About this calculator
How much energy does it take to heat a kettle of water? The answer is Q = m c ΔT: heat energy equals mass times specific heat capacity times the temperature change.
Fill in any three of the four boxes and leave the one you want to find empty. Masses are in kilograms, specific heat in joules per kilogram per kelvin, and heat in joules.
Worked examples
Real numbers, worked out by the same calculator. Press “Use these numbers” to try one above.
Heat 2 kg of water by 10 °C
- Heat energy (Q)
- 83,720 J
- Mass (m)
- 2 kg
- Specific heat capacity (c)
- 4,186 J/(kg·K)
- Temperature change (ΔT)
- 10 K (same size as °C)
Heat energy (Q) works out at 83,720 J.
Show the working
- Heat energy = mass × specific heat capacity × temperature change, Q = m c ΔT.
- Heat energy (Q) = 2 × 4,186 × 10 = 83,720 J.
Energy 83,720 J gives 2 kg of water what rise?
- Temperature change (ΔT)
- 10 K (same size as °C)
- Heat energy (Q)
- 83,720 J
- Mass (m)
- 2 kg
- Specific heat capacity (c)
- 4,186 J/(kg·K)
Temperature change (ΔT) works out at 10 K (same size as °C).
Show the working
- Heat energy = mass × specific heat capacity × temperature change, Q = m c ΔT.
- Temperature change (ΔT) = Heat energy (Q) ÷ (2 × 4,186) = 83,720 ÷ 8,372 = 10 K (same size as °C).
Which material? 900 J heats 0.5 kg by 2 K
- Specific heat capacity (c)
- 900 J/(kg·K)
- Heat energy (Q)
- 900 J
- Mass (m)
- 0.5 kg
- Temperature change (ΔT)
- 2 K (same size as °C)
Specific heat capacity (c) works out at 900 J/(kg·K).
Show the working
- Heat energy = mass × specific heat capacity × temperature change, Q = m c ΔT.
- Specific heat capacity (c) = Heat energy (Q) ÷ (0.5 × 2) = 900 ÷ 1 = 900 J/(kg·K).
How much water does 500 kJ heat by 40 °C?
- Mass (m)
- 2.986144 kg
- Heat energy (Q)
- 500,000 J
- Specific heat capacity (c)
- 4,186 J/(kg·K)
- Temperature change (ΔT)
- 40 K (same size as °C)
Mass (m) works out at 2.986144 kg.
Show the working
- Heat energy = mass × specific heat capacity × temperature change, Q = m c ΔT.
- Mass (m) = Heat energy (Q) ÷ (4,186 × 40) = 500,000 ÷ 167,440 = 2.986144 kg.
Typical specific heat capacities
- Water: 4,186 J/(kg·K)
- Ice: about 2,100
- Aluminium: about 900
- Iron or steel: about 450
- Copper: about 385
- Air: about 1,005
A high value means the material needs a lot of energy for each degree of warming. Water's value is unusually high, which is why it is used for heating systems and why the sea warms slowly.
Worked example
Heating 2 kg of water by 10 °C: Q = 2 × 4,186 × 10 = 83,720 J, or about 83.7 kJ. In energy-bill terms that is only 0.023 kWh, because 1 kWh is 3.6 million joules.
Limits of the formula
The formula ignores heat lost to the surroundings and assumes the material does not change state. Melting ice or boiling water needs extra energy (latent heat) that Q = mcΔT does not cover. A temperature change in °C is the same size as in kelvin, so you can use either.
Frequently asked questions
What is specific heat capacity?
The energy needed to raise 1 kilogram of a material by 1 degree. Water needs 4,186 joules.
Is ΔT in Celsius or Kelvin?
Either. A change of 1 °C equals a change of 1 K, so the number is the same.
What is ΔT?
The Greek letter delta means 'change in'. ΔT is the final temperature minus the starting temperature.
How do I find the specific heat of an unknown material?
Leave c blank and enter the energy, mass and temperature rise from your experiment. The calculator returns c.
Does it include heat loss?
No. Real kettles and heaters lose some heat, so you need more energy than the ideal figure.
Formulas tested against hand-worked answers. Last reviewed 29 September 2026. These calculators do arithmetic only; they are not financial, tax or legal advice.