is held in a flame. Some time later the other end of the rod has become warm.

Which statement describes how energy is conducted along the rod?

Show mark scheme

Marking point	Mark	Notes
Step 1Particles at the hot end have a greater mean kinetic energy. Through collisions (interactions) with neighbouring particles that have less kinetic energy they pass energy on, so energy travels along the rod although the particles stay in place.	✓ 1	Answer A

Answer: A · 1 stage of work, one mark

Every option, and why

ACorrect: conduction is the transfer of kinetic energy from particles with more to neighbouring particles with less, driven by the temperature difference along the rod.
BParticles in a solid vibrate about fixed positions and do not travel along the rod. Energy carried by moving matter is convection, which needs a fluid.
CParticles do not expand when heated; they vibrate with greater amplitude. Expansion of particles is not the mechanism of conduction.
DThere is no flow of 'cold'. The resultant energy transfer is from the hotter end to the cooler end.

Syllabus understandingB.1 — conduction in terms of the difference in the kinetic energy of particles Command term: Identify

8B-1A-08

Stefan–Boltzmann & Wien·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksIdentify

In each graph the solid line shows the variation with wavelength λ of the intensity I emitted by a black body at temperature T.

The body is then heated to a higher temperature. Which graph shows, as a dashed line, the emission of the body at the higher temperature?

Sketch graphs — not to scale

Show mark scheme

Marking point	Mark	Notes
Step 1A hotter black body emits more at every wavelength, so the new curve lies above the old one everywhere (the total power, the area under the curve, grows as T⁴).	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Wien's law, λ_maxT = 2.9 × 10⁻³ m K, puts the peak at a shorter wavelength.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

AThe peak has moved to a longer wavelength. By Wien's law a higher temperature gives a shorter peak wavelength.
BThe peak has stayed at the same wavelength. λ_max is inversely proportional to T, so the peak must move to a shorter wavelength.
CCorrect: the curve is higher at every wavelength and its peak is at a shorter wavelength.
DThe peak is lower than before. A hotter black body radiates more power at every wavelength, so the whole curve must rise.

9B-1A-09

Luminosity & apparent brightness·B.1 Thermal energy transfers

Paper 1AHard1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

A star of radius R behaves as a black body with surface temperature T. It is at a distance d from the Earth. σ is the Stefan–Boltzmann constant.

What is the apparent brightness of the star at the Earth?

Show mark scheme

Marking point	Mark	Notes
Step 1L = σAT⁴ with A = 4πR², so L = 4πR²σT⁴.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2b = L / 4πd² = 4πR²σT⁴ / 4πd² = σR²T⁴ / d².	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThe radiating area has been taken as πR², the area of the star's disc. The star radiates from its whole surface, 4πR².
BCorrect: the 4π in the surface area cancels the 4π in the inverse-square law, leaving σR²T⁴ / d².
CThe radiation has been spread over a hemisphere of area 2πd². It spreads over a whole sphere of area 4πd².
DThe luminosity has been divided by d² instead of by 4πd².

Syllabus understandingB.1 — 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 Command term: Determine

10B-1A-10

Mechanisms of thermal transfer·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

An electric heater stands on the floor of a closed room. A convection current is set up in the air of the room.

Which statement explains why the air just above the heater rises?

Show mark scheme

Marking point	Mark	Notes
Step 1Air near the heater is warmed and expands, so its density decreases. The denser, cooler air around it sinks and pushes the less dense warm air upwards, setting up a convection current.	✓ 1	Answer B

Answer: B · 1 stage of work, one mark

Every option, and why

A'Heat rises' is a misconception. Energy is carried upwards only because the warm air itself moves, and it moves because of a difference in density.
BCorrect: the warm air is less dense than the cool air around it, so it is pushed upwards while the denser air sinks to take its place.
CThe mass of each molecule does not change on heating. It is the density of the air — mass per unit volume — that falls, because the molecules spread further apart.
DThe molecules themselves do not expand. The air expands because its molecules move faster and spread further apart.

Syllabus understandingB.1 — qualitative description of thermal energy transferred by convection due to fluid density differences Command term: Identify

11B-1B-01

Specific heat capacity·B.1 Thermal energy transfers

Paper 1BMedium11 marks

Data-based question11 steps to full marksDetermine

A student determines the specific heat capacity c of brass. A brass cylinder of mass 0.745 ± 0.001 kg has two holes drilled in it: one holds an electric heater and the other a temperature probe connected to a data logger. The cylinder is wrapped in a layer of insulating foam.

The p.d. across the heater is constant at 11.9 ± 0.1 V and the current is constant at 1.52 ± 0.02 A. The heater is switched on at t = 0 and the temperature θ is recorded every 60 s. The table and the graph show the results.

t / s	0	60	120	180	240	300	360	420	480	540	600
θ / °C	19.6	21.0	23.4	26.8	30.0	33.5	37.1	40.3	43.9	47.4	50.7

Graph drawn to scale

(a)

(i)

Suggest why the temperature rises more slowly during the first 120 s than it does later.

(1)

(ii)

Draw the line of best fit for the data from t = 180 s onwards.

(1)

(b)

Determine the gradient of your line.

(2)

(c)

Determine c for brass.

(2)

(d)

The uncertainty in the gradient is ±3 %. Determine the absolute uncertainty in c and write down your answer to (c) with its uncertainty.

(3)

(e)

(i)

A data book gives c = 380 J kg⁻¹ K⁻¹ for brass. Comment on the student’s result.

(1)

(ii)

Suggest one change to the method that would reduce the systematic error you identified in (e)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
energy is first needed to heat the heater «and the brass near it»/the probe takes time to reach the temperature of the cylinder/energy takes time to conduct from the heater to the probe	✓ 1	OWTTE. Do not accept “the heater is warming up” without reference to energy or conduction.
Part (a)(ii)
single straight line through the points from 180 s to 600 s, with points scattered either side	✓ 1	The line must not be forced through the origin or through the first points.
Part (b)
large triangle or two well-separated points read from the line	✓ 1
gradient = 0.0573 K s⁻¹	✓ 1	Accept 0.055–0.060 K s⁻¹. Accept °C s⁻¹. Award [2] for a correct answer in the range.
Part (c)
P = VI = 11.9 × 1.52 = 18.1 W AND c = P / (m × gradient)	✓ 1	Must see P = VI and the rate form of Q = mcΔT.
c = 424 J kg⁻¹ K⁻¹	✓ 1	Allow ECF from (b). Accept 400–445 J kg⁻¹ K⁻¹.
Part (d)
percentage uncertainties: V 0.84 %, I 1.32 %, gradient 3 %	✓ 1	Mass contributes only 0.13 % and may be ignored.
total ≈ 5.3 % so Δc ≈ 22 J kg⁻¹ K⁻¹	✓ 1	Allow ECF from (c). Accept 5–6 %.
c = (420 ± 20) J kg⁻¹ K⁻¹	✓ 1	MP3 is for matching the precision of value and uncertainty. Accept (424 ± 22).
Part (e)(i)
380 lies outside the range 401–446 so the result is not consistent with the data-book value / there is a systematic error making c too large	✓ 1	Allow ECF from (d). Accept “energy is lost to the surroundings so the measured gradient is too small”.
Part (e)(ii)
thicker/better insulation OR start below room temperature and finish the same amount above it OR use a larger heater power/shorter time so that losses are a smaller fraction	✓ 1	The change must reduce energy exchange with the surroundings. Do not accept “repeat the experiment”.

Answers: (b) 0.0573 K s⁻¹ · (c) 424 J kg⁻¹ K⁻¹ · (d) (420 ± 20) J kg⁻¹ K⁻¹ (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; that power developed is the rate of energy transfer Command term: Determine

12B-1B-02

Latent heat & phase change·B.1 Thermal energy transfers

Paper 1BEasy9 marks

Data-based question9 steps to full marksDetermine

A beaker of water stands on a top-pan balance. An immersion heater in the water is supplied with a power of 150 W ± 2 %, measured with a joulemeter. When the water is boiling steadily, the balance reading m is recorded every 60 s.

t / s	0	60	120	180	240	300	360
m / g	486.3	482.8	478.8	475.2	471.2	467.8	464.0

Graph drawn to scale

(a)

State the absolute uncertainty in a reading of m.

(1)

(b)

Draw the line of best fit for the data.

(1)

(c)

Determine the rate at which water is boiled away, in kg s⁻¹.

(2)

(d)

Calculate the specific latent heat of vaporization L of water obtained from this experiment.

(1)

(e)

The uncertainty in the rate in (c) is ±3 %. Determine the absolute uncertainty in L and write down L with its uncertainty.

(2)

(f)

The accepted value of L is 2.26 × 10⁶ J kg⁻¹. Discuss, with reference to the uncertainty, the difference between the accepted value and the value from this experiment, and suggest the systematic error responsible for it.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
±0.1 g	✓ 1	Accept ±0.05 g.
Part (b)
single straight line with the points scattered either side of it	✓ 1
Part (c)
gradient from a large triangle on the line, e.g. mass lost ÷ time	✓ 1
6.2 × 10⁻⁵ kg s⁻¹	✓ 1	Accept 6.0 × 10⁻⁵ to 6.5 × 10⁻⁵ kg s⁻¹. Sign may be omitted. Award [1] max if left in g s⁻¹.
Part (d)
L = P / rate = 2.41 × 10⁶ J kg⁻¹	✓ 1	Allow ECF from (c).
Part (e)
percentage uncertainty = 2 % + 3 % = 5 % so ΔL ≈ 1.2 × 10⁵ J kg⁻¹	✓ 1	Allow ECF from (d).
L = (2.41 ± 0.12) × 10⁶ J kg⁻¹	✓ 1	Value and uncertainty must have matching precision. Accept (2.4 ± 0.1) × 10⁶.
Part (f)
the accepted value lies outside the range 2.29–2.53 × 10⁶ J kg⁻¹ so there is a systematic error «not only random error»	✓ 1	Allow ECF from (e).
energy is lost from the beaker/heater to the surroundings, so less water boils away than P alone would produce, the rate is too small and L is too large	✓ 1	The direction (too large) must be explained.

