Changes of state and specific latent heat — AQA GCSE Physics

SPEC 8463 • TOPIC 3.04

Specific Latent Heat

The energy to change state — without changing temperature. Two types (fusion & vaporisation), one equation (E = mL), one heating curve. Master this and you bank a guaranteed Paper 1 calculation.

The Big Idea

In the last lesson you used ΔE = mcΔθ — energy in, temperature up. But here's the puzzle: if you keep heating a beaker of ice at 0 °C, the temperature doesn't rise while it's melting. Where is the energy going?

The Answer

During a change of state, all the supplied energy is used to break the bonds between particles — not to make them move faster. The particles' potential energy increases, but their average kinetic energy (and so the temperature) stays the same.

This "hidden" energy has a name: latent heat. Latent means "hidden" — the energy is hidden in the bonds, not visible as a temperature rise. By the end of this lesson you'll be able to calculate it and spot it on a heating curve.

Mark-Scheme Definitions

Three definitions — the wording matters. AQA mark schemes credit the phrase "with no change in temperature" as a separate mark.

Specific Latent Heat

The amount of energy required to change the state of 1 kg of a substance with no change in temperature.

Specific Latent Heat of Fusion (L_f)

The specific latent heat for the change between solid and liquid — i.e. melting (or freezing).

Specific Latent Heat of Vaporisation (L_v)

The specific latent heat for the change between liquid and gas — i.e. boiling (or condensing).

The Equation

One equation, two contexts. It's on the equation sheet from Jun 2024 onwards, but you should know it cold.

E = m L

SymbolMeaningUnit

Eenergy supplied (or released) for the state changejoules (J)

mmass undergoing the state changekilograms (kg)

Lspecific latent heat (L_f or L_v)joules per kilogram (J/kg)

Unit Check

L is always in J/kg — not J/kg/°C. There's no Δθ in this equation because the temperature isn't changing. If you see a value in J/kg/°C, that's specific heat capacity, not specific latent heat.

The Heating Curve

Plot temperature against energy supplied as you heat a solid until it becomes a gas. The curve has two flat sections — those are the state changes.

Read The Curve

Slopes = temperature is changing → use ΔE = mcΔθ. Flat sections = state change → use E = mL. The boiling plateau is wider than the melting plateau because vaporisation needs more energy.

Cooling Curves — The Mirror Image

A cooling curve looks like a heating curve flipped upside-down: gas cools (slope) → condenses at b.p. (plateau, energy released = mL_v) → liquid cools (slope) → freezes at m.p. (plateau, energy released = mL_f) → solid cools. The same equations apply — energy is now given out to the surroundings, not absorbed.

SHC vs SLH — Distinguish Them

The spec specifically asks you to distinguish between specific heat capacity and specific latent heat. Examiner reports flag this as a top error — students confuse the two equations every year.

PropertySpecific Heat Capacity (c)Specific Latent Heat (L)

What it measuresEnergy to raise 1 kg by 1 °CEnergy to change state of 1 kg at constant temperature

EquationΔE = m c ΔθE = m L

Temperature change?Yes — Δθ is in the equationNo — temperature is constant

State change?No — substance stays in one stateYes — solid↔liquid or liquid↔gas

Energy goes into…Kinetic energy of particles (faster motion)Potential energy in bonds (bonds breaking)

Units of constantJ/kg °CJ/kg

On heating curveThe slopes (rising sections)The plateaus (flat sections)

Decision Rule

Read the question. Is the temperature changing? → mcΔθ. Is the state changing? → mL. A multi-stage problem uses both — see next slides.

Worked Example 1 — Melting

Calculate the energy needed to melt 2.0 kg of ice at 0 °C. The specific latent heat of fusion of water = 334 000 J/kg. 3 marks

E

Equation: E = m L_f

S

Substitute: E = 2.0 × 334 000

U

Units: joules (J)

A

Answer: E = 668 000 J ≈ 6.7 × 10⁵ J

Sanity Check

That's the energy from a 1 kW kettle running for over 11 minutes — just to melt the ice, with no warming at all. Latent heats are big numbers.

Worked Example 2 — Boiling

Calculate the energy needed to completely boil 0.50 kg of water at 100 °C. The specific latent heat of vaporisation of water = 2 260 000 J/kg. 3 marks

E

Equation: E = m L_v

S

Substitute: E = 0.50 × 2 260 000

U

Units: joules (J)

A

Answer: E = 1 130 000 J = 1.13 × 10⁶ J

Compare

To boil half the mass took nearly twice as much energy as melting 2 kg of ice. That's the L_v > L_f effect — held up in the next slides.

Multi-Stage Problems

The classic high-mark question: take a substance through a change of state and a temperature change. Split it into stages. Each stage uses one equation only.

Worked Pattern — 1 kg of ice at −20 °C → steam at 110 °C

Stage 1 — Heat ice from −20 °C to 0 °C: ΔE = mcΔθ (with c of ice)

Stage 2 — Melt ice at 0 °C: E = mL_f

Stage 3 — Heat water from 0 °C to 100 °C: ΔE = mcΔθ (with c of water)

Stage 4 — Boil water at 100 °C: E = mL_v

Stage 5 — Heat steam from 100 °C to 110 °C: ΔE = mcΔθ (with c of steam)

Total energy = sum of all 5 stages.

Exam Tip

If a problem mentions both a temperature range and a state change, write the stage list before plugging numbers. The mark scheme allocates marks per stage; miss a stage, lose a mark. The different c values (ice, water, steam) and L values will always be given in the question — you don't memorise them.

