Specific heat capacity links the energy transferred to a material with its mass and its temperature change: Q = m × c × Δθ. It appears in IGCSE Physics whenever a question heats or cools a known amount of water, metal or another material and asks for energy, mass, temperature change or c.
The hard part is rarely the algebra. It is using kg, J and °C together and not mixing them with grams, kilojoules or final temperatures.
How does the equation work, step by step?
Q is the energy transferred in joules (J), m is mass in kilograms (kg), c is specific heat capacity in J/kg °C, and Δθ is the temperature change in °C.
- List what you know and write each with its unit.
- Convert units now: g to kg, kJ to J, minutes to seconds if power is involved.
- Find the temperature change by subtracting: Δθ = final − initial.
- Rearrange if needed: Δθ = Q ÷ (m × c), m = Q ÷ (c × Δθ), or c = Q ÷ (m × Δθ).
- Calculate, then check the size: does the answer look sensible for the material and the amount of energy?
A useful habit is to work out m × c first. It is the energy needed for each 1 °C of change for this particular object, and it simplifies everything after it.
Worked example
A pan of 1.2 kg of water is heated from 18 °C to 38 °C. (Invented example data.) The specific heat capacity of water is 4200 J/kg °C. How much energy does the water gain?
Step 1, list: m = 1.2 kg, c = 4200 J/kg °C, initial 18 °C, final 38 °C.
Step 2, units: mass is already in kg.
Step 3, temperature change: Δθ = 38 − 18 = 20 °C.
Step 4, calculate: Q = 1.2 × 4200 × 20. First 1.2 × 4200 = 5040 J/°C. Then 5040 × 20 = 100 800 J.
Step 5, answer: Q = 100 800 J, or about 101 kJ.
Check in the other direction: 100 800 ÷ (1.2 × 4200) = 100 800 ÷ 5040 = 20 °C, which returns the original temperature change.
The mistake to watch for
A common slip is to use the mass in grams because that is how the question gave it.
Mistaken answer: Q = 1200 × 4200 × 20 = 100 800 000 J
The student typed 1200 g straight into an equation that needs kg. The answer is exactly 1000 times too big.
The correction is to convert before anything else: 1200 g = 1.2 kg. A quick size check catches it too. A pan of water does not need a hundred million joules to warm by 20 °C.
Check yourself
Give each answer with its unit.
1. How much energy is needed to raise 2.0 kg of water from 25 °C to 35 °C? (c = 4200 J/kg °C)
Show answer
Δθ = 35 − 25 = 10 °C. Q = 2.0 × 4200 × 10 = 84 000 J (84 kJ).
2. A 250 g aluminium block (c = 900 J/kg °C) absorbs 18 000 J. By how much does its temperature rise?
Show answer
m = 250 g = 0.250 kg. m × c = 0.250 × 900 = 225 J/°C. Δθ = 18 000 ÷ 225 = 80 °C.
3. A 0.40 kg piece of metal needs 12 kJ to rise by 15 °C. Calculate its specific heat capacity.
Show answer
12 kJ = 12 000 J. c = Q ÷ (m × Δθ) = 12 000 ÷ (0.40 × 15) = 12 000 ÷ 6.0 = 2000 J/kg °C.
Where this leads next
Once the equation is automatic, move on to interpreting an energy-temperature slope, which reads the same relationship from a graph. The whole topic is brought together in the heat calculations practice set.
Some students follow each step here but still slip on units when a longer question mixes several quantities. Our teachers look for exactly that pattern in online one-to-one Physics tuition.