Specific latent heat

Specific latent heat is the energy needed to change the state of 1 kg of a substance without changing its temperature. The specific latent heat equation is E = mL: energy = mass × specific latent heat. For melting or freezing it is called the specific latent heat of fusion, and for boiling or condensing the specific latent heat of vaporisation. Water's values are about 334 000 J/kg for fusion and 2 260 000 J/kg for vaporisation.

What is latent heat?

Heat a pan of ice and the temperature rises until it reaches 0 °C. Then it stops rising, even though the hob is still supplying energy, and stays at 0 °C until every bit of ice has melted. Only then does the water start to warm up. The same thing happens at 100 °C while the water boils.

The energy going in during those flat stretches is latent heat, from the Latin for hidden: it changes the state without showing up on a thermometer. It is used to break the bonds between particles, or to pull them further apart, which raises their potential energy rather than their kinetic energy. A thermometer reads how fast the particles are moving on average, and that does not change, so the reading stays put.

The specific latent heat definition used in UK exams is: the amount of energy required to change the state of one kilogram of a substance with no change in temperature.

The specific latent heat equation

The unit of specific latent heat is the joule per kilogram, J/kg, also written J kg⁻¹. Unlike specific heat capacity, there is no temperature in the unit, because the temperature does not change.

Latent heat of fusion and latent heat of vaporisation

Water's values come from HyperPhysics at Georgia State University (80 and 539 calories per gram). Exam questions give the value they want you to use. Vaporising water takes nearly seven times as much energy as melting the same mass of ice, because the particles must be pulled completely apart rather than just loosened. US textbooks spell it vaporization.

The Specific Heat Capacity Calculator also does latent heat: choose a change of state and it works out E = mL for you.

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Specific Heat Capacity Calculator

Work out heating energy and specific and latent heat

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Specific latent heat examples

Example 1: melting ice

How much energy is needed to melt 0.5 kg of ice at 0 °C? The specific latent heat of fusion of water is 334 000 J/kg.

The energy needed is 167 000 J, or 167 kJ.

Example 2: finding the specific latent heat

It takes 1.13 MJ to turn 0.5 kg of water at 100 °C into steam. Find the specific latent heat of vaporisation of water.

Convert first: 1.13 MJ = 1 130 000 J.

Example 3: a kettle left boiling

A 2.2 kW kettle keeps water boiling for 3 minutes after it reaches 100 °C. What mass of water boils away? Use L = 2 260 000 J/kg.

Energy supplied = power × time = 2200 × 180 = 396 000 J. Then:

About 0.175 kg of water boils away, assuming all the energy goes into the water.

Heating curves: latent heat and specific heat capacity together

A heating curve shows temperature against energy supplied. The sloping parts are where the temperature rises, worked out with ΔE = mcΔθ from specific heat capacity. The flat parts are changes of state, worked out with E = mL.

Here x is the energy supplied in kJ and y is the temperature in °C. The flat section is much longer than the sloping ones: melting the ice takes far more energy than warming it by 10 °C or warming the water by 20 °C.

Example 4: a full heating curve

How much energy turns 0.2 kg of ice at −10 °C into water at 20 °C? Use c = 2100 J/kg °C for ice, c = 4200 J/kg °C for water, and L = 334 000 J/kg.

Work in three stages, one for each part of the graph:

• Warm the ice to 0 °C: 0.2 × 2100 × 10 = 4200 J
• Melt it: 0.2 × 334 000 = 66 800 J
• Warm the water to 20 °C: 0.2 × 4200 × 20 = 16 800 J

Total: 4200 + 66 800 + 16 800 = 87 800 J.

Example 5: condensing steam (A-level)

0.01 kg of steam at 100 °C is bubbled into 0.3 kg of water at 20 °C. All the steam condenses. Find the final temperature T, ignoring losses. Use L = 2 260 000 J/kg and c = 4200 J/kg °C.

The steam gives out energy in two stages: condensing, then cooling as water from 100 °C to T. The cold water takes in the same amount.

So T = 39.9 °C. Just 10 g of steam warms 300 g of water by about 20 °C, mostly through its latent heat. That is why steam burns are so much worse than burns from boiling water.

Here is a quick check.

Common mistakes

• Using a temperature change in E = mL. There is no temperature in this equation. It applies only at the melting or boiling point.
• Using the wrong latent heat. Melting and freezing use fusion; boiling and condensing use vaporisation.
• Missing a stage. Ice at −10 °C must warm to 0 °C before it melts. Split the problem into stages.
• Units of MJ or kJ. Convert to joules first: 2.26 MJ/kg is 2 260 000 J/kg.
• Saying the temperature rises during melting. It stays constant until the change of state is complete.
• Forgetting that energy is released. Freezing and condensing give out the same latent heat that melting and boiling take in.

Is the specific latent heat equation on the equation sheet?

Yes, on AQA's GCSE sheet and in AQA's A-level booklet, checked in November 2026. AQA's GCSE Physics equations sheet gives thermal energy for a change of state = mass × specific latent heat, E = mL, next to the specific heat capacity equation. AQA's A-level Physics data booklet gives Q = ml under thermal physics. Other boards publish their own sheets, so check yours. You still need to know the definition, and to explain in particle terms why the temperature stays constant.

Practise thermal physics in a short timed paper

GCSE Physics Higher and A-level Physics: Year 2 each give you a short timed paper every day, with 8 questions worth 25 to 35 marks across the course. To watch particles melt and boil, try the States of Matter Simulator.

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