The same energy that warms 1 kg of copper by 10 °C only warms 1 kg of water by about 1 °C. Water has a much bigger specific heat capacity: the energy needed to raise the temperature of 1 kg of a substance by 1 °C.

change in thermal energy = mass × specific heat capacity × temperature change
ΔE = m c Δθ

Energy in J, mass in kg, c in J/kg°C, temperature change Δθ in °C. This one is on the equation sheet for AQA, Edexcel and OCR, and the paper gives you the value of c.

Some values: water 4200, olive oil about 2000, aluminium 900, concrete 880, iron 450, copper 385, all in J/kg°C.

1. A kettle heats 1.2 kg of water from 15 °C to 100 °C. c = 4200 J/kg°C.
2. Δθ = 100 − 15 = 85 °C
3. ΔE = 1.2 × 4200 × 85 = 428 400 J
4. = 428.4 kJ

Rearranging works the same way as any three-letter equation: Δθ = ΔE ÷ (m c), m = ΔE ÷ (c Δθ), c = ΔE ÷ (m Δθ).

Δθ is the change: final minus starting temperature. Putting in the final temperature is the classic lost mark.

You find c for a metal block that has two holes: one for an electric heater, one for a thermometer.

1. Weigh the block (balance), then wrap it in insulation to cut energy losses
2. Heater in one hole; thermometer in the other, with a drop of water for good contact
3. Record the starting temperature, switch on, start the stopwatch
4. Read the energy supplied (joulemeter, or V × I × t) and the temperature every minute for 10 minutes
5. c = ΔE ÷ (m Δθ), or plot temperature against energy and use the gradient

Independent variable: the energy supplied (or time). Dependent variable: the temperature. Control variables: the mass of the block, the heater's power.

With a heater of steady power P, the temperature-time graph is a straight line. Each second, P = m c × gradient, so c = P ÷ (m × gradient).

Your value almost always comes out higher than the real one. Some of the energy heats the air and the bench instead of the block, so the temperature rises less than it should.