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IGCSE Physics, Cambridge 0625, Malaysia

Specific Latent Heat of Vaporisation

Symbol: Lv · Unit: J/kg

Definition of specific latent heat of vaporisation for Cambridge IGCSE Physics 0625, with the formula and explanation of the flat section on a heating curve at boiling point.

The specific latent heat of vaporisation is the energy required to change 1 kg of a substance from liquid to gas at its boiling point, without any change in temperature.

The formula

Q = m Lv

where Q is the energy in joules (J), m is the mass in kilograms (kg), and Lv is the specific latent heat of vaporisation in joules per kilogram (J/kg).

Why temperature stays constant during boiling

At the boiling point, the energy supplied breaks the remaining bonds between particles in the liquid. The particles escape from the surface and become gas. This increases their potential energy (separation) but not their kinetic energy, so the temperature does not rise.

The temperature stays at the boiling point until all the liquid has boiled away.

On a heating curve

A heating curve shows a second flat section at the boiling point. This flat section is usually longer than the one at the melting point because the specific latent heat of vaporisation is larger.

Example

The specific latent heat of vaporisation of water is 2,260,000 J/kg.

To boil 0.2 kg of water already at 100 degrees Celsius: Q = 0.2 x 2,260,000 = 452,000 J = 452 kJ

Compare this with melting 0.2 kg of ice: Q = 0.2 x 334,000 = 66,800 J = 66.8 kJ

Boiling requires about 6.8 times more energy than melting for the same mass of water.

Why vaporisation needs more energy than fusion

In a liquid, particles are close together but can move past each other. To become a gas, particles must be completely separated against intermolecular forces. This requires more energy than simply loosening the solid structure during melting.

Steam burns

Steam at 100 degrees causes worse burns than water at 100 degrees because the steam releases its latent heat of vaporisation into the skin as it condenses, delivering a large amount of energy on top of the thermal energy from the high temperature.

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A 0625 specialist can explain this concept using the student's own questions and show how it appears in exam papers.