Core practical: specific heat capacity and melting ice

Core practical 14.11 investigates the properties of water: its specific heat capacity, and what happens to the temperature as ice melts.

Practise this lesson free (6 questions)

Measuring specific heat capacity

1. Put a known mass of water in an insulated beaker.
2. Heat it with an electric immersion heater. Measure the energy supplied with a joulemeter, or use a voltmeter, ammeter and stopwatch: E = I × V × t.
3. Record the temperature rise with a thermometer, stirring the water.
4. c = ΔQ ÷ (m × Δθ)
Insulate the beaker and use a lid to reduce energy lost to the surroundings. The measured value is usually a bit higher than the true value (4200 J/kg °C), because some energy heats the surroundings, not the water.

Melting ice

1. Put crushed ice (from a freezer, below 0 °C) in a beaker with a thermometer.
2. Heat gently (for example in a warm water bath), stirring, and record the temperature every 30 seconds.
3. Plot temperature against time.
The temperature rises to 0 °C, then stays flat while the ice melts, then rises again once all the ice has melted.

Explaining the results

Heating a substance increases its internal energy. This either raises its temperature or changes its state, not both at once.
During melting, the energy breaks the bonds between particles (increasing their potential energy), not their kinetic energy, so the temperature stays constant. This energy is the latent heat.
Worked example

A 12 V heater with a current of 4.0 A heats 0.50 kg of water for 600 s. The temperature rises by 14 °C. Calculate the specific heat capacity.

  1. E = I × V × t = 4.0 × 12 × 600 = 28 800 J
  2. c = 28 800 ÷ (0.50 × 14)
  3. = 4114 J/kg °C (to the nearest whole number)

Answer: About 4100 J/kg °C

Key idea

Measure energy in (E = IVt or a joulemeter), mass and temperature rise, then c = ΔQ ÷ (mΔθ). Insulate to reduce losses. When ice melts, the temperature stays at 0 °C because the energy breaks bonds instead of raising the temperature.

Key words
Specific heat capacity
The energy needed to raise the temperature of 1 kg of a substance by 1 °C. Unit J/kg °C.
Temperature
K = °C + 273. A change of 1 °C is the same as a change of 1 K.
Time
1 hour = 3600 s. Convert minutes and hours to seconds before using P = E ÷ t or Q = I × t.
Kinetic energy
Energy in the store of a moving object: KE = ½ × m × v^{2}.

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Quick check

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Extra practice

2 more questions

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  1. A heater runs at 6.0 V and 2.0 A for 600 s. How much energy does it transfer?

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    Answer: 7200 J

    E = V × I × t = 6.0 × 2.0 × 600 = 7200 J

  2. 12 600 J raises the temperature of 0.50 kg of water by 6 °C. Calculate the specific heat capacity in J/kg °C.

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    Answer: 4200 J/kg °C

    c = E ÷ (m × Δθ) = 12 600 ÷ (0.50 × 6) = 4200 J/kg °C

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Sources and references

Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.

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