How to Calculate Heat with the Specific Heat Formula

Calculates heat as Q = mass × specific heat × temperature change. Mass is in grams, specific heat in J/(g·K) (about 4.2 for water), the change in kelvins, and heat in joules.

Explanation

This works out the heat needed to warm something up. More mass, or a bigger temperature rise, means more heat.

Q = m c \Delta T

$Q$ is the heat, in joules (J)
$m$ is the mass, entered here in grams (g)
$c$ is the specific heat, in J/(g·K); water is about 4.2
$\Delta T$ is the temperature change, in kelvins (K)

Specific heat is the heat needed to raise one gram of a substance by one kelvin. Water's 4.2 J/(g·K) is high compared with most everyday materials, which is exactly why water is slow to heat up and slow to cool down.

Example

The defaults are 200 g of water, a specific heat of 4.2 J/(g·K) and a temperature change of 30 K.

Q = 200 \times 4.2 \times 30 = 25200\,\mathrm{J}

The heat is 25200 J. The calculator converts that as well: with 1 kcal = 4184 J, it comes to 25200 ÷ 4184 ≈ 6.02 kcal. It is the heat needed to take a 200 g cup of water from 20 °C to 50 °C.

Notes

The mass goes in grams, not kilograms, because the specific heat here is per gram. A kilogram of water is 1000 g. Enter kilograms by mistake and the answer comes out a thousand times too small.

$\Delta T$ is the change in temperature, not the temperature itself. Going from 20 °C to 50 °C means entering 30. A kelvin and a degree Celsius are the same size, so a temperature difference has the same number on either scale.

The formula covers heating and cooling with no change of state. While ice melts or water boils, heat keeps going in but the temperature stands still, because that heat goes into the change of state instead. So this will not tell you the heat needed to melt ice.

Mass and specific heat must be greater than zero. The temperature change may be negative, and then the heat comes out negative too: heat removed rather than added.