OCR GCSE Physics A: Changes of state, specific heat capacity and latent heat (J249) – 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 topic P1.2 (Changes of state, statements P1.2a to P1.2f) of OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026). None of the statements is Higher Tier only, so every question is suitable for both tiers. Where data are needed they are given in the question. Learn the ideas first in the study guide or the revision notes. The OCR GCSE Physics hub and the checklist show the rest of the course, and the free 10-minute diagnostics help you find weak topics.

Questions

1. A strong sealed flask contains iodine crystals. The flask and contents have a mass of 215.4 g. The flask is warmed gently and the crystals change directly into a purple gas, with no liquid forming.

(a) Name this change of state. [1] (b) State the mass of the flask and contents after the change and explain your answer. [2]

2. Classify each change as physical or chemical and give a reason for each.

(a) Molten candle wax sets hard again. [1] (b) The wax of a lit candle burns. [1] (c) An iron nail left in damp air becomes covered in rust. [1] (d) Solid carbon dioxide changes into carbon dioxide gas. [1]

3. A solid is heated steadily until it melts.

(a) Describe what happens to the energy supplied while the solid is warming below its melting point. [2] (b) Describe what happens to the energy supplied while the solid is melting, and state what happens to its temperature. [2]

4.

(a) Define specific heat capacity. [1] (b) Explain how specific latent heat of vaporisation differs from specific heat capacity. [2]

5. A sandbox contains 4.5 kg of dry sand. The sun warms the sand from 12 °C to 37 °C. The specific heat capacity of sand is 830 J/kg °C. Calculate the energy transferred to the sand. [3]

6. A student heats a 0.45 kg metal block with an electrical heater. A joulemeter shows 16 200 J has been supplied. The block warms from 20.0 °C to 44.0 °C.

(a) Calculate the specific heat capacity of the metal. [3] (b) The accepted value for this metal is lower than the student’s result. Suggest a cause and explain how it affects the result. [2]

7. A tank holds 1.6 kg of oil at 22 °C. The oil absorbs 96 000 J. Its specific heat capacity is 2000 J/kg °C. Calculate the final temperature. [3]

8. For water, the specific latent heat of vaporisation is taken as 2 260 000 J/kg.

(a) Calculate the energy needed to turn 0.28 kg of water at 100 °C into steam at 100 °C. [2] (b) It takes 148 500 J to melt 0.45 kg of a metal that is already at its melting point. Calculate the specific latent heat of fusion of the metal. [2]

9. A 0.20 kg sample of a solid is heated and the energy supplied is measured with a joulemeter.

Energy supplied (kJ)	0	6	12	32	52	58	64
Temperature (°C)	25	40	55	55	55	67.5	80

(a) State the melting point of the substance. [1] (b) Explain why the temperature stays constant between 12 kJ and 52 kJ although energy is still being supplied. [2] (c) Calculate the specific heat capacity of the solid. [2] (d) Calculate the specific latent heat of fusion. [2] (e) Calculate the specific heat capacity of the liquid. [1]

10. A sample of steam has a mass of 40 g at 100 °C. It condenses to water at 100 °C and the water then cools to 60 °C. Calculate the total energy released. For water, use L = 2 260 000 J/kg (vaporisation) and c = 4200 J/kg °C. [5]

11. A 250 g piece of ice at −12 °C is heated in an insulated container until it has all become water at 30 °C. The specific heat capacity of ice is 2100 J/kg °C, the specific latent heat of fusion of ice is 334 000 J/kg and water has c = 4200 J/kg °C. Calculate the total energy transferred to the sample. Give your answer to 3 significant figures. [8]

12. Two identical beakers each hold 1.5 kg of liquid at the same starting temperature. Each is given 63 000 J. One liquid is water (c = 4200 J/kg °C) and the other is oil (c = 2100 J/kg °C).

(a) Calculate the temperature rise of the water. [2] (b) Calculate the temperature rise of the oil. [1] (c) Explain, using specific heat capacity, why the two rises differ. [1]

Answers

1. (a) Sublimation [1] (b) 215.4 g [1]. The gas is still inside the sealed flask, so no particles have left the system and mass is conserved [1] Examiner insight: “The mass is the same” earns little without a reason; link it to the particles still being in the sealed container.

2. (a) Physical: the wax is recovered with the same properties [1] (b) Chemical: new substances (carbon dioxide and water vapour) are made and the wax cannot be recovered by reversing it [1] (c) Chemical: rust is a new substance, not iron [1] (d) Physical: the same substance changes state and can be condensed or frozen back with its original properties [1] Examiner insight: A bare “physical” or “chemical” is not enough when a reason is requested. Use the test from the specification: are the original properties recovered if it is reversed?

3. (a) The particles gain kinetic energy [1], so the temperature rises [1] (b) The energy works against the attraction between particles, so their potential energy rises [1]; the temperature stays constant [1] Examiner insight: Say what the energy is doing in each case. “It melts” does not explain where the energy goes.

