For melting or boiling at a constant temperature, the energy needed is Q = m × l, where l is the specific latent heat in J/kg. Whether latent heat calculations are required depends on your syllabus year and route, so check the current Cambridge IGCSE Physics 0625 syllabus page before relying on this lesson for your exam.
Latent heat questions often combine the two equations in a single problem, so the skill builds on using specific heat capacity and recognising a phase change.
How do I handle a question with more than one stage?
Split the process into stages and ask of each one: is the temperature changing, or is the state changing? Use m × c × Δθ for a temperature change and m × l for a change of state. Then add the stage energies.
- Sketch the stages as a simple list: for example, ice at 0 °C, melting, water warming to 30 °C.
- Write the equation beside each stage.
- Convert units (g to kg, kJ to J) before calculating anything.
- Calculate each stage separately and keep the units.
- Add the stages only if the question asks for a total, and state the answer with its unit.
Worked example
Take 0.40 kg of ice at 0 °C, which melts and is then warmed to 30 °C. (Invented example data.) Use specific latent heat of fusion of ice = 3.3 × 10⁵ J/kg and specific heat capacity of water = 4200 J/kg °C. Find the total energy.
Step 1, stages: melting at 0 °C (state change), then warming from 0 °C to 30 °C (temperature change).
Step 2, melting: Q₁ = m × l = 0.40 × 330 000 = 132 000 J.
Step 3, warming: Δθ = 30 − 0 = 30 °C. Q₂ = 0.40 × 4200 × 30. First 0.40 × 4200 = 1680. Then 1680 × 30 = 50 400 J.
Step 4, total: Q = 132 000 + 50 400 = 182 400 J, or about 1.8 × 10⁵ J.
Step 5, check: melting needs more energy than warming 30 °C here because 132 000 is greater than 50 400. That matches the idea that a change of state takes a lot of energy. Melting 0.40 kg uses as much energy as warming the same mass of water by about 79 °C, since 132 000 ÷ 1680 ≈ 78.6.
The mistake to watch for
A common slip is to use the temperature-change equation for the melting stage.
Mistaken answer: Q₁ = m × c × Δθ = 0.40 × 4200 × 0 = 0 J, so only the warming stage counts and the total is 50 400 J.
The temperature does not change during melting, so the equation gives zero. It is the wrong equation for that stage.
The correction is to use Q = m × l for the melting stage and Q = m × c × Δθ for the warming stage. The total is then 182 400 J. If an equation gives zero for a stage where energy is clearly being supplied, that is the signal that the stage needs a different equation.
Check yourself
Use l (vaporisation of water) = 2.3 × 10⁶ J/kg and l (fusion of ice) = 3.3 × 10⁵ J/kg.
1. How much energy boils away 0.20 kg of water already at 100 °C?
Show answer
Q = m × l = 0.20 × 2 300 000 = 460 000 J (4.6 × 10⁵ J).
2. A 2000 W heater melts 0.50 kg of ice at 0 °C with no energy lost. How long does it take?
Show answer
Q = 0.50 × 330 000 = 165 000 J. t = E ÷ P = 165 000 ÷ 2000 = 82.5 s.
3. It takes 69 000 J to melt 0.30 kg of a solid at its melting point. Calculate its specific latent heat of fusion.
Show answer
l = Q ÷ m = 69 000 ÷ 0.30 = 230 000 J/kg (2.3 × 10⁵ J/kg).
Where this leads next
Real experiments rarely match these ideal numbers, which is the subject of identifying energy losses as a limitation of a model. The heat calculations practice set mixes both equations.
Multi-stage problems reward a teacher who can watch you decide which equation fits which stage. That is a core part of online one-to-one Physics tuition.