Daily Physicist

Today is the birthday of

• Sir James Hall, 4th Baronet (1761-1832).

Today’s Problem

Show that the efficiency of an engine that absorbs heat \(Q_H\) from a reservoir at temperature \(T_H\) and expels heat \(Q_c\) to a reservoir at temperature \(T_C\) must always be less than or equal to the efficiency of a Carnot heat engine.

Answer

The work performed by the engine is \(W=Q_H-Q_C\) and the efficiency of the engine is

\[e=\frac{W}{Q_H}=\frac{Q_H-Q_C}{Q_H}=1-\frac{Q_C}{Q_H}\]

The entropy lost by the hot reservoir is \(Q_H/T_H\). For heat to flow from the hot reservoir to the engine, the engine must be at a temperature less than \(T_H\). Call this temperature \(T_{EH}<T_H\). The entropy gained by the engine is then \(Q_H/T_{EH}>Q_H/T_H\). The entropy gained by the cold reservoir is \(Q_C/T_C\). For heat to flow from the engine to the cold reservoir, the engine must be at a temperature greater than \(T_C\). Call this temperature \(T_{EC}>T_C\). The entropy lost by the engine is then \(Q_C/T_{EC}<Q_C/T_C\) but since the engine gains no net entropy this must equal the entropy gained by the engine, \(Q_C/T_{EC}=Q_H/T_{EH}\). So we have

\[\frac{Q_H}{T_H} < \frac{Q_H}{T_{EH}}=\frac{Q_C}{T_{EC}} < \frac{Q_C}{T_C}\]

or

\[\frac{Q_H}{T_H} < \frac{Q_C}{T_C}\]

\[\frac{Q_C}{Q_H} > \frac{T_C}{T_H}\]

Therefore, for the efficiency we must have

\[e \le 1-\frac{T_C}{T_H}\]

where we have equality only for a Carnot engine.