Carnot Engine Energy Accounting

For a reversible Carnot cycle the heats exchanged with the two reservoirs are tied to their absolute temperatures by $\frac{Q_C}{Q_H}=\frac{T_C}{T_H}$, a consequence of the entropy change being zero over the reversible cycle. Solving for the rejected heat gives $Q_C=Q_H\frac{T_C}{T_H}=1000\times \frac{300}{500}=600\text{ J}$. 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 $0.4\times 1000=400\text{ J}$ and the rejected heat is $1000-400=600\text{ J}$, 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.