OCR A Level Physics A: Temperature, kinetic model, internal energy, specific heat capacity and latent heat (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 5.1.1 to 5.1.3 of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. Revise first with the study guide or the revision notes. The course hub is OCR A Level Physics and the printable checklist lists each statement.

Questions

1. Temperature scales.

(a) Liquid nitrogen boils at 77 K. Give this temperature in °C. [1] (b) A room is at 20 °C. Give this temperature in kelvin. [1] (c) A sample warms from 280 K to 305 K. State its temperature rise in °C. [1]

2. Thermal equilibrium.

(a) State what is meant by two objects being in thermal equilibrium. [1] (b) A thermometer is placed in hot water and read at once. The reading is too low. Explain why. [2] (c) Explain why a scale based on the resistance of a thermistor is not an absolute scale of temperature. [1]

3. A liquid has a density 800 times that of its vapour at the same temperature. Estimate how many times further apart the molecules are in the vapour than in the liquid. [2]

4. Brownian motion.

(a) Describe what is seen when smoke in a glass cell is lit from the side and viewed through a microscope. [2] (b) Explain the cause of this motion. [2] (c) State what the observation shows about the molecules of air. [1]

5. Internal energy.

(a) Define the internal energy of a system. [2] (b) A copper block is cooled from 300 K to 100 K. State how its internal energy changes and explain this in terms of the molecules. [2] (c) State what is special about the internal energy of a substance at 0 K. [1]

6. A solid is heated at constant power and melts.

(a) Explain, in terms of energy, why the temperature stays constant during melting even though energy is supplied. [3] (b) The specific latent heat of vaporisation of a substance is much greater than its specific latent heat of fusion. Suggest why. [2]

7. A lead block of mass 1.80 kg has specific heat capacity 130 J kg⁻¹ K⁻¹. A total of 9.1 kJ is transferred to it.

(a) Calculate the temperature rise. [2] (b) The block starts at 17.0 °C. Find its final temperature. [1] (c) Find the time for a 60 W heater to supply this energy, with no losses. [1]

8. A student determines the specific heat capacity of a metal block of mass 0.950 kg. The heater has a potential difference of 11.8 V across it and a current of 2.60 A. A graph of block temperature against time has gradient 0.0350 K s⁻¹.

(a) Describe how the apparatus is set up, including thermal contact and how the electrical energy is found. [3] (b) Calculate the power of the heater. [1] (c) Calculate the specific heat capacity of the metal. [2] (d) The value obtained is larger than the data value for this metal. Give one reason and one improvement. [2]

9. A steel sample of mass 0.120 kg is heated to 95.0 °C and placed in 0.250 kg of water at 15.0 °C in an insulated cup. The final temperature is 19.0 °C. Take c for water as 4200 J kg⁻¹ K⁻¹.

(a) Calculate the specific heat capacity of the steel. [3] (b) State one source of error and its effect on your answer. [2]

10. A student finds the specific latent heat of fusion of ice. A heater with 12.0 V across it and 4.00 A through it is placed in crushed ice in a funnel for 300 s. The meltwater collected is 0.0512 kg. A control funnel with no heater collects 0.0083 kg in the same time.

(a) Explain why the control is needed. [2] (b) Find the specific latent heat of fusion of the ice. [4]

11. A student finds the specific latent heat of vaporisation of a liquid by boiling it for 420 s at a power of 70 W, then at 140 W. The mass of liquid lost is 0.0113 kg at 70 W and 0.0241 kg at 140 W.

(a) Explain why using two powers removes the effect of energy losses. [2] (b) Find the specific latent heat of vaporisation. [3] (c) The specific latent heat of fusion of the same liquid’s solid form is 3.34 × 10⁵ J kg⁻¹. Calculate the ratio of the two values, and explain in terms of molecules why the ratio is greater than 1. [3]

12. A 500 W heater, with no energy losses, heats 0.350 kg of ice at −20 °C until it is water at 40 °C. Use specific heat capacity of ice 2100 J kg⁻¹ K⁻¹, of water 4200 J kg⁻¹ K⁻¹, and specific latent heat of fusion 3.34 × 10⁵ J kg⁻¹.

(a) Describe the temperature-time graph. [2] (b) Find the total energy supplied. [4] (c) Find the time taken. [1] (d) Explain how the internal energy changes in the flat section. [2]

Answers

1. (a) 77 − 273 = −196 °C [1] (b) 20 + 273 = 293 K [1] (c) The change is the same in both scales: 25 °C [1] Examiner insight: Answers gain credit for adding 273 to go from °C to K; a temperature rise should never be converted by adding 273.

2. (a) There is no net transfer of energy between them (they are at the same temperature). [1] (b) The thermometer is initially cooler than the water, so energy flows from the water to the thermometer. [1] It is not yet in equilibrium, so it reads low. [1] (c) It depends on a property of a particular material (the thermistor), whereas an absolute scale does not depend on any substance. [1] Examiner insight: Explanations tend to gain credit when they mention net energy transfer, not just “same temperature”.

3. Volume per molecule is 800 times larger, and volume scales with spacing cubed, so ratio = 800^(1/3). [1] = 9.3 times further apart. [1] Examiner insight: A common slip is to give the volume ratio (800) as the spacing ratio.