Why L_v > L_f

For every substance, vaporisation needs more energy per kg than fusion. For water the ratio is about 7 : 1. The particle model explains why.

Melting (L_f)

Particles only need enough energy to loosen the bonds — they slide past each other but stay close together. Most inter-particle forces remain.

Boiling (L_v)

Particles need enough energy to break free completely — all inter-particle forces must be overcome so particles fly apart into a gas.

334 kJ/kg

L_f of water

2260 kJ/kg

L_v of water

~7×

L_v ÷ L_f

Examiner Alert

The #1 Mistake

Students apply ΔE = mcΔθ during a state change — but Δθ = 0 during melting and boiling, so this gives zero. "There is no temperature change during a phase transition, so c·Δθ doesn't apply — only E = mL applies" (Revision Guide, §3.04).

Wrong Mass In Lab Calculations

In lab questions where a substance boils or melts and mass is measured before and after, the value of m in E = mL is not either reading on its own — it is the change in mass Δm, the mass that actually changed state. Substituting the starting mass (or the final mass) is a frequently penalised error.

How To Avoid Both

(1) Read the question. If it says "change state at constant temperature" → E = mL, never mcΔθ. (2) If the question gives you two mass readings (before and after), use the difference, not either reading on its own.

Measuring L_f of Water — Lab Method

The spec specifies "perform an experiment to measure the latent heat of fusion of water" (AT 5). Worth knowing the apparatus and method, plus the two main sources of error.

Apparatus

Funnel containing crushed ice, immersion heater inserted into the ice, beaker beneath the funnel to collect meltwater, top-pan balance, joulemeter (or known-power supply + stopwatch).

Method

Place crushed ice (already at 0 °C) into the funnel; set up an identical "control" funnel without a heater.

Record the mass of the empty collecting beaker (m₁).

Switch on the heater; record the time the heater runs (t) and the energy delivered (E = P × t, or read off joulemeter).

After time t, measure the mass of the beaker + meltwater (m₂). Mass of meltwater from heater = m₂ − m₁.

From the control, measure mass of ice that melted just from the surroundings in the same time; subtract this from m₂ − m₁ to get the true Δm caused by the heater.

Calculate L_f = E ÷ Δm.

Two Sources of Error (Mark-Scheme Favourites)

(1) Ice melts from surroundings — fixed by running a control. (2) Heat is lost to surroundings from the heater — gives a calculated L_f that is too high, because some of the supplied energy didn't go into melting the ice. The same logic transfers to L_v experiments.

PEE — Explanation Question

Explain why the temperature of a beaker of water stays at 100 °C while it is boiling, even though the heater is still on. 3 marks

P

Point: During boiling, all the supplied energy is used to break the bonds between water molecules, not to raise their kinetic energy.

E

Evidence: Temperature is a measure of the average kinetic energy of the particles; here the energy is going into potential energy (bonds).

E

Explain: Because kinetic energy of the particles is unchanged, temperature stays constant at the boiling point until all the liquid has vaporised.

Weak vs Strong Answer

Same question — why does temperature stay constant during melting? Same student, two attempts. The difference is the particle-level mechanism.

Weak — 1 mark

"The energy is being used to melt the ice, so the temperature can't go up."

Strong — 3 marks

"During melting, the supplied energy is used to break the bonds between particles, increasing their potential energy. The kinetic energy of the particles is unchanged, so the temperature stays constant."

The Pattern

Mark-scheme language: "bonds between particles broken" + "potential energy increases" + "kinetic energy unchanged". Hit those three phrases and you've banked Level 3.

Common Errors Table

Students Often WriteCorrect VersionWhy It Matters

Uses ΔE = mcΔθ during meltingUse E = mL during a state change (Δθ = 0)Wrong equation — gives zero

L for water used when ice is meltingUse the L_f value for the substance changing stateWrong constant — wrong answer

Uses single mass reading for ΔmUse mass that changed state = m₁ − m₂Lab questions reward Δm, not absolute mass

Forgets the state-change stage in a multi-stage problemSum: heating + melting + heating + boiling + heatingUnderestimates total energy

"Particles move faster" to explain plateau"Bonds broken; potential energy increases; KE unchanged"Weak vs Level-3 phrasing

Quoting L units as J/kg/°CL units are J/kg onlyΔθ = 0 so no °C in units

Practice Question

A 1500 W kettle is used to boil away water that is already at 100 °C. The kettle runs for 180 s. The specific latent heat of vaporisation of water = 2.26 × 10⁶ J/kg.

Calculate the mass of water that boils away. Assume all the electrical energy is transferred to the water. 4 marks

E

Equations: E = P × t, then m = E ÷ L_v

S

Substitute: E = 1500 × 180 = 270 000 J. Then m = 270 000 ÷ 2.26 × 10⁶

U

Units: kilograms (kg)

A

Answer: m = 0.119 kg ≈ 0.12 kg (2 s.f.)

Lesson Summary

The Three Things You Must Know

Definition : SLH = energy to change state of 1 kg with no temperature change. Two types: L_f (fusion) and L_v (vaporisation).

Equation : E = mL — on the equation sheet. L always in J/kg.

Heating curve : slopes use mcΔθ, plateaus use mL. Boiling plateau is wider because L_v > L_f.

The Three Things You Must Avoid

Don't use mcΔθ during a state change — Δθ = 0.

Don't quote L units as J/kg/°C — it's J/kg.

Don't use a single mass reading for Δm — use the change in mass.

For Level-3 Marks

When explaining plateaus, hit the three mark-scheme phrases: "bonds between particles broken", "potential energy of particles increases", "kinetic energy of particles is unchanged so temperature stays constant".