The big idea: Put a pan of icy water on a hot ring: it sits stubbornly at 0 °C until the last lump of ice has melted, even though the flame keeps pouring in energy.
That energy goes into pulling the particles apart (breaking the bonds between them), not into making them move faster — so the temperature does not change while the state does.
The energy needed for this is called the latent heat ('latent' = hidden, because no temperature change shows up).
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Watch the graph build step by step in study mode.
Spot it on the curve: Sloping part = the temperature is changing (warming up).
Flat part = a state change (melting or boiling) at constant temperature — this is where latent heat is used.
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Specific latent heat L is the energy needed to change the state of 1 kg of a substance, with no temperature change. So the energy for a mass m is just mass × L:
- thermal energy transferred (J)
- mass changing state (kg)
- specific latent heat (J kg⁻¹)
Two different L values: Latent heat of fusion (Lf) = melting or freezing.
Latent heat of vaporisation (Lv) = boiling or condensing.
For the same substance Lv is much bigger than Lf — that's why the boiling plateau is the longer one.
For the sloping parts, where the temperature does change, you instead use specific heat capacity. Specific heat capacity c is the energy to warm 1 kg by 1 degree:
- thermal energy transferred (J)
- mass (kg)
- specific heat capacity (J kg⁻¹ K⁻¹)
- temperature change (K or °C)
| Part of the curve | What is happening | Equation to use |
|---|---|---|
| Sloping | temperature changing (warming/cooling) | Q = mc ΔT |
| Flat | state changing at constant temperature | Q = mL |
A block of ice of mass 0.50 kg is already at its melting point, 0 °C. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg⁻¹. Find the energy needed to melt it completely.
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How this is tested — latent heat almost always appears with energy conservation (calorimetry / mixtures):
Paper 1A
- A one-mark calculation — the mass of ice melted by some warm water.
- Or a ratio of c to L read off a constant-rate heating run.
Paper 2
- A 'show that' on an equilibrium temperature when a hot and a cold sample are mixed (sometimes one of them melts).
The classic trap: At a state change the temperature is constant, so use Q = mL (no ΔT) — and add a separate Q = mc ΔT term for every part where the temperature actually moves.
The calorimetry rule: When two things are mixed and no energy escapes:
energy lost by the hot thing = energy gained by the cold thing.
Build one Q-term for each step (warm, melt, warm again …) and set the two totals equal.
0.20 kg of water at 50 °C is poured onto ice that is already at 0 °C. The water cools to 0 °C and some of the ice melts. Take c(water) = 4.2 × 10³ J kg⁻¹ K⁻¹ and L(fusion of ice) = 3.3 × 10⁵ J kg⁻¹. Assuming no energy is lost to the surroundings, find the mass of ice that melts.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.