Specific heat calculator: Q = mcΔT

The energy needed to change a substance's temperature is Q = mcΔT: mass times specific heat capacity times temperature change. Specific heat capacity, c, is the energy to raise 1 kg by 1 °C (or 1 K). Heating 1.5 kg of water (c = 4200 J/kg °C) by 80 °C takes 1.5 × 4200 × 80 = 504,000 J. This specific heat calculator solves Q = mcΔT for any value, handles melting and boiling with Q = mL, and finds the final temperature when two substances mix.

It opens on that water example. Open the full specific heat calculator over this page to try your own.

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Specific Heat Capacity Calculator

Work out heating energy and specific and latent heat

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How to use the specific heat calculator

1. Under Calculate, choose Heating, Latent heat or Mixing.
2. In Heating, choose what to find: Energy, Mass, Capacity (specific heat capacity) or Change in T.
3. Type the values you know. For the specific heat capacity you can pick a material, such as water, ice, aluminium, copper or iron, instead of typing a number.
4. Read the answer and steps. Energy mode also gives the object's heat capacity and how long a 2 kW kettle would take to supply that energy.
5. To practise, press Random problem in the Practice section.

Specific heat formula

Q is in joules, m in kilograms, c in J/kg °C (the same as J/kg K) and ΔT is the change in temperature. A change of 1 °C is the same size as a change of 1 K, so ΔT is the same in either. Water's specific heat is high: OpenStax lists 4186 J/kg °C, and school questions usually round it to 4200, which is the value the calculator uses.

Worked examples you can check in the calculator

Example 1: heating water

Q = 1.5 × 4200 × 80 = 504,000 J (504 kJ). The calculator adds that a 2 kW kettle would supply this in 252 s, ignoring heat lost to the surroundings.

Example 2: finding the specific heat capacity

27,000 J raises a 2 kg block by 30 °C. Choose Capacity: c = 27,000 ÷ (2 × 30) = 450 J/kg °C, the value the calculator lists for iron. Measuring c like this, with a heater and a thermometer, is a common school experiment.

Example 3: finding the temperature change

4812.5 J is supplied to 0.25 kg of copper (c = 385 J/kg °C). Choose Change in T: ΔT = 4812.5 ÷ (0.25 × 385) = 50 °C. The same energy would warm 0.25 kg of water by under 5 °C, because water's specific heat is so much bigger.

Example 4: melting ice (latent heat)

Changing state takes energy without changing temperature: Q = mL, where L is the specific latent heat. Melting 0.5 kg of ice at 0 °C, with L = 334,000 J/kg, takes 0.5 × 334,000 = 167,000 J. Boiling the same mass of water away takes 0.5 × 2,260,000 = 1,130,000 J, almost seven times as much. Choose Latent heat and pick the material.

Example 5: mixing two substances

A 0.5 kg aluminium block at 100 °C is dropped into 1 kg of water at 20 °C. The heat lost by the block equals the heat gained by the water, so the final temperature is:

The mixture settles at about 27.7 °C, assuming no heat escapes. Choose Mixing to check.

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Common mistakes

• Using a temperature instead of a change. ΔT is final minus starting temperature.
• Grams instead of kilograms. With c in J/kg °C, mass must be in kg, or choose grams as the unit.
• Using Q = mcΔT during a change of state. While ice melts or water boils, the temperature stays the same: use Q = mL.
• Mixing up specific heat capacity and heat capacity. Heat capacity is for a whole object, mc, in J/°C.
• Forgetting heat losses. Real experiments lose energy to the surroundings, so measured values of c often come out too high.

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