OCR GCSE Physics A: Changes of state, specific heat capacity and latent heat (J249) – Revision Notes

These notes condense 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). Every statement here is for both tiers. For full explanations see the study guide. Test yourself with the practice questions, check progress on the checklist, see the OCR GCSE Physics hub and take a free 10-minute diagnostic.

Key ideas at a glance

Statement	Remember
P1.2a	Mass is conserved in melting, freezing, evaporating, condensing and sublimating
P1.2b	Physical change: original properties return if reversed. Chemical change: they do not
P1.2c	Heating raises internal energy: temperature rises, or the state changes
P1.2d	c: energy per kg per °C. L: energy per kg for a change of state
P1.2e	E = m × c × Δθ
P1.2f	E = m × L

P1.2a: Mass conservation

• The same particles are present before and after. Only their arrangement and motion change.
• Total mass of a closed system stays constant.
• A balance reading that falls means particles have left the pan as a gas, not that mass has been destroyed.
• Evaporation then condensation in closed distillation apparatus collects the same mass of liquid that evaporated.
• Sublimation: solid to gas with no liquid stage, mass unchanged.

P1.2b: Physical or chemical?

Question to ask	Physical	Chemical
Is a new substance made?	No	Yes
Reverse it: are the original properties back?	Yes	No
Examples	Ice melting, steam condensing, iodine subliming	Wood burning, iron rusting, a cake baking

P1.2c: What heating does

• Heating transfers energy into the system, so internal energy goes up.
• Below a change of state: the energy raises particle kinetic energy and the temperature rises.
• At a change of state: the energy increases potential energy as particles separate. Temperature stays constant.
• Temperature graph for steady heating: slopes show warming, flat parts show melting and boiling.
• Flat boiling section is longer than the flat melting section for the same sample.

P1.2d: Definitions

• Specific heat capacity: the energy that must be transferred to 1 kg of the substance to warm it by 1 °C. Unit J/kg °C.
• Specific latent heat: the energy that must be transferred to change 1 kg of the substance from one state to another, with the temperature constant. Unit J/kg.
• Specific latent heat of fusion: solid and liquid (melting, freezing).
• Specific latent heat of vaporisation: liquid and gas (boiling, condensing).

Equations

Equation	Symbols and units	Use when
E = m × c × Δθ	energy (J), mass (kg), specific heat capacity (J/kg °C), temperature change (°C)	Temperature changes, no change of state
E = m × L	energy (J), mass (kg), specific latent heat (J/kg)	State changes, temperature constant

Rearrangements: c = E / (m × Δθ); Δθ = E / (m × c); m = E / (c × Δθ); L = E / m; m = E / L.

Method in steps

1. Write what happens at each stage (warming, melting, warming, boiling).
2. Choose E = mcΔθ for a sloping section and E = mL for a flat section.
3. Convert the mass to kg and find Δθ as final minus initial.
4. Calculate each stage on its own line with units.
5. Add the stages. Give the answer to a sensible number of significant figures.
6. Check the size: a latent heat stage is often much larger than a warming stage.

Worked reminders

Warming only. 0.80 kg of steel (c = 490 J/kg °C) is heated by 60 °C.

E = 0.80 × 490 × 60 = 23 520 J ≈ 24 000 J (2 s.f.)

State change only. Energy needed to vaporise 0.12 kg of water at 100 °C (L = 2 260 000 J/kg).

E = 0.12 × 2 260 000 = 271 200 J

Two stages. 0.15 kg of ice at 0 °C becomes water at 20 °C (L = 334 000 J/kg, c = 4200 J/kg °C).

Melting: 0.15 × 334 000 = 50 100 J
Warming: 0.15 × 4200 × 20 = 12 600 J
Total = 62 700 J

Practical (as the specification suggests)

• Specific heat capacity: heat a metal block or water with an electrical heater and a joulemeter. Plot energy against temperature rise. The gradient is m × c. Lag the apparatus: heat lost to the surroundings makes c look too high.
• Specific latent heat: measure the energy needed to boil or melt a known mass (for example water or stearic acid). Use the mass that changed state.

Explaining a heating graph

Use this pattern for a temperature-time graph of a substance heated at a steady rate.