4. (a) Small bright specks (smoke particles) are seen. [1] They move in random, irregular zig-zag paths. [1] (b) Air molecules move rapidly and randomly and collide with the smoke particle. [1] The impacts from different sides are unbalanced at any instant, giving a resultant force that changes direction randomly. [1] (c) They are in continuous random motion (and are moving rapidly). [1] Examiner insight: Answers lose credit by calling the visible specks molecules.

5. (a) The sum of the kinetic energies [1] and potential energies of the molecules, which are randomly distributed. [1] (b) Internal energy decreases. [1] The average kinetic energy of the atoms falls as temperature falls. [1] (c) It is the minimum internal energy the substance can have. [1] Examiner insight: Definitions score best when they name both kinetic and potential energy of the molecules.

6. (a) Energy supplied increases the internal energy. [1] It goes into the potential energy of the molecules as bonds are overcome and they separate. [1] The average kinetic energy does not change, so the temperature is constant. [1] (b) In vaporisation the molecules move from close together to far apart, [1] which needs more work against the attractive forces than loosening the structure in melting. [1] Examiner insight: “No energy is needed” is wrong; credit comes from saying where the energy goes.

7. (a) Δθ = E / (mc) [1] = 9100 / (1.80 × 130) = 38.9, so 39 K [1] (b) 17.0 + 38.9 = 56 °C [1] (c) t = E / P = 9100 / 60 = 152, so 1.5 × 10² s [1] Examiner insight: Rearrange before substituting; keep unrounded values between parts.

8. (a) Heater and thermometer sit in holes in the insulated block, with oil in each for thermal contact. [1] An ammeter is in series with the heater and a voltmeter is across it. [1] Energy supplied is VIt. [1] (b) P = VI = 11.8 × 2.60 = 30.7 W [1] (c) c = P / (m × gradient) [1] = 30.68 / (0.950 × 0.0350) = 9.2 × 10² J kg⁻¹ K⁻¹ [1] (d) Reason: energy was lost to the surroundings, so the temperature rise was smaller than expected. [1] Improvement: add more insulation. [1] Examiner insight: A reason is stronger when it links the loss to a smaller temperature rise.

9. (a) Energy gained by water = 0.250 × 4200 × 4.0 = 4200 J [1] Energy lost by steel = 0.120 × c × 76.0 [1] c = 4200 / 9.12 = 4.6 × 10² J kg⁻¹ K⁻¹ [1] (b) Energy is lost from the water and cup to the surroundings. [1] The final temperature is lower than it should be, so the calculated c is too small. [1] Examiner insight: State the effect on the calculated value, not only the source of the error.

10. (a) Some ice melts because of the warm room, not the heater. [1] The control measures this mass so that it can be subtracted. [1] (b) E = 12.0 × 4.00 × 300 = 14 400 J [1] Mass melted by the heater = 0.0512 − 0.0083 = 0.0429 kg [1] l = E / m [1] = 14 400 / 0.0429 = 3.36 × 10⁵ J kg⁻¹ [1] Examiner insight: Subtract the control mass before dividing; dividing by the full 0.0512 kg is the usual slip.

11. (a) The liquid is boiling at the same temperature in both runs, so the rate of energy loss is the same. [1] The extra energy supplied at the higher power only vaporises the extra mass. [1] (b) Extra energy = (140 − 70) × 420 = 29 400 J [1] Extra mass = 0.0241 − 0.0113 = 0.0128 kg [1] l = 29 400 / 0.0128 = 2.30 × 10⁶ J kg⁻¹ [1] (c) Ratio = 2.30 × 10⁶ / 3.34 × 10⁵ = 6.9 [1] In vaporisation molecules go from close contact to far apart, gaining more potential energy. [1] In fusion they stay close and are only loosened. [1] Examiner insight: The same-temperature argument is the key idea in part (a); a vague “losses cancel” is weaker.

12. (a) Temperature rises steadily from −20 °C to 0 °C, [1] a flat section at 0 °C, then a less steep rise to 40 °C (water has the larger specific heat capacity). [1] (b) Warm ice: 0.350 × 2100 × 20 = 14 700 J [1] Melt: 0.350 × 3.34 × 10⁵ = 116 900 J [1] Warm water: 0.350 × 4200 × 40 = 58 800 J [1] Total = 1.90 × 10⁵ J [1] (c) t = 190 400 / 500 = 381 s [1] (d) The internal energy increases. [1] The potential energy rises as molecules move apart while the average kinetic energy stays the same. [1] Examiner insight: Treat each stage separately and list the three energies before adding.

Where marks are usually lost

• Subtracting 273 when converting from °C to K.
• Treating the specks in Brownian motion as the molecules themselves.
• Saying no energy is transferred during a change of phase.
• Using m c Δθ across a change of phase.
• Dividing by the whole mass collected, not the mass above the control.
• Giving J instead of J kg⁻¹ for a specific latent heat.

Next steps

• Re-read the revision notes.
• Return to the study guide.
• Visit the OCR A Level Physics hub and the checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), first assessment 2017, published by OCR: sections 5.1.1, 5.1.2 and 5.1.3.