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

This guide covers topic P1.2 (Changes of state), statements P1.2a to P1.2f, of the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026). None of these statements is printed in bold, so all of them are for both Foundation and Higher Tier. It teaches the ideas from scratch and works through the two calculations the specification names under PM1.2i and PM1.2ii.

Related pages: revision notes, practice questions, the wider Matter study guide, the OCR GCSE Physics hub, the printable checklist and the free 10-minute diagnostics.

What you need to be able to do

Statement	What you must be able to do
P1.2a	Describe how mass is conserved when substances melt, freeze, evaporate, condense or sublimate
P1.2b	Describe how physical changes differ from chemical changes: the original properties return when the change is reversed
P1.2c	Describe how heating a system changes the energy stored in it and raises its temperature or produces changes of state
P1.2d	Define specific heat capacity and distinguish it from specific latent heat (of fusion and of vaporisation)
P1.2e	Use the relationship between change in internal energy, mass, specific heat capacity and temperature change
P1.2f	Use the relationship between specific latent heat and mass to find the energy change in a change of state

P1.2a: Mass is conserved in a change of state

The changes you need are melting (solid to liquid), freezing (liquid to solid), evaporating (liquid to gas), condensing (gas to liquid), and sublimating (solid straight to gas). The reverse of sublimation, gas straight to solid, is also a change of state.

In every one of these, the particles are the same particles before and after. They are only arranged differently and move differently. Nothing is made and nothing is destroyed, so the mass of the whole system stays the same.

A change in the reading on a balance does not mean mass has changed. It means some of the substance has left the pan, usually as a gas. Seal the substance in a container and weigh the container and contents together, and the total stays constant.

Example. A sealed bag holds a pellet of dry ice (solid carbon dioxide). The bag and pellet together have a mass of 64.2 g. The bag slowly inflates as the pellet sublimates. The bag and gas still have a mass of 64.2 g. The pellet has not vanished: the carbon dioxide is now a gas filling the bag.

An open dish of water that loses mass over a week has not destroyed any water. Water particles have escaped into the air by evaporation. The mass of the dish, the water and the air above it together has not changed.

P1.2b: Physical changes and chemical changes

A change of state is a physical change. The substance is still the same substance, so when you reverse the change you get the original material back with its original properties. Melt a block of wax and let it set again: you have wax. Boil water and condense the steam: you have water.

A chemical change makes a new substance. You cannot simply reverse it and recover the starting material with its original properties. Burn the wax in a candle and you get carbon dioxide and water vapour, not wax.

	Physical change	Chemical change
New substance formed?	No	Yes
Original material back if reversed?	Yes, with its original properties	Not by simply reversing it
Examples	Melting, boiling, condensing, freezing, sublimation	Burning a fuel, rusting, cooking an egg

The test to use is the one in the specification: if you reverse the change, does the material recover its original properties? If yes, it was a physical change.

P1.2c: What heating does to a system

A system has energy stored in the motion of its particles (kinetic energy) and in the forces between them (potential energy). The total is the internal energy. Heating a system transfers energy into it, so its internal energy rises. That energy can do one of two things.

1. Raise the temperature. Particles gain kinetic energy and move faster. This happens while a substance stays in one state.
2. Produce a change of state. The energy goes into overcoming the forces between particles, so they move further apart. This increases potential energy. The temperature stays constant while the change is happening.

On a graph of temperature against time for steady heating, the sloping sections are single states warming and the flat sections are the changes of state. The flat section for boiling is normally longer than the one for melting, because separating the particles completely takes more energy per kilogram.

Temperature and heat are different things. Temperature tells you about the average kinetic energy of the particles. Energy transferred by heating is measured in joules.

P1.2d: Two quantities that must not be mixed up

Specific heat capacity, c, is the energy that must be transferred to 1 kg of a substance to warm it by 1 °C. Its unit is J/kg °C. It is about a temperature change within one state. A high value means the substance needs a lot of energy for each degree, so it warms slowly.

