Clausius theorem

Remarks

Split the cycle into small steps, where step $i$ takes heat $\delta Q_i$ at temperature $T_i$. Supply each $\delta Q_i$ through a reversible Carnot engine that runs between one fixed reservoir at temperature $T_0$ and temperature $T_i$. That engine draws heat $T_0 \, \delta Q_i / T_i$ from the fixed reservoir. The system and the engines together run in a cycle whose only effect is to take heat $T_0 \sum_i \delta Q_i / T_i$ from one reservoir and turn it into work. The Kelvin–Planck statement forces this to be at most $0$; for a reversible cycle, running it backwards gives the opposite inequality as well. As a consequence, $\int \delta Q / T$ along a reversible path depends only on its end points; this is how entropy is defined. For a Carnot cycle that takes $Q_h > 0$ at $T_h$ and gives out $Q_c > 0$ at $T_c$, equality reads $Q_h / T_h = Q_c / T_c$.

Going deeper

For a student

The theorem is the bridge from the classical second law to entropy. Equality for reversible cycles shows that $\delta Q_{\mathrm{rev}} / T$ is the differential of a state function, the entropy $S$. The inequality for irreversible processes then gives the Clausius inequality $dS \ge \delta Q / T$.