Heat Capacity - Thermodynamics and Statistical Mechanics Step by Step

Heat capacity C is the energy needed to raise a body's temperature by one kelvin, written C = Q / Delta T. It is a property of the whole object, not of the material alone, so a bigger sample of the same stuff has a larger C. The ratio is exact only for the stated heat input and temperature change; real measurements lose heat to surroundings and depend on thermometer calibration.

Heat capacity says how many joules are needed for each kelvin of temperature change. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Read heat capacity

A heat capacity of 10 joules per kelvin means each kelvin of warming needs 10 joules.

C = 10\ \text{J/K}

Multiply by the temperature change

For a 3 kelvin change, use 3 copies of that energy.

Q = C\Delta T

Compute the heat

The transferred heat is 10 times 3, or 30 joules.

Q = 10\ \text{J/K}\cdot 3\ \text{K} = 30\ \text{J}

Capacity scales the heat directly

Hold the temperature change at 3 kelvin. The diagram shows the middle row; more joules per kelvin means more total heat.

\begin{array}{c|c|c}C&\Delta T&Q\\5\ \text{J/K}&3\ \text{K}&15\ \text{J}\\10\ \text{J/K}&3\ \text{K}&30\ \text{J}\\15\ \text{J/K}&3\ \text{K}&45\ \text{J}\\\end{array}

Temperature change also scales heat directly

Now hold the heat capacity at 10 joules per kelvin. More kelvin of warming needs more joules.

\begin{array}{c|c|c}C&\Delta T&Q\\10\ \text{J/K}&1\ \text{K}&10\ \text{J}\\10\ \text{J/K}&3\ \text{K}&30\ \text{J}\\10\ \text{J/K}&5\ \text{K}&50\ \text{J}\\\end{array}