4. (a) The energy that must be transferred to 1 kg of a substance to warm it by 1 °C [1] (b) Specific latent heat of vaporisation is the energy per kilogram to change liquid to gas [1] with no change in temperature, whereas specific heat capacity involves a temperature change [1] Examiner insight: A strong answer states the three quantities in each definition: energy, 1 kg, and what happens to temperature. Leaving out “1 kg” or “per kilogram” is a common loss.

5. Δθ = 37 − 12 = 25 °C [1]; E = 4.5 × 830 × 25 [1]; E = 93 375 J ≈ 93 400 J [1] Examiner insight: Write the substitution before the answer so that your method is visible, and give the unit.

6. (a) Δθ = 44.0 − 20.0 = 24.0 °C [1]; c = E / (mΔθ) = 16 200 / (0.45 × 24.0) [1]; c = 1500 J/kg °C [1] (b) Energy escapes from the block to the air around it (poor lagging) [1], so the block warms less than expected for the energy measured and c is calculated too high [1] Examiner insight: When asked how an error affects the result, give the direction (too high or too low) with a reason, not just the source of the error.

7. Δθ = E / (m c) = 96 000 / (1.6 × 2000) [1]; Δθ = 30 °C [1]; final temperature = 22 + 30 = 52 °C [1] Examiner insight: A common slip is giving the temperature rise (30 °C) as the final temperature. Read the last line of the question.

8. (a) E = 0.28 × 2 260 000 [1]; 632 800 J (about 633 000 J) [1] (b) L = E / m = 148 500 / 0.45 [1]; 330 000 J/kg [1] Examiner insight: No temperature change appears in E = mL. Using a temperature change here is a sign the wrong equation was chosen.

9. (a) 55 °C [1] (b) The supplied energy separates the particles, raising their potential energy, instead of raising their average kinetic energy [1]; the solid is melting, so the temperature is constant until it has all melted [1] (c) Solid warms from 25 to 55 °C with 12 kJ: Δθ = 30 °C [1]; c = 12 000 / (0.20 × 30) = 2000 J/kg °C [1] (d) Energy for melting = 52 − 12 = 40 kJ = 40 000 J [1]; L = 40 000 / 0.20 = 200 000 J/kg [1] (e) Liquid: 12 kJ for 25 °C rise: c = 12 000 / (0.20 × 25) = 2400 J/kg °C [1] Examiner insight: In table questions, take differences between readings. Using the total energy from zero for each stage is a typical error.

10. m = 40 g = 0.040 kg [1]; condensing: E₁ = 0.040 × 2 260 000 = 90 400 J [1]; cooling: Δθ = 100 − 60 = 40 °C [1]; E₂ = 0.040 × 4200 × 40 = 6720 J [1]; total = 90 400 + 6720 = 97 120 J ≈ 97 100 J [1] Examiner insight: Compare the size of the two stages as a check: the state change dominates. Forgetting the unit conversion changes the answer by a factor of 1000.

11. m = 250 g = 0.250 kg [1]; ice warms: Δθ = 12 °C, E₁ = 0.250 × 2100 × 12 = 6300 J [2]; melting: E₂ = 0.250 × 334 000 = 83 500 J [2]; water warms: E₃ = 0.250 × 4200 × 30 = 31 500 J [2]; total = 6300 + 83 500 + 31 500 = 121 300 J = 121 000 J (3 s.f.) [1] Examiner insight: Three labelled stages, each on its own line with units, make the working easy to follow. Missing the ice-warming stage is the usual loss of marks.

12. (a) Δθ = E / (m c) = 63 000 / (1.5 × 4200) [1]; 10 °C [1] (b) 63 000 / (1.5 × 2100) = 20 °C [1] (c) Oil has the lower specific heat capacity, so less energy is needed per kilogram for each degree and the temperature rises further [1] Examiner insight: Link the answer back to the definition of c, not just the numbers.

Where marks are usually lost

• Leaving mass in grams in E = mcΔθ or E = mL.
• Using E = mcΔθ across a flat section of a graph.
• Using the wrong latent heat for melting or boiling.
• Missing a stage in a multi-stage problem.
• Giving temperature change when the final temperature was asked for.
• Saying energy “stops” being supplied at a flat section of a heating graph.
• Not explaining a conservation of mass answer in terms of the particles remaining in the system.
• Omitting the unit, or rounding too early in a multi-step calculation.

Next steps

• Re-read the revision notes and the study guide.
• Go to the OCR GCSE Physics hub and tick off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Gateway Science Physics A (J249), Version 5.0 (July 2026), first assessment 2018, published by OCR. Topic P1.2 Changes of state, statements P1.2a to P1.2f, PM1.2i and PM1.2ii.