 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2014/paper-2/section-c/c10/b/solution

The reversible working substance changes entropy by

Q_{h} / T_{h}

on its hot isotherm and

- Q_{c} / T_{c}

on its cold isotherm, with zero change on the two reversible adiabats. Closure requires

Q_{h} / T_{h} = Q_{c} / T_{c}

. The substance has zero entropy change over a cycle, while the reservoirs have changes

- Q_{h} / T_{h}

and

Q_{c} / T_{c}

. Thus $Δ S_{universe} = 0$ for the complete Carnot cycle.

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