B.4.12 (HL)—Heat engine efficiency

Efficiency

For a heat engine,

$$\eta=\frac{W_{\mathrm{useful}}}{Q_H}=1-\frac{Q_C}{Q_H}$$

where $$Q_H$$ is energy absorbed from the hot reservoir and $$Q_C$$ is energy rejected to the cold reservoir.

Use a complete cycle

Over one cycle, $$\Delta U=0$$, so net work is related to net thermal transfer. On a PV diagram, useful net work is the signed enclosed area; input energy is the heat taken from the hot reservoir.

Sanity checks

Efficiency is dimensionless and lies between 0 and 1 for a physical engine. Use the same energy units for numerator and denominator and allow method marks by showing the substituted values.

Worked example from local Question Bank row 28983

An engine produces $300\,\mathrm{J}$ of work and rejects $900\,\mathrm{J}$ per cycle. Energy conservation gives $Q_H=W+Q_C=1200\,\mathrm{J}$, so

$$\eta=\frac{W}{Q_H}=\frac{300}{1200}=0.25=25\%$$

The rejected energy is part of the input bookkeeping, not the denominator by itself.

Common trap

Do not divide useful work by rejected heat. The denominator is the energy input from the hot reservoir.