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Physics · O-Level · Exam Technique · Thermal Physics

How to Solve Specific Heat Capacity and Latent Heat Questions

2 September 2026 · Dojo Education · 4 min read

How to Solve Specific Heat Capacity and Latent Heat Questions

Thermal properties of matter is one of the most predictable topics in the Pure Physics (6091) paper. Almost every year there is a calculation involving specific heat capacity, a heating or cooling curve to interpret, or a structured question on melting and boiling. The good news is that the whole topic runs on two equations and a very small set of exam phrases. Once you drill those, these become some of the easiest marks on Paper 2.

The Two Equations You Must Never Confuse

Specific heat capacity

E = mcΔθ

Use this whenever the temperature is changing.

Latent heat

E = mL

Use this whenever there is a change of state at constant temperature.

The single biggest error we see in marking practice papers is students putting a Δθ into a melting stage. If the thermometer reading is not moving, there is no Δθ, so mcΔθ does not apply.

Definitions the mark scheme wants

Examiners are strict here. Learn these word for word:

The phrase "without a change in temperature" is often worth its own mark. Do not drop it.

Bringing In Electrical Energy

Most exam calculations supply energy with an immersion heater, so you combine:

E = Pt = VIt

Then set that equal to mcΔθ or mL.

Worked example

A 60 W heater is used to heat 0.50 kg of water from 25 °C to 45 °C in 8.0 minutes. Find the specific heat capacity obtained from this experiment.

  1. Energy supplied: E = Pt = 60 × (8.0 × 60) = 28 800 J
  2. Δθ = 45 − 25 = 20 °C
  3. c = E / (mΔθ) = 28 800 / (0.50 × 20) = 2880 J/(kg·°C)

Then the follow-up question: why is this larger than the accepted value of 4200 J/(kg·°C)? Wait, it is smaller. Notice the trap. If some thermal energy is lost to the surroundings, less energy actually reaches the water, so the calculated value of c comes out larger than the true value, because you assumed all 28 800 J went into the water. Always reason it through rather than memorising a direction.

In this example the number is below 4200, which tells you the question is testing whether you can spot that the heater may not have been on for the full time, or that the data is fictional. In the real exam, state your reasoning clearly: "Thermal energy is lost to the surroundings, so the energy absorbed by the water is less than Pt, and the calculated c is therefore higher than the true value."

Multi-Stage Problems

When ice at −10 °C becomes steam at 110 °C, break it into stages and add:

  1. Ice warming: mc(ice)Δθ
  2. Ice melting at 0 °C: mL(fusion)
  3. Water warming: mc(water)Δθ
  4. Water boiling at 100 °C: mL(vaporisation)
  5. Steam warming: mc(steam)Δθ

Write each stage on its own line with its own working. Method marks are awarded per stage, so even a slip in stage 3 will not cost you stages 1, 2, 4 and 5.

Reading Heating and Cooling Curves

A temperature against time graph is a gift if you know what to say:

The explanation mark almost always hinges on the words "potential energy" and "separation between molecules". Saying only "the energy is used to melt it" is a circular answer and scores nothing.

Quick Checklist Before You Move On

Drill five past-paper questions across the 2015 to 2023 papers using this structure and you will find the topic becomes almost mechanical.

Want a tutor who sat these papers recently to walk you through a multi-stage question live? Start a free trial class with Dojo on WhatsApp.

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