Specific Heat

Specific Heat Equation

Q = mc Delta T

Specific heat separates the material property from the mass and the 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

Use mass and specific heat

The sample has mass 2 kilograms. Its specific heat is 5 joules per kilogram per kelvin.

m=2 kgc=5 J/(kg K)m = 2\ \text{kg}\qquad c = 5\ \text{J/(kg K)}
Specific heat caseThe heat bar is bound to mass, specific heat, and temperature change.30 Jheat300 Kstart303 Kend

Specific heat scales by mass

Specific heat is per kilogram, so multiply by the mass and by the temperature change.

Q=mcΔTQ = mc\Delta T

Compute the heat

Multiply 2 kilograms by 5 joules per kilogram per kelvin by 3 kelvin. The result is 30 joules.

Q=2 kg⋅5 J/(kg K)⋅3 K=30 JQ = 2\ \text{kg}\cdot 5\ \text{J/(kg K)}\cdot 3\ \text{K} = 30\ \text{J}
Specific heat caseThe heat bar is bound to mass, specific heat, and temperature change.30 Jheat300 Kstart303 Kend

Mass is one direct factor

Hold the material and temperature change fixed. The diagram shows the middle row; more kilograms require more heat.

mcΔTQ1 kg5 J/(kg K)3 K15 J2 kg5 J/(kg K)3 K30 J4 kg5 J/(kg K)3 K60 J\begin{array}{c|c|c|c}m&c&\Delta T&Q\\1\ \text{kg}&5\ \text{J/(kg K)}&3\ \text{K}&15\ \text{J}\\2\ \text{kg}&5\ \text{J/(kg K)}&3\ \text{K}&30\ \text{J}\\4\ \text{kg}&5\ \text{J/(kg K)}&3\ \text{K}&60\ \text{J}\\\end{array}
Specific heat caseThe middle table row is the checked diagram.30 Jheat300 Kstart303 Kend

Material choice changes joules per kilogram

Now hold mass and warming fixed. Higher specific heat means each kilogram needs more joules for the same temperature change.

mcΔTQ2 kg3 J/(kg K)3 K18 J2 kg5 J/(kg K)3 K30 J2 kg7 J/(kg K)3 K42 J\begin{array}{c|c|c|c}m&c&\Delta T&Q\\2\ \text{kg}&3\ \text{J/(kg K)}&3\ \text{K}&18\ \text{J}\\2\ \text{kg}&5\ \text{J/(kg K)}&3\ \text{K}&30\ \text{J}\\2\ \text{kg}&7\ \text{J/(kg K)}&3\ \text{K}&42\ \text{J}\\\end{array}
Specific heat caseThe middle table row is the checked diagram.30 Jheat300 Kstart303 Kend

Temperature change is the third direct factor

Finally hold mass and material fixed. More kelvin of warming requires more transferred energy.

mcΔTQ2 kg5 J/(kg K)1 K10 J2 kg5 J/(kg K)3 K30 J2 kg5 J/(kg K)5 K50 J\begin{array}{c|c|c|c}m&c&\Delta T&Q\\2\ \text{kg}&5\ \text{J/(kg K)}&1\ \text{K}&10\ \text{J}\\2\ \text{kg}&5\ \text{J/(kg K)}&3\ \text{K}&30\ \text{J}\\2\ \text{kg}&5\ \text{J/(kg K)}&5\ \text{K}&50\ \text{J}\\\end{array}
Specific heat caseThe middle table row is the checked diagram.30 Jheat300 Kstart303 Kend