Temperature and Heat

Heat as Energy Transfer

warming a body

Heat is energy transferred because of a temperature difference, not a substance stored inside an object. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Heat changes a thermal state

The body warms from 300 kelvin to 303 kelvin, a change of 3 kelvin.

ΔT=3 K\Delta T = 3\ \text{K}
Heating state changeThe thermometers and heat bar are computed from quantities.30 Jheat300 Kstart303 Kend

Use heat capacity

This body needs 10 joules for each kelvin of warming.

C=10 J/KC = 10\ \text{J/K}

Compute transferred energy

Multiply 10 joules per kelvin by 3 kelvin. The heat transfer is 30 joules.

Q=CΔT=10 J/K⋅3 K=30 JQ = C\Delta T = 10\ \text{J/K}\cdot 3\ \text{K} = 30\ \text{J}
Heating state changeThe thermometers and heat bar are computed from quantities.30 Jheat300 Kstart303 Kend

More heat capacity needs more heat

Hold the temperature change at 3 kelvin. The diagram shows the middle row; larger heat capacity means more joules for the same warming.

CΔTQ5 J/K3 K15 J10 J/K3 K30 J15 J/K3 K45 J\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}
Heating state changeThe middle table row is the checked diagram.30 Jheat300 Kstart303 Kend

More warming needs more heat

Now hold the heat capacity at 10 joules per kelvin. A larger temperature change requires more transferred energy.

CΔTQ10 J/K1 K10 J10 J/K3 K30 J10 J/K5 K50 J\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}
Heating state changeThe middle table row is the checked diagram.30 Jheat300 Kstart303 Kend