Answers: (c) 6.2 × 10⁻⁵ kg s⁻¹ · (d) 2.41 × 10⁶ J kg⁻¹ · (e) (2.41 ± 0.12) × 10⁶ J kg⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers with the use of specific latent heat of vaporization as given by Q = mL; that a phase change represents a change in particle behaviour arising from a change in energy at constant temperature Command term: Determine

13B-1B-03

Stefan–Boltzmann & Wien·B.1 Thermal energy transfers

Paper 1BHard11 marks

Data-based question11 steps to full marksDetermine

A student uses a published catalogue of eclipsing binary stars to test the Stefan–Boltzmann law. For these stars the radius R, the luminosity L and the surface temperature T have been found by independent methods. R is given in units of the solar radius and L in units of the solar luminosity.

If every star radiates as a black body, L = 4πσR²T⁴, and in these units lg(L/R²) = 4 lg T − 4 lg T_☉, where T_☉ is the surface temperature of the Sun.

For star D the catalogue does not give T but gives the peak wavelength of its spectrum, λ_max = 4.12 × 10⁻⁷ m. The uncertainty in each value of lg(L/R²) is ±0.04; the uncertainty in lg T is too small to show. The graph shows the data for the other five stars.

Star	T / K	R / R_☉	L / L_☉	lg(T / K)	lg(L/R²)
A	4480	0.74	0.207	3.651	−0.42
B	5290	0.93	0.592	3.723	−0.16
C	6180	1.24	2.06	3.791	0.13
D	— (see text)	1.58	5.41
E	8350	1.83	14.1	3.922	0.62
F	9720	2.31	44.2	3.988	0.92

Graph drawn to scale

(a)

Show that the surface temperature of star D is about 7000 K.

(1)

(b)

Calculate the two missing values for star D in the table.

(1)

(c)

Plot the point for star D on the graph and draw the line of best fit for all six stars.

(2)

(d)

(i)

Determine the gradient of your line.

(2)

(ii)

By drawing lines of maximum and minimum gradient, determine the uncertainty in the gradient.

(2)

(e)

State whether the data support the Stefan–Boltzmann law for these stars.

(1)

(f)

Using your line, determine T_☉.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
T = 2.9 × 10⁻³ / 4.12 × 10⁻⁷ = 7039 K	✓ 1	Must see the substitution into Wien’s law OR an answer to at least 3 s.f.
Part (b)
lg T = 3.848 AND lg(L/R²) = 0.34	✓ 1	Both needed. Accept 3.845–3.848 «from T = 7000 K» and 0.33–0.34.
Part (c)
star D plotted correctly to within half a small square	✓ 1	Allow ECF from (b).
single straight line passing through all six error bars	✓ 1
Part (d)(i)
gradient from a large triangle on the line	✓ 1
gradient = 4.0	✓ 1	Accept 3.8–4.2. No unit.
Part (d)(ii)
steepest and shallowest lines drawn through all the error bars	✓ 1	Expect gradients close to 4.2 and 3.8.
uncertainty = ½(max − min) ≈ ±0.2	✓ 1	Accept ±0.1 to ±0.3. Allow ECF.
Part (e)
yes: the law predicts a gradient of 4, which lies within the range of the gradient «AND all six error bars are crossed by one straight line»	✓ 1	Allow ECF from (d). Reference to the uncertainty is needed.
Part (f)
L/R² = 1 «in solar units» for a star at T_☉, so read lg T where the line crosses lg(L/R²) = 0, ≈ 3.76	✓ 1	Or use lg T_☉ = −intercept/gradient.
T_☉ = 10^3.761 ≈ 5800 K	✓ 1	Accept 5500–6100 K. Allow ECF.

Answers: (a) 7039 K · (b) 3.848 and 0.34 · (d)(i) 4.0 · (d)(ii) ±0.2 · (f) 5800 K (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; the determination of the temperature of the body using Wien's displacement law Command term: Determine

14B-2-01

Specific heat capacity·B.1 Thermal energy transfers

Paper 2Easy12 marks

Short answer & extended response12 steps to full marksCalculate

A car of mass 1250 kg travels along a level road at 24.0 m s⁻¹. The driver brakes and the car comes to rest. During the stop, 85 % of the kinetic energy of the car is transferred to the four steel brake discs, which have a total mass of 28.0 kg.

Specific heat capacity of steel = 450 J kg⁻¹ K⁻¹.

(a)

(i)

Calculate the kinetic energy of the car before it brakes.

(1)

(ii)

Show that the temperature of the discs rises by about 24 K.

(2)

(iii)

Before the stop the discs were at 35 °C. State their temperature after the stop, in kelvin.

(1)

(b)

(i)

Immediately after the stop, the discs are hotter than the surrounding air. State the direction of the net thermal energy transfer between the discs and the air, and what determines it.

(1)

(ii)

Outline two mechanisms by which thermal energy is transferred from the discs to the surroundings.

(2)

(c)

Later, the car travels down a long hill.

(i)

It moves at a constant speed and loses 120 m in height. The driver uses only the brakes to keep the speed constant. Assume that 85 % of the gravitational potential energy lost by the car is transferred to the discs and ignore air resistance. Calculate the energy transferred to the discs during the descent.

(2)

(ii)

The discs are at 35 °C at the top of the hill. Determine whether they would reach 150 °C by the bottom of the hill if they lost no energy to the surroundings.

(2)

(iii)

In practice the discs reach a lower temperature than your answer to (c)(ii) suggests. Suggest one reason why.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E_k = ½ × 1250 × 24.0² = 3.6 × 10⁵ J	✓ 1
Part (a)(ii)
energy to the discs = 0.85 × 3.6 × 10⁵ = 3.06 × 10⁵ J	✓ 1
ΔT = Q/(mc) = 3.06 × 10⁵ / (28.0 × 450) = 24.3 K	✓ 1	Must see full substitution OR answer to at least 3 s.f. Award [1] max if the 85 % is omitted (gives 28.6 K).
Part (a)(iii)
35 + 24.3 = 59.3 °C, so T = 332 K	✓ 1	Accept 332–333 K. Allow ECF from (a)(ii); accept 59 + 273.
Part (b)(i)
from the discs to the air, because the discs are at the higher temperature	✓ 1	Both needed. Accept 'the temperature difference determines the direction: hotter to colder'.
Part (b)(ii)
a first mechanism with a brief correct outline, e.g. convection: air next to the disc is heated, becomes less dense and rises / is carried away and replaced by cooler air	✓ 1	Names alone (e.g. 'convection and radiation') score [1] max.
a second, different mechanism outlined, e.g. radiation: the hot disc emits infrared electromagnetic radiation; OR conduction: energy passes by particle collisions to the hub/pads/wheel in contact	✓ 1
Part (c)(i)
ΔE_p = mgΔh = 1250 × 9.81 × 120 = 1.47 × 10⁶ J	✓ 1
energy to discs = 0.85 × 1.47 × 10⁶ = 1.25 × 10⁶ J	✓ 1	Award [2] for CNA. Award [1] max for 1.47 × 10⁶ J (85 % omitted).
Part (c)(ii)
ΔT = 1.25 × 10⁶ / (28.0 × 450) = 99 K	✓ 1	Allow ECF from (c)(i).
final temperature ≈ 35 + 99 = 134 °C, which is below 150 °C, so the discs would not reach 150 °C	✓ 1	MP2 needs a comparison and a conclusion consistent with the candidate's value.
Part (c)(iii)
energy is transferred from the discs to the air (by convection/radiation) throughout the long descent; OR air resistance does part of the work, so less than 85 % of the lost potential energy reaches the discs	✓ 1	Do not accept 'energy is lost' without saying how or to where.

Answers: (a)(i) 3.6 × 10⁵ J · (a)(ii) 24.3 K · (a)(iii) 332 K · (c)(i) 1.25 × 10⁶ J · (c)(ii) about 134 °C — no (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; that the change in temperature of a system is the same when expressed with the Kelvin or Celsius scales; that temperature difference determines the direction of the resultant thermal energy transfer between bodies; that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer (synoptic link: A.3 — conservation of energy, ΔE_p = mgΔh) Command term: Calculate

15B-2-02

Latent heat & phase change·B.1 Thermal energy transfers

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDeduce

Paraffin wax is used in hand warmers. A sample of solid wax of mass 0.150 kg, initially at 20 °C, is heated by an electric heater that supplies energy to it at a constant rate of 60.0 W. The graph shows how the temperature θ of the wax varies with time t. Assume that all the energy supplied by the heater is absorbed by the wax.

Graph drawn to scale

(a)

State the melting point of the wax in kelvin.

(1)

(b)

Show that the specific latent heat of fusion of the wax is about 2 × 10⁵ J kg⁻¹.

(1)

(c)

Calculate the specific heat capacity of the solid wax.

(2)

(d)

Deduce, using the gradients of the graph, whether the specific heat capacity of the liquid wax is greater or smaller than that of the solid wax.

(2)

(e)

Explain, with reference to the molecules, why the temperature of the wax is constant between t = 180 s and t = 720 s.

(2)

(f)

A second sample of the same wax, of mass 0.300 kg, is heated from 20 °C by the same heater. Sketch, on the graph, how the temperature of this sample varies with time for the first 900 s.

