Carnot Engine Energy Accounting

Question

A Carnot engine working between 500 K and 300 K absorbs 1000 J of heat from the hot reservoir in each cycle. How much heat does it reject to the cold reservoir per cycle?

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For a reversible Carnot cycle the heats exchanged with the two reservoirs are tied to their absolute temperatures by $\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H}$, a consequence of the entropy change being zero over the reversible cycle. Solving for the rejected heat gives $Q_C = Q_H\dfrac{T_C}{T_H} = 1000	imes\dfrac{300}{500} = 600	ext{ J}$. The value 400 J is actually the net work $W = Q_H - Q_C = 1000 - 600$, not the rejected heat. The value 500 J would require equal temperatures or a wrong ratio. The value 300 J mistakes the cold temperature in kelvin for the heat in joules. We can cross-check using efficiency: $\eta = 1 - T_C/T_H = 1 - 300/500 = 0.4$, so the work is $0.4	imes1000 = 400	ext{ J}$ and the rejected heat is $1000 - 400 = 600	ext{ J}$, matching exactly. This entropy balance, expressed as $Q_H/T_H = Q_C/T_C$, states that the entropy gained by the gas from the hot reservoir is exactly returned to the cold reservoir, so the total entropy change over the reversible cycle is zero. Any real irreversible engine between the same reservoirs would reject more than 600 J and thus do less work, increasing total entropy as the Second Law demands.

Carnot Engine Energy Accounting

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