Waste Heat - Thermodynamics and Statistical Mechanics Step by Step

Waste heat is the heat input an engine fails to turn into work, Q_waste = Q_in - W_out, the energy dumped to the cold reservoir as the first-law budget closes. The arithmetic is exact only for the stated heat input, work, and reservoir temperatures. Real systems add heat losses, non-equilibrium paths, calibration error, material variation, finite reservoirs, and measurement uncertainty.

Waste heat is the part of the heat input that must be rejected rather than converted to useful work. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Not all heat becomes work

The heat input is 30 joules, but useful work is only 12 joules.

Q_{\text{in}} = 30\ \text{J}\qquad W_{\text{out}} = 12\ \text{J}

Find the waste heat

Waste heat is the input heat minus the useful work.

Q_{\text{waste}} = Q_{\text{in}} - W_{\text{out}}

Compute the rejected heat

Subtract 12 joules from 30 joules. The waste heat is 18 joules.

Q_{\text{waste}} = 30\ \text{J} - 12\ \text{J} = 18\ \text{J}

More work leaves less waste at fixed input

Hold the heat input at 30 joules. The diagram shows the middle row; converting more to work leaves less waste heat.

\begin{array}{c|c|c}Q_{\text{in}}&W_{\text{out}}&Q_{\text{waste}}\\30\ \text{J}&6\ \text{J}&24\ \text{J}\\30\ \text{J}&12\ \text{J}&18\ \text{J}\\30\ \text{J}&18\ \text{J}&12\ \text{J}\\\end{array}

More input leaves more waste if work is fixed

Now hold useful work at 12 joules. A larger heat input must send the extra energy somewhere, so the waste row grows.

\begin{array}{c|c|c}Q_{\text{in}}&W_{\text{out}}&Q_{\text{waste}}\\24\ \text{J}&12\ \text{J}&12\ \text{J}\\30\ \text{J}&12\ \text{J}&18\ \text{J}\\36\ \text{J}&12\ \text{J}&24\ \text{J}\\\end{array}