Today is the birthday of
- Edward Teller (1908-2003).
Today’s Problem
For a Carnot heat engine operating between a hot reservoir at temperature \(T_H\) and a cold reservoir at temperature \(T_C\) show that efficiency of the engine is equal to
\[\frac{W}{Q_H} = \frac{T_H-T_C}{T_H} = 1-\frac{T_C}{T_H}\]
where \(W\) is the work performed by the engine and \(Q_H\) is the heat absorbed from the hot reservoir.
Answer
In a Carnot heat engine there is no net change in entropy. This means the entropy gained from the hot reservoir must equal the entropy expelled to the cold reservoir or
\[\frac{Q_H}{T_H}=\frac{Q_C}{T_C}=\Delta S\]
There is also no net change in internal energy , \(\Delta U=0\), so the work performed by the engine is equal to the difference between the heat absorbed from the hot reservoir and the heat expelled to the cold reservoir or \(W=Q_H-Q_C=(T_H-T_C)\Delta S=(T_H-T_C)Q_H/T_H\) so we arrive at
\[\frac{W}{Q_H} = \frac{T_H-T_C}{T_H} = 1-\frac{T_C}{T_H}\]
© 2026 Stefan Hollos and Richard Hollos