(3)

(g)

In a real experiment the wax loses some energy to the surroundings. Suggest how this affects the value of the specific latent heat found in (b).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
56 + 273 = 329 K	✓ 1	Accept 329.15 K.
Part (b)
L = Pt/m = 60.0 × (720 − 180) / 0.150 = 2.16 × 10⁵ J kg⁻¹	✓ 1	Must see the melting time of 540 s read from the graph and full substitution OR answer to at least 3 s.f.
Part (c)
c = Pt/(mΔθ) = 60.0 × 180 / (0.150 × 36)	✓ 1
= 2.0 × 10³ J kg⁻¹ K⁻¹	✓ 1	Accept 1.9–2.1 × 10³ J kg⁻¹ K⁻¹. Award [2] for CNA.
Part (d)
gradient for the solid = 36/180 = 0.20 K s⁻¹; for the liquid = 24/150 = 0.16 K s⁻¹	✓ 1	Accept equivalent readings.
same P and m, so c = P/(m × gradient) is inversely proportional to the gradient: the liquid has the smaller gradient, so its specific heat capacity is greater (≈ 2.5 × 10³ J kg⁻¹ K⁻¹)	✓ 1	ALTERNATIVE: calculating c for the liquid and comparing with (c) scores both marks. Do not accept a bare 'greater'.
Part (e)
the energy supplied increases the intermolecular potential energy: work is done separating the molecules against the attractive forces between them (breaking the solid structure)	✓ 1
the mean random kinetic energy of the molecules does not change, and temperature is a measure of the mean kinetic energy, so θ stays constant	✓ 1	OWTTE.
Part (f)
straight line from (0, 20 °C) with half the original gradient, reaching 56 °C at about 360 s	✓ 1	Accept 340–380 s.
then horizontal at the same temperature, 56 °C	✓ 1
the horizontal section continues to the end of the graph at 900 s (melting would last until 1440 s)	✓ 1	MP3: the line must not rise before 900 s.
Part (g)
during the melting some of the energy supplied is lost to the surroundings, so the energy actually used to melt the wax is less than 60.0 × 540 J / the melting takes longer than it would without losses	✓ 1
so the calculated value of L is too large (an overestimate)	✓ 1	MP2 only scores if MP1 scores.

Answers: (a) 329 K · (b) 2.16 × 10⁵ J kg⁻¹ · (c) 2.0 × 10³ J kg⁻¹ K⁻¹ · (d) liquid: greater (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 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: Deduce

16B-2-03

Conduction·B.1 Thermal energy transfers

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksDetermine

The outer wall of a mountain hut has an area of 15 m². It is made of two layers in close contact: a timber board 20 mm thick on the inside and a layer of mineral wool 80 mm thick on the outside. The inner surface of the wall is at 20 °C and the outer surface is at −12 °C. The wall has reached a steady state.

Thermal conductivity of the timber = 0.15 W m⁻¹ K⁻¹. Thermal conductivity of the mineral wool = 0.040 W m⁻¹ K⁻¹.

Ignore the steel bolts shown in the diagram until (b).

Diagram NOT accurately drawn

(a)

(i)

Explain why, in the steady state, the rate of thermal energy transfer through the timber is equal to the rate of thermal energy transfer through the mineral wool.

(2)

(ii)

Show that the temperature at the boundary between the timber and the mineral wool is 18 °C.

(2)

(iii)

Calculate the rate of thermal energy transfer through the wall.

(2)

(b)

The two layers are held together by 40 steel bolts that pass right through the wall. Each bolt has a cross-sectional area of 1.0 × 10⁻⁴ m², and its ends are at the temperatures of the two surfaces of the wall. Thermal conductivity of steel = 50 W m⁻¹ K⁻¹. Determine whether the bolts increase the rate of thermal energy transfer through the wall by more than 10 %.

(3)

(c)

Explain, in terms of particles, why steel conducts thermal energy far better than mineral wool, which consists of fine fibres with air trapped between them.

(2)

(d)

Suggest one change to the design of the wall, other than using thicker mineral wool, that would reduce the rate of thermal energy transfer through it.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
in the steady state the temperature at every point of the wall, including the boundary, is constant, so the internal energy of the layers does not change	✓ 1
so, by conservation of energy, the energy transferred per second into the boundary through the timber must all leave it through the mineral wool	✓ 1	OWTTE. Do not accept 'the layers are in series' without an energy argument.
Part (a)(ii)
0.15 × (20 − θ)/0.020 = 0.040 × (θ + 12)/0.080 «A cancels»	✓ 1	Accept the equation with A on both sides.
7.5(20 − θ) = 0.50(θ + 12) so 8.0θ = 144, θ = 18 °C	✓ 1	Must see the equation solved. ALTERNATIVE: substitution of 18 °C into both sides to show both give 15 W m⁻² scores [2].
Part (a)(iii)
ΔQ/Δt = 0.040 × 15 × (18 + 12)/0.080 OR 0.15 × 15 × (20 − 18)/0.020	✓ 1
= 225 W	✓ 1	Accept 220–230 W. Award [2] for CNA. Award [1] max for 240 W or 3.6 × 10³ W (the whole temperature difference of 32 K used across one layer).
Part (b)
rate through one bolt = 50 × 1.0 × 10⁻⁴ × 32 / 0.100 = 1.6 W	✓ 1	Length of bolt = 0.100 m (both layers) must be used.
rate through 40 bolts = 64 W	✓ 1	Allow ECF from MP1.
64/225 = 28 % «increase», which is more than 10 %, so yes	✓ 1	Allow ECF from (a)(iii). MP3 needs a comparison and a consistent conclusion.
Part (c)
steel contains free / delocalised electrons that move quickly through the metal and carry kinetic energy from hotter to colder regions «as well as the vibrations passed between closely packed ions»	✓ 1
in mineral wool energy must pass through thin fibres touching at few points and through air, whose molecules are far apart, so collisions transfer kinetic energy only slowly	✓ 1	OWTTE.
Part (d)
replace the steel bolts by fixings of low thermal conductivity (e.g. wood/plastic) OR use bolts that do not pass right through the wall OR use bolts of smaller cross-sectional area / fewer bolts	✓ 1	Accept a thicker timber board or a material of lower thermal conductivity. Do not accept 'thicker wool'.

Answers: (a)(ii) 18 °C · (a)(iii) 225 W · (b) 64 W, a 28 % increase — yes (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 temperature difference determines the direction of the resultant thermal energy transfer between bodies (with A.3 — conservation of energy) Command term: Determine

17B-2-04

Luminosity & apparent brightness·B.1 Thermal energy transfers

Paper 2Hard13 marks

Short answer & extended response13 steps to full marksDetermine

A space probe moves around the Sun in a circular orbit of radius 0.40 AU. It carries a flat sunshade that is always perpendicular to the Sun's rays.

Luminosity of the Sun = 3.85 × 10²⁶ W. 1 AU = 1.50 × 10¹¹ m. The Earth orbits the Sun at 1.00 AU with a period of 365 days.

Axes for (b)(iii)

(a)

(i)

Show that the intensity of the Sun's radiation at a distance of 1.00 AU from the Sun is about 1.4 kW m⁻².

(1)

(ii)

Calculate the intensity of the Sun's radiation at the probe.

(1)

(b)

The sunshade absorbs all the radiation incident on its front face. It emits as a black body from both its front face and its back face.

(i)

Show that the equilibrium temperature of the sunshade is about 520 K.

(2)

(ii)

Deduce the equilibrium temperature that the same sunshade would have at a distance of 1.00 AU from the Sun.

(1)

(iii)

Sketch, on the axes, a graph to show how the equilibrium temperature T of the sunshade varies with its distance r from the Sun, for r from 0.20 AU to 1.00 AU.

(2)

(c)

The front face of the sunshade is given a coating that reflects 70 % of the incident solar radiation. Both faces still emit as black bodies at infrared wavelengths. Determine the new equilibrium temperature of the sunshade.

(3)

(d)

(i)

Determine, using Kepler's third law, the orbital period of the probe in days.

(2)

(ii)

Explain why the orbital period of the probe does not depend on its mass.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
b = L/4πd² = 3.85 × 10²⁶ / (4π × (1.50 × 10¹¹)²) = 1.36 × 10³ W m⁻²	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (a)(ii)
b ∝ 1/d²: 1.36 × 10³ / 0.40² = 8.5 × 10³ W m⁻²	✓ 1	Accept 8.5–8.8 × 10³ W m⁻² (using 1.4 kW m⁻² gives 8.75 × 10³).
Part (b)(i)
power absorbed by the front face = power emitted from both faces: IA = 2σAT⁴	✓ 1	The factor 2 must be seen.
T = (8.51 × 10³ / (2 × 5.67 × 10⁻⁸))^¼ = 523 K	✓ 1	Must see full substitution OR answer to at least 3 s.f.
Part (b)(ii)
T ∝ (intensity)^¼ ∝ r^−½: T = 523 × √0.40 = 331 K	✓ 1	Accept 328–334 K. ALTERNATIVE: T = (1.36 × 10³/(2 × 5.67 × 10⁻⁸))^¼. Allow ECF from (b)(i).
Part (b)(iii)
a decreasing curve whose gradient becomes less steep as r increases «not a straight line»	✓ 1	Do not accept a straight line.
passing through about (0.40 AU, 523 K) and (1.00 AU, 331 K)	✓ 1	Allow ECF from (b)(ii). The curve should reach about 740 K at 0.20 AU «T ∝ 1/√r»; ± half a large square.
Part (c)
absorbed intensity = 0.30 × 8.5 × 10³ = 2.6 × 10³ W m⁻²	✓ 1	Allow ECF from (a)(ii).
0.30I = 2σT⁴ / T ∝ (absorbed intensity)^¼	✓ 1
T = 387 K	✓ 1	Accept 380–390 K. Award [3] for CNA. Award [2] max for 479 K (0.70 used as the absorbed fraction).
Part (d)(i)
T² ∝ r³ so T = 365 × 0.40^3/2	✓ 1
T = 92 days	✓ 1	Accept 92–93 days. Award [2] for CNA.
Part (d)(ii)
the gravitational force on the probe and the centripetal force it needs (mv²/r) are both proportional to its mass, so the mass cancels	✓ 1	OWTTE.

