Consider a full cycle of a Carnot engine, in which the hot reservoir loses entropy -Qh/Th and the cold reservoir gains entropy +Qc/Tc. Since Qh/Th equals Qc/Tc for a Carnot engine, what is the total change in entropy for the complete reversible cycle?

1. Over a full cycle, the hot reservoir's entropy change is -Qh/Th, since heat leaves it, and the cold reservoir's entropy change is +Qc/Tc, since heat enters it. 2. Adding these gives ΔS_tot = -Qh/Th + Qc/Tc. 3. Because Qh/Th equals Qc/Tc for a Carnot engine, the two terms cancel exactly, leaving ΔS_tot = 0. 4. This is stated as a general result: the total change in entropy for a system in any reversible process is zero. 5. This rules out options A and B, which assign the total a fixed sign, and it can in fact be determined here directly from the given Carnot relationship, ruling out option C. _Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy_