Latent Heat
Latent Heat
- Keep heating water at 100 °C and the thermometer stops moving — but the energy keeps going in. Where it goes is a three-mark explanation and one new formula.
The heating curve
- On the flat sections the substance is changing state at constant temperature. The energy supplied there is called — latent means hidden, because no thermometer sees it.
- Where it goes: melting and boiling pull molecules apart, so the energy raises the molecules' potential energy (separation increases against attractive forces) while their kinetic energy stays the same — and temperature tracks kinetic energy, so temperature stays constant (9702/41/O/N/25 Q2(b)(ii)).
Definitions and the formula
- : the energy per unit mass to change a substance from solid to liquid at constant temperature. For vaporisation , replace solid→liquid with liquid→gas. Both the “per unit mass” and the “at constant temperature / without temperature change” phrases score (9702/41/M/J/25 Q3(a)) (9702/42/O/N/24 Q3(a)).
Symbols
- = energy supplied during the change of state (J)
- = mass that changes state (kg)
- = specific latent heat (fusion or vaporisation) (J kg⁻¹)
- No appears — there is no temperature change to multiply by. For water: and — boiling costs roughly seven times more than melting, exactly the plateau ratio in the figure.
- Why (a banked 3-marker): melting only loosens the structure — molecular separation barely grows — while boiling separates the molecules completely, a much larger rise in potential energy; and the gas expands greatly, so work is also done pushing back the atmosphere (9702/41/M/J/25 Q3(b)) (9702/42/O/N/24 Q3(c)).
Measuring and using L
- Vaporisation: boil water with an electrical heater on a balance; the mass falls as steam leaves. Then with the mass lost. A 2024 paper extended this chain with the work done by the expanding steam (9702/42/O/N/24 Q3(b)).
- Fusion: drop ice at 0 °C into warm water and account for every joule — the method a 2025 paper examined (9702/41/M/J/25 Q3(c)). The energy the water gives up melts the ice and warms the melt-water up to the final temperature.
Worked example
Ice-and-water method for Lf
46 g of ice at 0 °C is added to 250 g of water at 25.0 °C in an insulated cup. The mixture settles at 8.7 °C. Find .
- Energy released by the warm water: .
- Energy that warmed the melted ice from 0 to 8.7 °C: .
- The remainder melted the ice: , so .
Answer
Common mistake
Forgetting the second step — the melted ice still has to warm from 0 °C to the final temperature. Miss it and comes out too large. Two-stage questions work the same way in reverse: warm the water with first, then boil part of it with , and add (9702/42/F/M/23 Q2(c)).
Worked example
Two stages: warm, then boil
A kettle holds 0.25 kg of water at 20 °C. How much energy warms it to 100 °C and then boils away 30 g?
- Warming: .
- Boiling: .
- Total: .
Answer
Boiling away just 30 g costs nearly as much as warming the whole 250 g by eighty degrees — latent heats are big.
Evaporation below the boiling point
A puddle dries at 20 °C: the fastest molecules escape from the surface, taking more than their share of energy, so the liquid left behind cools. Same physics as boiling — molecules gaining enough potential energy to leave — but only at the surface and at any temperature. A 2023 paper linked this to the first law of thermodynamics (9702/42/O/N/23 Q3(b)(ii)) — the link is made in the next chapter.