Answers: (a)(i) 1.36 × 10³ W m⁻² · (a)(ii) 8.5 × 10³ W m⁻² · (b)(i) 523 K · (b)(ii) 331 K · (c) 387 K · (d)(i) 92 days (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; 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 (synoptic links: B.2 — the conservation of energy; D.1 — Kepler's three laws of orbital motion) Command term: Determine

18B-2-13

Density & the molecular model·B.1 Thermal energy transfers

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksExplain

Liquid nitrogen boils at −196 °C at atmospheric pressure. At this temperature the density of liquid nitrogen is 808 kg m⁻³. At room temperature (20 °C) and atmospheric pressure, the density of nitrogen gas is 1.16 kg m⁻³.

A technician pours 1.00 × 10⁻³ m³ of liquid nitrogen into an open vacuum flask in a laboratory at 20 °C. Over the following hours all of the liquid boils away.

(a)

State the boiling point of nitrogen in kelvin.

(1)

(b)

Calculate the average kinetic energy of the molecules of nitrogen gas at its boiling point.

(1)

(c)

Show that, when the gas has warmed to room temperature, its volume is about 700 times the volume of the liquid from which it formed.

(1)

(d)

Estimate the ratio of the mean separation of the molecules in the gas at room temperature to the mean separation of the molecules in the liquid.

(2)

(e)

Describe the differences between liquid nitrogen and nitrogen gas in terms of the arrangement, the forces and the motion of their molecules.

(3)

(f)

Explain how the internal energy of the nitrogen changes as the liquid boils at its boiling point and as the gas then warms to room temperature.

(3)

(g)

The flask is not being heated. Explain why the nitrogen nevertheless boils.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
−196 + 273 = 77 K	✓ 1	Accept 77.15 K.
Part (b)
E_k = 3/2 k_BT = 1.5 × 1.38 × 10⁻²³ × 77 = 1.59 × 10⁻²¹ J	✓ 1	Accept 1.59–1.60 × 10⁻²¹ J. Allow ECF from (a).
Part (c)
V_gas/V_liquid = ρ_liquid/ρ_gas = 808 / 1.16 = 697	✓ 1	Must see the ratio of densities (or mass 0.808 kg and volume 0.697 m³) OR answer to at least 3 s.f.
Part (d)
the volume occupied by each molecule is about 700 times larger, and the separation is proportional to the cube root of the volume per molecule	✓ 1
ratio ≈ 700^1/3 = 8.9 ≈ 9	✓ 1	Accept 8.8–9. Award [1] max for 700 or 26 (√700).
Part (e)
arrangement: in the liquid the molecules are close together (almost touching) in a disordered arrangement; in the gas they are far apart	✓ 1
forces: intermolecular forces are significant in the liquid but negligible in the gas (except during collisions)	✓ 1
motion: in the liquid the molecules vibrate and slide past one another; in the gas they move freely and randomly at high speed in straight lines between collisions	✓ 1
Part (f)
internal energy = total intermolecular potential energy + total random kinetic energy of the molecules	✓ 1	May be implied by the answers for the two stages.
boiling: the potential energy increases because the molecules are separated against the attractive forces, while the kinetic energy stays the same (constant temperature)	✓ 1
warming: the random kinetic energy increases because the temperature rises, while the potential energy stays (almost) the same — so the internal energy increases in both stages	✓ 1
Part (g)
the laboratory/surroundings are at a higher temperature than the nitrogen, so there is a net thermal energy transfer into the nitrogen, which is used to boil it	✓ 1

Answers: (a) 77 K · (b) 1.59 × 10⁻²¹ J · (c) 697 · (d) 8.9 (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — molecular theory in solids, liquids and gases; density ρ as given by ρ = m/V; that Kelvin and Celsius scales are used to express temperature; that Kelvin temperature is a measure of the average kinetic energy of particles as given by E_k = 3/2 k_BT; that the internal energy of a system is the total intermolecular potential energy … plus the total random kinetic energy of the molecules Command term: Explain

19B-2-14

Mechanisms of thermal transfer·B.1 Thermal energy transfers

Paper 2Hard14 marks

Short answer & extended response14 steps to full marksDetermine

A steel panel radiator is filled with hot water at 70 °C. It transfers thermal energy to a room at a total rate of 1.6 kW. The outer surface of the radiator has a total area of 1.8 m² and an emissivity of 0.90. The steel wall of the radiator is 1.2 mm thick and the thermal conductivity of steel is 50 W m⁻¹ K⁻¹. The air and the walls of the room are at 20 °C.

Diagram NOT accurately drawn

(a)

(i)

Show that the temperature difference across the steel wall of the radiator is less than 0.1 K.

(1)

(ii)

Hence state the temperature of the outer surface of the radiator.

(1)

(b)

(i)

Calculate the power emitted as radiation by the outer surface of the radiator.

(2)

(ii)

The net rate of energy transfer by radiation from the radiator is given by eσA(T_r⁴ − T_w⁴), where T_r is the temperature of the radiator and T_w the temperature of the walls of the room. Explain why this is smaller than your answer to (b)(i).

(1)

(iii)

Determine whether more than half of the energy transferred from the radiator to the room is transferred by radiation.

(4)

(c)

(i)

Annotate the diagram to show the convection current that the radiator sets up in the room.

(2)

(ii)

Explain how the radiator sets up this convection current.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
ΔT = (ΔQ/Δt)Δx/(kA) = 1600 × 1.2 × 10⁻³ / (50 × 1.8) = 0.021 K, which is less than 0.1 K	✓ 1	Must see full substitution OR the value 0.021 K.
Part (a)(ii)
70 °C / 343 K (the same as the water, to within 0.1 K)	✓ 1
Part (b)(i)
P = eσAT⁴ = 0.90 × 5.67 × 10⁻⁸ × 1.8 × 343⁴	✓ 1	Kelvin temperature must be used.
= 1.27 × 10³ W	✓ 1	Accept 1.27–1.28 × 10³ W. Award [2] for CNA.
Part (b)(ii)
the radiator also absorbs radiation emitted by the walls of the room (at 293 K), which partly offsets what it emits	✓ 1	OWTTE.
Part (b)(iii)
net radiation = 0.90 × 5.67 × 10⁻⁸ × 1.8 × (343⁴ − 293⁴)	✓ 1
= 5.9 × 10² W	✓ 1	Accept 590–600 W.
fraction by radiation = 5.9 × 10² / 1600 = 0.37	✓ 1	Accept 37 %.
less than half, so no: the remaining ≈ 1.0 kW (about two-thirds) is transferred by convection (the name 'radiator' is misleading)	✓ 1	MP4 needs a conclusion consistent with the candidate's value.
Part (c)(i)
arrows rising from the radiator to the ceiling, then across the ceiling towards the far wall	✓ 1
arrows down the far side of the room and back along the floor to the radiator, forming a closed loop	✓ 1	Ignore small loops near the window.
Part (c)(ii)
air next to the radiator is heated and expands	✓ 1
so its density decreases and it rises (buoyancy)	✓ 1
cooler, denser air sinks elsewhere and flows in to replace it, so a continuous circulation is set up	✓ 1	OWTTE.

Answers: (a)(i) 0.021 K · (a)(ii) 70 °C · (b)(i) 1.27 × 10³ W · (b)(iii) 5.9 × 10² W, 37 % — no (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer; quantitative analysis of rate of thermal energy transfer by conduction … as given by ΔQ/Δt = kAΔT/Δx; qualitative description of thermal energy transferred by convection due to fluid density differences; quantitative analysis of energy transferred by radiation … L = σAT⁴ (with B.2 — emissivity) Command term: Determine

20B-1A-23

Internal energy & the molecular model·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

Which statement about the internal energy of a substance is correct?

Show mark scheme

Marking point	Mark	Notes
Step 1Internal energy is the sum of the total random kinetic energy of the molecules and the total intermolecular potential energy arising from the forces between them.	✓ 1	Answer A

Answer: A · 1 stage of work, one mark

Every option, and why

ACorrect: internal energy has both a kinetic part, which the temperature measures, and a potential part, which depends on the separations.
BThis leaves out the potential energy stored in the separations between molecules, which is what changes during melting and boiling.
CTemperature measures the average kinetic energy per molecule. Two samples at the same temperature can have very different internal energies if they differ in mass or in phase.
DInternal energy is about the random motion of the molecules. Motion of the body as a whole is ordinary kinetic energy, not internal energy.

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 Command term: Identify

21B-1A-24

Density & the molecular model·B.1 Thermal energy transfers

Paper 1AHard1 mark

Multiple choice · 1 mark2 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 the ratio (mean separation of the molecules in steam) / (mean separation of the molecules in liquid water)?

Show mark scheme

Marking point	Mark	Notes
Step 1The molecules have the same mass in both phases, so the volume occupied per molecule is inversely proportional to the density: steam has 958/0.598 ≈ 1600 times the volume per molecule.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Volume per molecule ∝ (separation)³, so the separation ratio is 1600^1/3 ≈ 12.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThe ratio has been inverted. The molecules in steam are much further apart than those in the liquid.
BCorrect: the volume per molecule is about 1600 times larger in steam, and the cube root of 1600 is about 12.
CA square root has been taken. Volume goes as the cube of the separation, so a cube root is needed.
DThis is the ratio of the densities, which gives the ratio of the volumes per molecule, not of the separations.

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

22B-1A-25

Specific heat capacity·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

When an energy Q is supplied to block X, its temperature rises by ΔT. Block Y has twice the mass of X, and the specific heat capacity of Y is 1.5 times that of X.

The same energy Q is supplied to Y. No energy is lost. What is the rise in temperature of Y?

Show mark scheme

Marking point	Mark	Notes
Step 1ΔT = Q / (mc), so for the same Q the rise is inversely proportional to mc.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2mc for Y is 2 × 1.5 = 3 times that for X, so the rise is ΔT/3.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: Y's thermal capacity mc is 2 × 1.5 = 3 times larger, so the same energy gives a third of the rise.
BOnly the doubled mass has been taken into account. The larger specific heat capacity reduces the rise further.
COnly the specific heat capacity has been taken into account (1/1.5 = 2/3); the doubled mass has been ignored.
DThe proportionality has been inverted. A larger mc gives a smaller temperature rise for the same energy, not a larger one.

Syllabus understandingB.1 — quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c as given by Q = mcΔT Command term: Determine

23B-1A-26

Latent heat & phase change·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

The water in an electric kettle is boiling at 100 °C. The heating element supplies energy at a rate of 2.0 kW, and all of this energy is transferred to the water.

Specific latent heat of fusion of water = 3.34 × 10⁵ J kg⁻¹. Specific latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹.

Which row gives the mass of water turned into steam in 60 s, and the change in the intermolecular potential energy of the water that boils?

Mass turned into steamIntermolecular potential energy

Show mark scheme

Marking point	Mark	Notes
Step 1Energy supplied in 60 s: Q = Pt = 2000 × 60 = 1.2 × 10⁵ J.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2m = Q/L = 1.2 × 10⁵ / 2.26 × 10⁶ = 0.053 kg ≈ 53 g. The temperature stays at 100 °C, so the mean kinetic energy is unchanged: the energy separates the molecules and increases their potential energy.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThe mass is right, but the energy supplied must go somewhere. At constant temperature the mean kinetic energy is unchanged, so the intermolecular potential energy increases as the molecules are pulled apart.
BCorrect: 1.2 × 10⁵ J boils 1.2 × 10⁵ / 2.26 × 10⁶ = 0.053 kg, and this energy increases the intermolecular potential energy at constant temperature.
CTwo errors: the latent heat of fusion has been used for a change from liquid to vapour, and the energy has not been assigned to the intermolecular potential energy, which increases.
DThe latent heat of fusion has been used (1.2 × 10⁵ / 3.34 × 10⁵ = 0.36 kg). Boiling is a change from liquid to vapour, so the latent heat of vaporization applies.

24B-1A-27

Conduction·B.1 Thermal energy transfers

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two rods of the same length carry the same temperature difference between their ends. Rod X has four times the thermal conductivity of rod Y and twice the cross-sectional area.

The rate of thermal energy transfer along X is how many times that along Y?

Show mark scheme

Marking point	Mark	Notes
Step 1ΔQ/Δt = kAΔT/Δx. The length and the temperature difference are the same for both rods, so the rate is proportional to the product kA.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2k is four times larger and A is twice as large, and these are independent factors that multiply.	—
Step 3Ratio = 4 × 2 = 8.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the rate is proportional to kA, so the ratio is 4 × 2 = 8.
BThe two factors have been added. In ΔQ/Δt = kAΔT/Δx they multiply.
CThis uses the conductivity ratio only, ignoring the difference in cross-sectional area.
DThis uses the area ratio only. The conductivity also differs, and both factors multiply the rate.

25B-1A-28

Stefan–Boltzmann & Wien·B.1 Thermal energy transfers

Paper 1AHard1 mark

Multiple choice · 1 mark4 steps to full marksDetermine

Star P has twice the radius of star Q, but only half its surface temperature. Both behave as black bodies.

The luminosity of P is how many times that of Q?

Show mark scheme

Marking point	Mark	Notes
Step 1L = σ(4πr²)T⁴, so the luminosity depends on r² and on T⁴.	—	All 4 steps must be completed — there is no mark for a part-answer.
Step 2Twice the radius gives 2² = 4 times the surface area.	—
Step 3Half the temperature gives (½)⁴ = 1/16 of the power radiated per square metre.	—
Step 4Ratio = 4 × 1/16 = 0.25, so P is only a quarter as luminous as Q despite being the larger star.	✓ 1	Answer A

Answer: A · 4 stages of work, one mark

Every option, and why

ACorrect: the area is 4 times larger but each square metre radiates only (½)⁴ = 1/16 as much, and 4 × 1/16 = 0.25.
BThe temperature ratio has been used to the first power. The Stefan-Boltzmann law has T⁴.
CThis uses the area ratio only and ignores the difference in temperature.
DThis doubles the temperature instead of halving it. P is the cooler star.

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 Command term: Determine

26B-1A-29

Luminosity & apparent brightness·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

Two stars have the same luminosity, but one of them is three times as far from the Earth as the other.

What is the ratio (apparent brightness of the more distant star) / (apparent brightness of the nearer star)?

Show mark scheme

Marking point	Mark	Notes
Step 1b = L / 4πd² and the luminosities are equal, so b ∝ 1/d².	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2Three times the distance gives 1/3² = 1/9 of the brightness.	✓ 1	Answer B

Answer: B · 2 stages of work, one mark

Every option, and why

AThe distance ratio has been cubed. The radiation spreads over a spherical surface, whose area goes as d², not d³.
BCorrect: b ∝ 1/d², so three times the distance gives 1/3² = 1/9 of the brightness.
CBrightness falls as the square of the distance, not in simple inverse proportion to it.
DThe ratio has been inverted. The more distant star must appear fainter, so the ratio is less than 1.

Syllabus understandingB.1 — the concept of apparent brightness b; luminosity L of a body as given by b = L / 4πd² Command term: Determine

27B-1A-30

Mechanisms of thermal transfer·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark1 step to full marksIdentify

Energy reaches the Earth from the Sun across empty space.

By which mechanism is it transferred?

Show mark scheme

Marking point	Mark	Notes
Step 1Only radiation transfers energy through a vacuum, because electromagnetic waves need no medium. Conduction and convection both require matter.	✓ 1	Answer D

Answer: D · 1 stage of work, one mark

Every option, and why

AConduction needs particles in contact to pass energy along. There are almost none in the space between the Sun and the Earth.
BConvection needs a fluid whose density differences can drive bulk movement. There is no such fluid in space.
CEvaporation is a phase change at a liquid surface, not one of the three mechanisms of thermal energy transfer, and it also needs matter.
DCorrect: radiation is the transfer of energy by electromagnetic waves, which need no medium at all.

Syllabus understandingB.1 — that conduction, convection and thermal radiation are the primary mechanisms for thermal energy transfer Command term: Identify

28B-1A-31

Stefan–Boltzmann & Wien·B.1 Thermal energy transfers

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

The surface of the Sun behaves as a black body at a temperature of about 5800 K. Wien's displacement law constant is 2.9 × 10⁻³ m K.

At what wavelength does the Sun's emission spectrum peak?

Show mark scheme

Marking point	Mark	Notes
Step 1Wien's law rearranged: λ_max = 2.9 × 10⁻³ / T.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2λ_max = 2.9 × 10⁻³ / 5800 = 5.0 × 10⁻⁷ m, which is 500 nm, in the visible region.	✓ 1	Answer D

Answer: D · 2 stages of work, one mark

Every option, and why

AThe equation has been inverted: 5800 / 2.9 × 10⁻³ = 2.0 × 10⁶.
BWien's constant has been multiplied by the temperature instead of divided by it: 2.9 × 10⁻³ × 5800 = 17.
CThe factor 10⁻³ in Wien's constant has been dropped: 2.9 / 5800 = 5.0 × 10⁻⁴ m, a thousand times too long (far infrared).
DCorrect: λ_max = 2.9 × 10⁻³ / 5800 = 5.0 × 10⁻⁷ m, or 500 nm — near the middle of the visible range.

29B-1B-11

Method of mixtures·B.1 Thermal energy transfers

Paper 1BEasy8 marks

Data-based question8 steps to full marksDetermine

Steel ball bearings are left in boiling water for 10 minutes and then transferred quickly into water in an insulated polystyrene cup. The mixture is stirred and the highest temperature reached is recorded. Masses are measured with a balance of resolution 0.1 g and temperatures with a thermometer read to ±0.5 °C.

The specific heat capacity of water is 4180 J kg⁻¹ K⁻¹. Assume that no energy is transferred to the cup or to the air.

Quantity	Value
mass of steel balls	215.2 g
mass of water in cup	150.0 g
temperature of boiling water	99.5 °C
initial temperature of water in cup	18.5 °C
highest temperature of mixture	29.5 °C

(a)

State the absolute uncertainty in the temperature rise of the water in the cup.

(1)

(b)

Calculate the specific heat capacity c of steel.

(2)

(c)

Determine the absolute uncertainty in c.

(2)

(d)

The accepted value for this steel is 490 J kg⁻¹ K⁻¹. Comment on the student’s value.

(1)

(e)

Identify the measurement that contributes most to the uncertainty in c, and suggest how its percentage uncertainty could be reduced.

(1)

(f)

The balls cool slightly in the air while being transferred. State and explain the effect of this on the calculated value of c.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
±1.0 °C / ±1.0 K	✓ 1	Two readings of ±0.5 °C are added. Accept ±1 °C.
Part (b)
m_sc(99.5 − 29.5) = m_w × 4180 × (29.5 − 18.5)	✓ 1	Energy lost by the steel = energy gained by the water. Masses may be in g on both sides.
c = 458 J kg⁻¹ K⁻¹	✓ 1	Accept 460.
Part (c)
percentage uncertainties: ΔT_water 9.1 %, ΔT_steel 1.4 % «masses negligible», total ≈ 11 %	✓ 1	Allow ECF from (a).
Δc ≈ 50 J kg⁻¹ K⁻¹	✓ 1	Accept 48–50. Allow ECF from (b).
Part (d)
490 lies within the range 410–510 J kg⁻¹ K⁻¹, so the value is consistent with the accepted value	✓ 1	Allow ECF from (c).
Part (e)
the temperature rise of the water «≈ 9 %»; use a thermometer of finer resolution / a digital probe OR use less water «or more steel» so that its temperature rise is larger	✓ 1	Both the identification and a matching suggestion are needed.
Part (f)
the steel enters the cup below 99.5 °C, so its actual temperature fall is smaller than the value used; the calculated c is too small	✓ 1	Direction and reason both needed. OWTTE.

Answers: (a) ±1.0 °C · (b) 458 J kg⁻¹ K⁻¹ · (c) ±50 J kg⁻¹ K⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — that temperature difference determines the direction of the resultant thermal energy transfer between bodies; quantitative analysis of thermal energy transfers as given by Q = mcΔT Command term: Determine

30B-1B-12

Latent heat & phase change·B.1 Thermal energy transfers

Paper 1BHard12 marks

Data-based question12 steps to full marksDetermine

Crushed ice at 0 °C fills a funnel and surrounds an electric heater. Meltwater drips into a beaker. For each heater power P, once the dripping is steady, the meltwater is collected for 300 s and its mass m is measured. The ice also gains energy from the room, so some ice melts even when the heater is off.

From repeat runs, the uncertainty in each mass is ±1.5 g. The melting rate is R = m / 300 s. The graph shows R against P with error bars.

P / W	m / g	R / 10⁻⁵ kg s⁻¹
12.0	13.7	4.57
20.0	19.3	6.43
28.0	27.4
36.0	34.9	11.6
44.0	40.3	13.4
52.0	48.2	16.1

Graph drawn to scale

(a)

Calculate the missing value of R for P = 28.0 W, giving your answer to an appropriate number of significant figures.

(1)

(b)

Draw the line of best fit for the data.

(1)

(c)

Determine the gradient of your line, giving its unit.

(2)

(d)

Show how the gradient is related to the specific latent heat of fusion L of ice, and hence determine L.

(2)

(e)

By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in L. Hence state whether the result is consistent with the accepted value of 3.34 × 10⁵ J kg⁻¹.

(3)

(f)

State what is represented by the intercept of the line on the R axis.

(1)

(g)

Explain why the energy gained from the room does not affect the value of L found from the gradient.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
9.13 «× 10⁻⁵ kg s⁻¹»	✓ 1	= 27.4 × 10⁻³ / 300. Must be given to 3 s.f., like the other values in the column.
Part (b)
single straight line through all the error bars, not forced through the origin	✓ 1
Part (c)
gradient from a large triangle	✓ 1
0.289 × 10⁻⁵ kg J⁻¹ OR 2.9 × 10⁻⁶ kg s⁻¹ W⁻¹	✓ 1	Unit needed. Accept 0.275–0.304 × 10⁻⁵.
Part (d)
P + P_room = RL, so R = P/L + P_room/L and gradient = ΔR/ΔP = 1/L	✓ 1	OWTTE.
L = 1/gradient = 3.46 × 10⁵ J kg⁻¹	✓ 1	Allow ECF from (c). Accept 3.3 × 10⁵–3.6 × 10⁵.
Part (e)
max and min lines through all error bars «gradients ≈ 0.308 and 0.270 × 10⁻⁵» giving L between ≈ 3.2 × 10⁵ and 3.7 × 10⁵ J kg⁻¹	✓ 1
ΔL = ½(3.7 × 10⁵ − 3.2 × 10⁵) ≈ 2 × 10⁴ J kg⁻¹	✓ 1	Accept 1 × 10⁴ to 3 × 10⁴. Allow ECF.
the accepted value lies within the range so the result is consistent	✓ 1	Allow ECF from their range.
Part (f)
the rate of melting due to energy from the room / when P = 0 «≈ 0.9 × 10⁻⁵ kg s⁻¹»	✓ 1	OWTTE.
Part (g)
the energy «rate» gained from the room is the same for every value of P	✓ 1	OWTTE.
so it shifts the line up by a constant amount / changes only the intercept, not the gradient «ΔR/ΔP»	✓ 1

Answers: (a) 9.13 × 10⁻⁵ kg s⁻¹ · (c) 0.289 × 10⁻⁵ kg J⁻¹ · (d) 3.46 × 10⁵ J kg⁻¹ · (e) ±2 × 10⁴ J kg⁻¹ (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 the use of specific latent heat of fusion as given by Q = mL Command term: Determine

31B-1B-13

Luminosity & apparent brightness·B.1 Thermal energy transfers

Paper 1BMedium12 marks

Data-based question12 steps to full marksDetermine

A small filament lamp is treated as a point source that radiates uniformly in all directions. A radiation sensor measures the intensity b of the radiation at a distance d from the lamp. d is measured with a metre rule from a mark on the front edge of the lamp holder to the face of the sensor. The room is dark and the sensor is zeroed before the lamp is switched on. Each reading of b has an uncertainty of ±2 %.

The student plots b^−½ against d. The graph shows six of the seven points.

d / m	b / W m⁻²	b^−½ / W^−½ m
0.100	12.3	0.285
0.150	5.99	0.409
0.200	3.65	0.523
0.250	2.39
0.300	1.73	0.760
0.400	0.983	1.009
0.500	0.651	1.239

Graph drawn to scale

(a)

Explain why a graph of b^−½ against d is expected to be a straight line through the origin.

(2)

(b)

Calculate the missing value of b^−½ and plot it on the graph.

(1)

(c)

State the percentage uncertainty in each value of b^−½.

(1)

(d)

Draw the line of best fit and use it to determine the luminosity L of the lamp.

(4)

(e)

The line does not pass through the origin. Deduce what this shows about the measurement of d, and estimate the size of the effect.

(2)

(f)

Another student uses only the reading at d = 0.100 m to calculate L. Calculate her value and explain why it is smaller than yours.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
b = L/4πd² so b^−½ = √(4π/L) × d	✓ 1
√(4π/L) is constant, so b^−½ ∝ d: a straight line through the origin with gradient √(4π/L)	✓ 1
Part (b)
0.647 «W^−½ m» AND correctly plotted	✓ 1	Accept 0.65.
Part (c)
±1 %	✓ 1	Half the percentage uncertainty in b, because of the power −½.
Part (d)
single straight line of best fit through the points, not forced through the origin	✓ 1
gradient = 2.39 «W^−½»	✓ 1	Accept 2.32–2.46.
L = 4π/gradient²	✓ 1
L = 2.2 W	✓ 1	Accept 2.1–2.3 W. Allow ECF.
Part (e)
the line meets the d axis at a negative value / has a positive intercept on the b^−½ axis, so the true distance from the source is greater than d; systematic error «the filament is behind the mark»	✓ 1
by about 2.0 cm «= −intercept on d axis»	✓ 1	Accept 1.5–2.5 cm. Accept (y-intercept ÷ gradient) = 0.048 / 2.39.
Part (f)
L = 4π × 0.100² × 12.3 = 1.5 W	✓ 1
the true distance is larger than 0.100 m by the systematic error; the gradient is not affected by the offset but the single reading is	✓ 1	OWTTE.

Answers: (b) 0.647 W^−½ m · (c) ±1 % · (d) L = 2.2 W · (e) ≈ 2.0 cm · (f) 1.5 W (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 Command term: Determine

32B-1B-20

Conduction·B.1 Thermal energy transfers

Paper 1BHard12 marks

Data-based question12 steps to full marksDetermine

A student measures the thermal conductivity k of acrylic. A square electric heater plate of side 10.0 ± 0.1 cm rests on a stack of acrylic sheets of the same area. The stack stands on a metal block kept at 0.0 °C by iced water, and a thermostat keeps the lower face of the heater at 50.0 °C. The temperature difference across the stack is 50.0 ± 1.0 K. The top and sides of the heater are insulated.

For stacks of 1 to 5 sheets the total thickness x is measured with vernier calipers, and the electrical power P supplied to the heater is measured once the temperatures are steady. The uncertainty in each P is ±0.5 W.

x / mm	(1/x) / m⁻¹	P / W
2.9	345	33.6
6.0	167	16.6
9.1	110	11.8
12.0	83	8.8
15.1	66	7.6

Graph drawn to scale

(a)

Outline how the student could confirm that a steady state had been reached before recording P.

(1)

(b)

Draw the line of best fit for the data.

(1)

(c)

Determine k.

(3)

(d)

By drawing lines of maximum and minimum gradient, determine the percentage uncertainty in the gradient.

(2)

(e)

Determine the absolute uncertainty in k. Hence comment on whether the result agrees with the literature value of 0.19 W m⁻¹ K⁻¹.

(3)

(f)

The line does not pass through the origin. Suggest a reason for this and explain why it does not affect the value of k.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)
the power reading/the temperature of the cold block «and heater» no longer changes over several minutes	✓ 1	OWTTE.
Part (b)
single straight line through all error bars, not forced through the origin	✓ 1
Part (c)
gradient ≈ 0.094 W m	✓ 1	Accept 0.089–0.098 W m.
P = kAΔT × (1/x) «+ constant» so k = gradient / (AΔT)	✓ 1
k = 0.0937 / (0.0100 × 50.0) = 0.19 W m⁻¹ K⁻¹	✓ 1	Allow ECF from gradient. Unit needed.
Part (d)
max and min gradients ≈ 0.0969 and 0.0910 W m	✓ 1
percentage uncertainty = ½(max − min)/best ≈ 3 %	✓ 1	Accept 2–6 %. Allow ECF.
Part (e)
ΔA/A = 2 × 1 % = 2 % AND Δ(ΔT)/ΔT = 2 %	✓ 1
total ≈ 7 % so k = (0.187 ± 0.013) W m⁻¹ K⁻¹	✓ 1	Allow ECF from (c) and (d). Matching precision needed.
0.19 lies within the range so the result agrees	✓ 1	Allow ECF from their range.
Part (f)
energy is lost from the heater through the insulation «at the same rate for every stack» / the thermostat and wires use some power	✓ 1	OWTTE.
this adds a constant to P, changing the intercept but not the gradient, which alone is used for k	✓ 1

Answers: (c) 0.19 W m⁻¹ K⁻¹ · (d) ≈ 3 % · (e) (0.187 ± 0.013) W m⁻¹ K⁻¹ (the remaining parts are explanations — see the table above)

33B-2-19

Density & the molecular model·B.1 Thermal energy transfers

Paper 2Easy10 marks

Short answer & extended response10 steps to full marksExplain

Gallium is a metal that melts at 29.8 °C. A student holds a 25.0 g bead of solid gallium, initially at 20.0 °C, in her closed hand. The skin of her hand is at 34.0 °C.

Specific heat capacity of solid gallium = 370 J kg⁻¹ K⁻¹. Specific latent heat of fusion of gallium = 8.0 × 10⁴ J kg⁻¹. At its melting point the density of solid gallium is 5.91 × 10³ kg m⁻³ and the density of liquid gallium is 6.10 × 10³ kg m⁻³.

(a)

State the melting point of gallium in kelvin.

(1)

(b)

Outline why there is a net transfer of thermal energy from the hand to the gallium, and how this energy is transferred by conduction.

(2)

(c)

(i)

Show that about 2.1 kJ must be transferred to the bead to melt it completely.

(1)

(ii)

The hand transfers energy to the bead at a mean rate of 2.5 W. Calculate the time taken for the bead to melt completely, in minutes.

(1)

(d)

(i)

Calculate the change in the volume of the bead when it melts.

(2)

(ii)

Most substances expand when they melt. Explain this, using the molecular model of solids and liquids.

(2)

(iii)

Suggest what your answer to (d)(i) shows about the arrangement of the atoms in solid gallium.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
29.8 + 273 = 303 K	✓ 1	Accept 302.95 K.
Part (b)
the hand is at a higher temperature than the gallium «and the net transfer is always from higher to lower temperature»	✓ 1
particles of the skin have a greater mean kinetic energy; in collisions at the surface of contact they pass kinetic energy to the gallium atoms, which pass it on to neighbouring atoms «free electrons also carry energy through the metal»	✓ 1	OWTTE. Do not accept 'heat particles' or 'cold flows into the hand'.
Part (c)(i)
Q = mcΔθ + mL = 0.0250 × 370 × 9.8 + 0.0250 × 8.0 × 10⁴ = 90.7 + 2000 = 2.09 × 10³ J	✓ 1	Must see full substitution OR answer to at least 3 s.f. Both terms must be seen.
Part (c)(ii)
t = 2.09 × 10³/2.5 = 836 s ≈ 14 min	✓ 1	Accept 13–14 min. Allow ECF from (c)(i).
Part (d)(i)
V = m/ρ: solid 4.230 × 10⁻⁶ m³, liquid 4.098 × 10⁻⁶ m³	✓ 1
the volume decreases by 1.3 × 10⁻⁷ m³ «0.13 cm³, about 3 %»	✓ 1	A decrease must be stated or clear. Award [1 max] for the value without the direction.
Part (d)(ii)
in a solid the particles vibrate about fixed positions in a closely packed, ordered arrangement	✓ 1
in a liquid the particles can move past one another in a disordered arrangement, so their mean separation is slightly larger and the same mass occupies a larger volume «lower density»	✓ 1	OWTTE.
Part (d)(iii)
the atoms in solid gallium are arranged in an open «loosely packed» structure, so on average they are further apart than in the liquid	✓ 1	Accept 'the atoms are closer together in the liquid than in the solid'.

Answers: (a) 303 K · (c)(i) 2.09 × 10³ J · (c)(ii) ≈ 14 min · (d)(i) −1.3 × 10⁻⁷ m³ (the remaining parts are explanations — see the table above)

Syllabus understandingB.1 — molecular theory in solids, liquids and gases; density ρ as given by ρ = m/V; that Kelvin and Celsius scales are used to express temperature; that temperature difference determines the direction of the resultant thermal energy transfer between bodies; conduction in terms of the difference in the kinetic energy of particles; 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 L as given by Q = mcΔT and Q = mL Command term: Explain

34B-2-20

Continuous-flow heating·B.1 Thermal energy transfers

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksDetermine

In a nuclear power station the fission of uranium-235 in the reactor releases energy at a rate of 3.2 GW. The electrical power output of the station is 1.1 GW. The rest of the energy is removed in the condenser by cooling water that is taken from a river and returned to it at a higher temperature.

Energy released per fission = 200 MeV. Mass of a uranium-235 nucleus = 235 u. Specific heat capacity of water = 4180 J kg⁻¹ K⁻¹.

(a)

(i)

Calculate the overall efficiency of the power station.

(1)

(ii)

Determine the number of fissions that occur in the reactor each second.

(2)

(iii)

Estimate the mass of uranium-235 that undergoes fission in one day.

(2)

(b)

Regulations allow the station to take no more than 5.0 × 10⁴ kg of water per second from the river, and the water may be returned no more than 8.0 K warmer than when it was taken.

(i)

Determine whether the station can operate at full power without breaking either regulation.

(4)

(ii)

Outline how the control rods can be used to reduce the power of the reactor.

(2)

(c)

The warm water returned to the river tends to remain near the surface instead of mixing with the cooler water below it. Explain why.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
η = 1.1/3.2 = 0.34 OR 34 %	✓ 1	Accept 0.344 or 34.4 %.
Part (a)(ii)
energy per fission = 200 × 1.60 × 10⁻¹³ = 3.2 × 10⁻¹¹ J	✓ 1
number per second = 3.2 × 10⁹/3.2 × 10⁻¹¹ = 1.0 × 10²⁰ s⁻¹	✓ 1	Award [2] for CNA.
Part (a)(iii)
mass of one nucleus = 235 × 1.66 × 10⁻²⁷ = 3.90 × 10⁻²⁵ kg	✓ 1
mass per day = 1.0 × 10²⁰ × 86 400 × 3.90 × 10⁻²⁵ ≈ 3.4 kg	✓ 1	Accept 3.3–3.4 kg. Allow ECF from (a)(ii).
Part (b)(i)
power to be removed = 3.2 − 1.1 = 2.1 GW «2.1 × 10⁹ W»	✓ 1
greatest power that can be removed = 5.0 × 10⁴ × 4180 × 8.0	✓ 1
= 1.67 × 10⁹ W «1.7 GW»	✓ 1
this is less than 2.1 × 10⁹ W, so the station cannot run at full power	✓ 1	ALTERNATIVE 1: flow needed for an 8.0 K rise = 2.1 × 10⁹/(4180 × 8.0) = 6.3 × 10⁴ kg s⁻¹ ✓✓, more than 5.0 × 10⁴ kg s⁻¹ ✓. ALTERNATIVE 2: rise at the greatest flow = 2.1 × 10⁹/(5.0 × 10⁴ × 4180) = 10 K ✓✓, more than 8.0 K ✓. MP4 requires a comparison of calculated values.
Part (b)(ii)
control rods absorb neutrons «without undergoing fission»	✓ 1
inserting them further means fewer neutrons go on to cause fissions, so the fission rate «and the power» decreases	✓ 1	OWTTE.
Part (c)
warm water has expanded, so its density is lower than that of the cooler river water	✓ 1
the less dense water is buoyed up «resultant upward force» and stays above the denser water, so convection does not carry it downwards	✓ 1	OWTTE.

Answers: (a)(i) 34 % · (a)(ii) 1.0 × 10²⁰ s⁻¹ · (a)(iii) 3.4 kg · (b)(i) 2.1 × 10⁹ W needed, 1.67 × 10⁹ W possible: no (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; qualitative description of thermal energy transferred by convection due to fluid density differences (linked: E.4 — that energy is released in spontaneous and neutron-induced fission; the role of control rods, moderators, heat exchangers and shielding in a nuclear power plant; A.3 — efficiency η in terms of energy transfer or power) Command term: Determine

35B-2-21

Conduction·B.1 Thermal energy transfers

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksCalculate

On a winter night a layer of ice forms on the surface of a pond. The upper surface of the ice is at −6.0 °C. The lower surface of the ice is in contact with the water of the pond and is at 0 °C. Assume that all of the energy conducted upwards through the ice is released by water freezing on the lower surface of the ice.

Thermal conductivity of ice = 2.2 W m⁻¹ K⁻¹. Density of ice = 920 kg m⁻³. Specific latent heat of fusion of ice = 3.34 × 10⁵ J kg⁻¹.

Axes for (c)

(a)

Energy is removed from the water that freezes on the lower surface of the ice. Explain, in terms of particles, why the temperature of this water does not fall below 0 °C while it freezes.

(2)

(b)

At one moment the ice is 4.0 cm thick.

(i)

Calculate the rate of thermal energy transfer through each square metre of the ice.

(2)

(ii)

Show that the mass of water that freezes on the lower surface of each square metre of ice in one hour is about 3.6 kg.

(1)

(iii)

Calculate the increase in the thickness of the ice in one hour.

(1)

(c)

On the axes, sketch a graph to show how the temperature varies with depth through the ice when it is 4.0 cm thick.

(2)

(d)

State and explain how the rate at which the ice thickens changes as the ice becomes thicker, if the temperature of its upper surface stays at −6.0 °C.

