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 \b[$\Delta S_{\rm universe}=0$] for the complete <Carnot cycle>.

Solution

Past exam of the physics B course of the University of Cambridge / 2014 / Paper 2 / Section C / C10 / b / Solution

 Solution (source code)