Section	What is happening	Energy goes into	Equation
Rising slope, solid	Solid warming	Kinetic energy of particles	E = mcΔθ
First flat part	Melting	Potential energy (particles separate slightly)	E = mL, fusion
Rising slope, liquid	Liquid warming	Kinetic energy of particles	E = mcΔθ
Second flat part	Boiling	Potential energy (particles fully separate)	E = mL, vaporisation
Rising slope, gas	Gas warming	Kinetic energy of particles	E = mcΔθ

Cooling reverses the order. The flat parts are then condensing and freezing, and energy is released to the surroundings.

For a steady heater, the time spent on a flat section is proportional to the energy needed, so a longer flat part means a larger mL. A steeper slope means a smaller c for the same mass, because the temperature rises faster for each joule.

Two points to include in a written explanation: say that energy is still being supplied, and say what the energy is doing to the particles.

Units and conversions

Quantity	Unit	Watch for
Mass	kg	Divide grams by 1000
Energy	J	Multiply kJ by 1000
Specific heat capacity	J/kg °C	Not the same unit as latent heat
Specific latent heat	J/kg	No temperature in the unit
Temperature change	°C	Final minus initial

Size check: water has c = 4200 J/kg °C, so 1 kg warmed by 1 °C needs 4200 J. Compare that with the latent heat of vaporisation of 2 260 000 J/kg: boiling away 1 kg takes over 500 times as much energy as warming it by 1 °C.

Must-know distinctions

• Temperature is not the same as energy transferred. Temperature is in °C, energy is in J.
• c needs a temperature change. L never does.
• Fusion: melting and freezing. Vaporisation: boiling and condensing.
• Freezing and condensing release energy; melting and boiling take it in.
• A physical change can be reversed to give back the original material. A chemical change cannot.

Quick self-test

1. 0.30 kg of water is heated from 20 °C to 65 °C. Calculate the energy transferred. (c = 4200 J/kg °C)
2. A joulemeter shows 7500 J was supplied to a 0.50 kg block, which warmed by 30 °C. Find c.
3. 21 000 J is supplied to 0.50 kg of water (c = 4200 J/kg °C). Find the temperature rise.
4. It takes 85 000 J to melt 0.25 kg of a solid at its melting point. Find the specific latent heat of fusion.
5. How much water at 100 °C can be turned into steam by 565 000 J? (L = 2 260 000 J/kg)
6. State why the temperature of boiling water stays constant when you keep heating it.
7. A student weighs a sealed, strong flask of iodine crystals before and after the iodine all sublimates when warmed. Predict the readings and explain.
8. Wax melts and then sets again with the same properties. Is this a physical or chemical change? Give the reason.
9. Convert 250 g to kg, then find the energy to heat a 250 g copper cup by 40 °C (c = 385 J/kg °C).
10. Why does a steam burn usually hurt more than a burn from boiling water at the same temperature?

Answers

1. Δθ = 45 °C. E = 0.30 × 4200 × 45 = 56 700 J.
2. c = 7500 / (0.50 × 30) = 500 J/kg °C.
3. Δθ = 21 000 / (0.50 × 4200) = 10 °C.
4. L = 85 000 / 0.25 = 340 000 J/kg.
5. m = 565 000 / 2 260 000 = 0.25 kg.
6. The energy supplied goes into separating the particles (increasing potential energy), not into their kinetic energy, so the temperature does not rise.
7. The two readings are the same. The particles are still inside the sealed flask, only now as a gas, so mass is conserved.
8. Physical, because the original properties are recovered when the change is reversed.
9. 0.250 kg. E = 0.250 × 385 × 40 = 3850 J.
10. Steam condenses on the skin and releases its latent heat (energy per kilogram of vaporisation) before it cools further, so more energy is transferred to the skin.

Where marks are usually lost

• Using grams instead of kilograms in E = mcΔθ.
• Using E = mcΔθ for a stage where the substance is changing state.
• Forgetting a stage in a two-stage problem, or adding the stages without a sketch to check.
• Quoting the wrong latent heat: fusion for melting, vaporisation for boiling.
• Saying “heat” or “temperature” when the answer needs “energy” or “internal energy”.
• Explaining a flat graph section with “the substance has stopped heating”. The energy is still being supplied. It is being used to separate particles.
• Writing “the mass is lost” for evaporation instead of “the particles escape into the air”.
• Dropping units, or leaving a 6-figure answer with no rounding when significant figures are requested.

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.