(2)

(e)

Suggest why a layer of snow lying on top of the ice slows the growth of the ice.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
the energy removed reduces the intermolecular potential energy «as bonds form between the molecules»	✓ 1
the average «random» kinetic energy of the molecules does not change, so the temperature stays at 0 °C until all the water has frozen	✓ 1	OWTTE. Do not accept “the energy is used to freeze the water” alone.
Part (b)(i)
ΔQ/Δt = 2.2 × 1.0 × 6.0/0.040	✓ 1
= 330 W «per m²»	✓ 1	Award [1 max] for 3.3 W «thickness not converted to m».
Part (b)(ii)
m = 330 × 3600/3.34 × 10⁵ = 3.56 kg	✓ 1	Must see full substitution OR answer to at least 3 s.f. Allow ECF from (b)(i).
Part (b)(iii)
Δx = 3.56/(920 × 1.0) = 3.9 × 10⁻³ m «3.9 mm»	✓ 1	Accept 3.9–4.0 mm. Allow ECF from (b)(ii).
Part (c)
straight line with a positive gradient	✓ 1	Do not accept a curve.
from −6.0 °C at depth 0 to 0 °C at 4.0 cm	✓ 1
Part (d)
the rate of thickening decreases	✓ 1	No mark for a bare statement without reasoning. MP1 only scores if MP2 scores.
the same temperature difference acts across a greater thickness, so the temperature gradient ΔT/Δx and the rate of conduction are smaller, so less water freezes each second	✓ 1	OWTTE.
Part (e)
snow «containing trapped air» is a poor conductor, so the upper surface of the ice is warmer than the air/closer to 0 °C, giving a smaller temperature gradient in the ice	✓ 1	OWTTE.

Answers: (b)(i) 330 W · (b)(ii) 3.56 kg · (b)(iii) 3.9 × 10⁻³ m (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 latent heat L as given by Q = mL Command term: Calculate

36B-2-22

Latent heat & phase change·B.1 Thermal energy transfers

Paper 2Medium13 marks

Short answer & extended response13 steps to full marksExplain

A cyclist rides at a constant speed of 8.0 m s⁻¹ against a total resistive force of 30 N. The cyclist's muscles convert chemical energy from food into useful mechanical work with an efficiency of 20 %; the rest of the energy released becomes thermal energy in the body.

Mass of cyclist = 68 kg. Mean specific heat capacity of the human body = 3500 J kg⁻¹ K⁻¹. Specific latent heat of vaporization of sweat at skin temperature = 2.4 × 10⁶ J kg⁻¹.

(a)

(i)

Calculate the useful mechanical power developed by the cyclist.

(1)

(ii)

Show that thermal energy is produced in the cyclist's body at a rate of about 960 W.

(2)

(b)

During a 2.0 hour ride, 75 % of the thermal energy produced is removed by the evaporation of sweat.

The cyclist drinks 1.5 kg of water during the ride. Determine whether this replaces the water lost by evaporation.

(4)

(c)

(i)

Estimate the time it would take for the cyclist's body temperature to rise by 2.0 K if none of the thermal energy produced could leave the body.

(2)

(ii)

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

(3)

(iii)

Suggest why cycling on a hot, humid day is more likely to cause overheating than cycling on a hot, dry day.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
P = Fv = 30 × 8.0 = 240 W	✓ 1
Part (a)(ii)
input power = 240/0.20 = 1200 W	✓ 1
thermal power = 1200 − 240 = 960 W	✓ 1	Award [0] for 0.80 × 240 = 192 W. Must see both steps.
Part (b)
power removed by evaporation = 0.75 × 960 = 720 W	✓ 1	Allow ECF from (a)(ii).
rate of evaporation = 720/2.4 × 10⁶ = 3.0 × 10⁻⁴ kg s⁻¹	✓ 1
mass evaporated in 2.0 h = 3.0 × 10⁻⁴ × 7200 = 2.2 kg	✓ 1	ALTERNATIVE: energy removed = 720 × 7200 = 5.2 × 10⁶ J ✓, mass = 5.2 × 10⁶/2.4 × 10⁶ = 2.2 kg ✓.
2.2 kg is more than 1.5 kg, so the water drunk does not replace the water lost	✓ 1	MP4 requires a comparison with a calculated value.
Part (c)(i)
t = mcΔT/P = 68 × 3500 × 2.0/960	✓ 1
t ≈ 500 s «about 8 minutes»	✓ 1	Accept 470–500 s «use of 1 kW gives 476 s».
Part (c)(ii)
only the fastest/most energetic molecules have enough energy to overcome the intermolecular forces and escape from the liquid surface	✓ 1
so the average kinetic energy of the molecules remaining in the liquid decreases	✓ 1
temperature is a measure of the average kinetic energy, so the sweat cools and energy is transferred to it from the skin	✓ 1
Part (c)(iii)
in humid air the rate of evaporation is lower «molecules return to the liquid almost as fast as they leave», so less energy is removed each second «sweat drips off without evaporating»	✓ 1	OWTTE.

Answers: (a)(i) 240 W · (b) 2.2 kg lost: no · (c)(i) ≈ 500 s (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 (linked: A.3 — that power developed P is the rate of work done, or the rate of energy transfer, as given by P = ΔW/Δt = Fv; efficiency η in terms of energy transfer or power) Command term: Explain

37B-2-23

Stefan–Boltzmann & Wien·B.1 Thermal energy transfers

Paper 2Medium12 marks

Short answer & extended response12 steps to full marksSketch

A potter monitors the temperature inside a kiln by analysing the radiation that emerges from a small peephole in the kiln wall. The peephole has an area of 4.0 cm² and behaves as a black body at the temperature of the kiln interior.

The graph shows the spectrum of the radiation from the peephole at one moment during a firing.

Graph drawn to scale

(a)

(i)

State what is meant by a black body.

(1)

(ii)

Using the graph, determine the temperature of the kiln interior, in °C.

(2)

(iii)

The kiln later cools. Sketch, on the graph, the spectrum of the radiation from the peephole when the temperature of the kiln interior is lower.

(3)

(b)

Calculate the power radiated through the peephole at the temperature found in (a)(ii).

(2)

(c)

(i)

Calculate the energy of a photon at the peak wavelength of the spectrum.

(2)

(ii)

The interior of the kiln appears bright orange to the potter. Explain this observation, given that the peak of the spectrum is in the infrared.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
a body that absorbs all the electromagnetic radiation that falls on it «and so emits the maximum possible radiation at each wavelength for its temperature»	✓ 1	OWTTE.
Part (a)(ii)
λ_max = 2.0 μm read from the graph AND T = 2.9 × 10⁻³/2.0 × 10⁻⁶ = 1450 K	✓ 1	Accept λ_max from 1.9 to 2.1 μm.
T = 1450 − 273 ≈ 1180 °C	✓ 1	Accept 1110–1250 °C. Award [1 max] for an answer left in kelvin.
Part (a)(iii)
peak at a longer wavelength than 2.0 μm	✓ 1
peak lower than the original peak	✓ 1
curve below the original at every wavelength, with a similar shape	✓ 1	Curves must not cross.
Part (b)
P = σAT⁴ = 5.67 × 10⁻⁸ × 4.0 × 10⁻⁴ × 1450⁴	✓ 1	Allow ECF from (a)(ii).
P = 1.0 × 10² W	✓ 1	Award [1 max] for 44 W «temperature in °C used». Award [1 max] if the area is not converted to m².
Part (c)(i)
E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸/2.0 × 10⁻⁶	✓ 1
E = 9.9 × 10⁻²⁰ J «0.62 eV»	✓ 1	Allow ECF from the value of λ_max read in (a)(ii).
Part (c)(ii)
the spectrum is continuous and extends into the visible region, so some visible light is emitted	✓ 1
within the visible region the intensity is greatest at the red «long-wavelength» end and very small at the blue end, so the light looks orange-red	✓ 1	OWTTE.

Answers: (a)(ii) 1450 K ≈ 1180 °C · (b) 1.0 × 10² W · (c)(i) 9.9 × 10⁻²⁰ J (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 (linked: E.1 — E = hf; C.2 — the nature of electromagnetic waves) Command term: Sketch

38B-2-35

Mechanisms of thermal transfer·B.1 Thermal energy transfers

Paper 2Easy11 marks

Short answer & extended response11 steps to full marksAnnotate

The figure shows a cross-section of a vacuum flask. The flask has a double glass wall with a vacuum between the two layers of glass. The glass surfaces facing the vacuum are silvered. The flask is closed by a hollow plastic stopper.

The flask is filled with 0.45 kg of tea at 88 °C. After 8.0 hours the temperature of the tea is 71 °C. Specific heat capacity of tea = 4180 J kg⁻¹ K⁻¹.

Diagram NOT accurately drawn

(a)

Annotate the figure to explain how the vacuum, the silvered surfaces and the stopper each reduce the rate of thermal energy transfer from the tea.

(3)

(b)

Determine the mean rate of thermal energy transfer from the tea during the 8.0 hours.

(3)

(c)

Explain why thermal energy can be transferred across a vacuum by radiation but not by conduction or convection.

(2)

(d)

State and explain how the rate of thermal energy transfer from the tea changes as the tea cools.

(2)

(e)

Suggest, with a reason, which part of the flask is responsible for most of the energy transferred from the tea.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)
vacuum: contains «almost» no particles, so there is no conduction or convection across the gap	✓ 1	Annotations must be linked to the correct feature.
silvered surfaces: poor emitters and good reflectors of infrared radiation, so transfer by radiation is reduced	✓ 1
stopper: a poor conductor «containing trapped air» that reduces conduction AND stops warm air rising out of the flask «convection» or evaporation	✓ 1
Part (b)
Q = 0.45 × 4180 × (88 − 71)	✓ 1
Q = 3.2 × 10⁴ J	✓ 1
rate = 3.2 × 10⁴/(8.0 × 3600) = 1.1 W	✓ 1	Allow ECF from MP2. Award [3] for CNA.
Part (c)
conduction and convection need particles «that collide or that move as a fluid» to carry the energy	✓ 1
radiation is energy carried by electromagnetic waves, which need no medium	✓ 1
Part (d)
the rate decreases	✓ 1	No mark for a bare statement without reasoning.
because the temperature difference between the tea and the surroundings decreases	✓ 1
Part (e)
the stopper/neck of the flask: there is no vacuum there, so energy is conducted through the stopper and the glass where the two walls join	✓ 1	OWTTE.

Answers: (b) 1.1 W (the remaining parts are explanations — see the table above)

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; quantitative analysis of thermal energy transfers Q with the use of specific heat capacity c as given by Q = mcΔT Command term: Annotate

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

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