Melting, Boiling and Specific Latent Heat

The specific latent heat content here is Extended (Supplement) only. The $E = mL$ calculation appears on Papers 2 and 4, not the Core papers. Examiners test it because it exposes a deep misconception: that adding energy always raises temperature. During melting and boiling, it does not.

Why does temperature stay constant during melting and boiling?

Energy supplied during a change of state does not increase the particles’ kinetic energy. Instead it increases their potential energy, working against the forces between particles to break the structure. Temperature measures average kinetic energy, so temperature stays constant until the change of state finishes. Pure ice melts at exactly 0 °C; pure water boils at 100 °C at standard atmospheric pressure.

A heating curve shows this as flat sections. Sloped sections mean temperature rise (use $\Delta E = mc\Delta\theta$); flat sections mean change of state (use $E = mL$).

What is specific latent heat and what is the equation?

Specific latent heat is the energy required per kilogram to change the state of a substance without any change in temperature. In words: $\text{energy} = \text{mass} \times \text{specific latent heat}$. In symbols: $E = mL$.

Quantity	Symbol	Unit
Energy transferred	$E$	joule (J)
Mass changed state	$m$	kilogram (kg)
Specific latent heat	$L$	J/kg

Two versions exist. Latent heat of fusion covers melting and freezing; for ice it is about 330 000 J/kg. Latent heat of vaporisation covers boiling and condensing; for water it is about 2 260 000 J/kg, nearly seven times larger, because vaporisation must separate the particles completely.

Boiling and evaporation differ too. Boiling happens at one fixed temperature, throughout the liquid, with bubbles. Evaporation happens at any temperature, only at the surface, as the most energetic particles escape. Escaping particles take above-average energy with them, so the remaining liquid cools. That is why sweating works.

Worked Exam Question

A freezer turns 0.20 kg of water at 0 °C into ice at 0 °C in 250 s. The specific latent heat of fusion of ice is 330 000 J/kg. Calculate (a) the energy removed from the water, and (b) the average rate of energy removal. [4]

Worked solution:

1. Equation: $E = mL$
2. Substitute: $E = 0.20 \times 330\,000$
3. (a) $E = 66\,000\ \text{J} = 66\ \text{kJ}$ (2 significant figures)
4. (b) Rate: $P = E \div t = 66\,000 \div 250 = 264\ \text{W} \approx 260\ \text{W}$

Mark scheme:

• M1: $E = mL$ stated or used
• A1: $66\,000\ \text{J}$ (unit required)
• M1: candidate’s energy $\div\ 250$
• A1: 264 W (accept 260 W; error carried forward allowed)

Common Mistakes

• Including a temperature change in $E = mL$. Latent heat questions involve no $\Delta\theta$. If the temperature changes as well, the question needs $mc\Delta\theta$ added as a separate step.
• Using the fusion value for boiling, or vice versa. Read which change of state the question describes before picking $L$.
• Saying particles “stop moving” on the flat part of a heating curve. They keep their kinetic energy; the added energy becomes potential energy.
• Describing evaporation as “the liquid boiling slowly”. Mark schemes want surface, any temperature, most energetic particles escape.
• Forgetting that condensing and freezing release the same energy that boiling and melting absorb.

Exam Technique Tip

For combined questions (ice at −10 °C to water at 20 °C) draw the heating curve first and label each segment. Each sloped segment is one $mc\Delta\theta$ calculation; each flat segment is one $mL$ calculation. Add the energies at the end. Marking your structure on a sketch graph prevents the classic error of merging steps, and it pairs naturally with graph-interpretation marks.

How This Is Examined

This is Extended territory: Paper 2 multiple choice and Paper 4 theory. Paper 4 loves three-step staircase calculations worth 5-6 marks, and 3-mark “explain why temperature stays constant” questions answered through kinetic versus potential energy. Core candidates only need the qualitative picture of melting, boiling and evaporation for Papers 1 and 3. On Papers 5 and 6, cooling-curve experiments with stearic acid or wax appear, testing whether you can read the flat section as the melting point. The staircase calculations follow one fixed recipe.