Specific heat capacity: formula, examples and the practical

1. Energy: heating 0.5 kg of water from 20 °C to 60 °C. $\Delta\theta = 40$ °C, so $\Delta E = 0.5 \times 4200 \times 40 = 84{,}000$ J.
2. Specific heat capacity: 27,000 J heats a 1.5 kg block by 20 °C. $c = 27000 \div (1.5 \times 20) = 900$ J/kg °C.
3. Temperature rise: 12,600 J goes into 0.3 kg of water. $\Delta\theta = 12600 \div (0.3 \times 4200) = 10$ °C.

Water's high specific heat capacity (about 4,200 J/kg °C) means it takes a lot of energy to heat up and releases a lot as it cools. That's why it's used in central heating and car cooling systems, and why coastal places have milder temperatures than inland ones. Metals have much lower values: aluminium is about 900 J/kg °C and copper about 390 J/kg °C.

1. Measure the mass of a metal block with holes for a heater and a thermometer.
2. Record the starting temperature.
3. Switch on the heater and record the energy supplied (with a joulemeter, or power × time).
4. Record the final temperature, then calculate $c = \dfrac{\Delta E}{m \Delta\theta}$.

Why the result is usually too high: some energy heats the air and the bench, not the block. The energy supplied is larger than the energy the block gains, so the calculated $c$ is bigger than the true value. Wrapping the block in insulation and using a thin layer of oil in the thermometer hole for good contact improve accuracy.

Heat capacity, also called thermal capacity in some syllabuses and older textbooks, is the energy needed to raise the temperature of a whole object by 1 °C, in J/°C. Specific heat capacity is the same idea per kilogram, so the two are linked by the mass:

Example: 2 kg of water has a heat capacity of $2 \times 4200 = 8400$ J/°C, so it needs 8,400 J for each degree it warms. A question that gives "thermal capacity" in J/°C wants $\Delta E = \text{heat capacity} \times \Delta\theta$, with no mass in the equation.

When a substance melts or boils, energy is transferred but its temperature doesn't change until the change of state is complete. The energy needed to change the state of 1 kg is the specific latent heat, in J/kg (in the AQA specification, $E = mL$). On a heating curve, these are the flat sections.

• Using the final temperature instead of the change.
• Using grams instead of kilograms.
• Confusing specific heat capacity (temperature change) with specific latent heat (change of state).

1. How much energy is needed to heat 0.2 kg of water by 50 °C? (c = 4,200 J/kg °C)
2. 9,000 J is transferred to 1 kg of aluminium (c = 900 J/kg °C). What is the temperature rise?
3. 4,200 J heats 0.1 kg of water. By how much does its temperature rise?

Answers: 42,000 J; 10 °C; 10 °C.