Clausius inequality and entropy production

For a reversible cycle, the integral of heat absorbed divided by the equilibrium temperature is zero. That path-independence result permits a state function S whose differential is $dS=\delta Q_{\mathrm{rev}}/T$. Entropy has units of joules per kelvin. To find its difference between equilibrated states, choose any suitable reversible path connecting them, whether or not the actual physical process followed that path. The reversible comparison is a computational device, not a description of a rapid irreversible trajectory.

For an irreversible process, substituting the actual heat directly into $\Delta S=Q/T$ generally fails. A thermally insulated free expansion of an ideal gas has no heat input but a positive entropy change. The endpoint entropy can instead be computed with a reversible isothermal expansion between the same volumes. That imagined comparison has heat input and boundary work, although the actual expansion has neither. State-function equality of entropy changes does not imply equality of the path-dependent transfers.

Suppose a closed system receives signed heats $Q_j$ from ideal reservoirs at fixed temperatures $T_j$. Each reservoir changes entropy by $-Q_j/T_j$ because its lost heat enters the system. If those reservoirs, the system, and ideal work receivers form an otherwise isolated combination, their total entropy change is nonnegative. Define this total as entropy production for the complete interaction. Work transfers energy without carrying entropy in this ideal bookkeeping, although dissipating that work later can generate entropy elsewhere.

This gives $S_{\mathrm{prod}}=\Delta S_{\mathrm{sys}}-\sum_jQ_j/T_j\ge0$. It includes irreversible contacts as well as internal system dissipation under the chosen combined boundary. A more local balance can use the boundary temperature where heat crosses and assign production to individual regions, but the allocation changes when the boundary moves. The total isolated-combination result remains consistent. Always state which reservoirs and interfaces are included before attaching an entropy-production number to one component.

Transfer a positive heat magnitude Q directly from a hot reservoir to a colder one. The hot side loses entropy $Q/T_h$, while the cold side gains $Q/T_c$. The gain exceeds the loss because the colder temperature has the larger reciprocal. Energy is conserved, but entropy increases. The same heat amount has different entropy-transfer magnitudes at different temperatures, which is why entropy is not simply another name for energy or for a quantity of heat stored in matter.

Entropy production vanishes in the reversible limit of infinitesimal thermal differences and no other dissipative effects. For a system that returns to its initial state, its entropy change is zero even while the reservoirs gain positive total entropy each cycle. For an isolated system alone, all external heats vanish and its entropy change equals its production. These special cases are consequences of one balance, not competing versions of the second law, and they provide useful checks on signs and boundary choices.