Specific latent heat, L, is the energy that must be transferred to change 1 kg of a substance from one state to another with no change in temperature. Its unit is J/kg. The specification names two versions:

• specific latent heat of fusion: for melting (and freezing, when the same energy is given out)
• specific latent heat of vaporisation: for boiling or evaporating (and condensing, when energy is given out)

	Specific heat capacity	Specific latent heat
Symbol, unit	c, J/kg °C	L, J/kg
Happens during	Temperature rise or fall	A change of state
Temperature	Changes	Stays constant
Needs a temperature change in the equation?	Yes	No

P1.2e: Energy to change temperature

change in thermal energy (J) = mass (kg) × specific heat capacity (J/kg °C) × change in temperature (°C)
E = m × c × Δθ

The change in temperature is final minus initial. A fall in temperature means energy is transferred out of the substance. Convert grams to kilograms before you substitute (divide by 1000).

Worked example 1: finding energy. A copper block of mass 1.8 kg is heated from 15 °C to 65 °C. The specific heat capacity of copper is 385 J/kg °C.

Δθ = 65 − 15 = 50 °C
E = 1.8 × 385 × 50
E = 34 650 J ≈ 35 000 J (2 significant figures)

Worked example 2: finding c. A 600 g sample of metal absorbs 21 600 J and warms by 80 °C. Find its specific heat capacity.

m = 600 g = 0.60 kg
c = E / (m × Δθ) = 21 600 / (0.60 × 80)
c = 450 J/kg °C

Worked example 3: finding a temperature change. A tank holds 2.4 kg of oil at 35 °C. The oil absorbs 144 000 J. Its specific heat capacity is 2000 J/kg °C. Find the final temperature.

Δθ = E / (m × c) = 144 000 / (2.4 × 2000) = 30 °C
final temperature = 35 + 30 = 65 °C

Finding c by experiment

The specification suggests investigating the specific heat capacity of different metals or water using an electrical heater and a joulemeter. A block of known mass, with a thermometer in a hole beside the heater, is heated while the joulemeter records the energy supplied. Plot energy against temperature rise. The gradient is m × c, so c is the gradient divided by the mass.

Lagging the block cuts down energy lost to the surroundings. If energy is lost, the block warms less than it should for the energy measured, so the calculated c comes out too high.

P1.2f: Energy to change state

thermal energy for a change in state (J) = mass (kg) × specific latent heat (J/kg)
E = m × L

There is no temperature change in this equation. Use the fusion value for melting or freezing and the vaporisation value for boiling or condensing.

Worked example 4: vaporising. How much energy is needed to turn 0.35 kg of water at 100 °C into steam at 100 °C? For water, the specific latent heat of vaporisation is taken as 2 260 000 J/kg.

E = m × L = 0.35 × 2 260 000
E = 791 000 J

Worked example 5: condensing. Steam of mass 0.050 kg condenses to water at 100 °C. How much energy does it release?

E = 0.050 × 2 260 000 = 113 000 J

Worked example 6: finding L. It takes 138 000 J to melt 0.60 kg of a metal already at its melting point. Find the specific latent heat of fusion.

L = E / m = 138 000 / 0.60
L = 230 000 J/kg

Two stages in one problem

Many questions go from below the melting point to a liquid, or from a liquid to a gas. Split the problem into stages, find each with the right equation, then add.

Worked example 7. A 1.2 kg block of lead starts at 20 °C. How much energy is needed to heat it to its melting point of 327 °C and then melt it completely? For lead, c = 130 J/kg °C and the specific latent heat of fusion is 23 000 J/kg.

Stage 1 (warming): Δθ = 327 − 20 = 307 °C
E₁ = 1.2 × 130 × 307 = 47 892 J
Stage 2 (melting): E₂ = 1.2 × 23 000 = 27 600 J
Total = 47 892 + 27 600 = 75 492 J ≈ 75 000 J (2 significant figures)

Sketch the temperature-time graph before you start. Each sloping section needs E = mcΔθ and each flat section needs E = mL.

Common errors

• Using E = mcΔθ across a change of state. The temperature does not change while a substance melts, so Δθ would be zero.
• Leaving the mass in grams.
• Using the wrong latent heat: fusion for melting, vaporisation for boiling.
• Saying that the particles “gain heat” or that the temperature rises during a change of state. The energy changes the arrangement of the particles, not their average speed.
• Writing that mass is lost when a liquid evaporates, without saying it has left the system as a gas.
• Calling melting a chemical change because the substance looks different.

Next steps

Work through the revision notes, then try the practice questions. Tick off what you can do on the OCR Physics checklist, and use the free 10-minute diagnostics to find weak spots.

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 and mathematical learning outcomes PM1.2i and PM1.2ii.