(Solved): 6. Thermodynamics: The Carnot efficiency. Derive the efficiency of a thermodynamically reversibl ...

6. Thermodynamics: The Carnot efficiency. Derive the efficiency $?$ of a thermodynamically reversible (Carnot) heat pump system using the following information. All temperatures are absolute $(\mathrm{K})$. (Note: For a real-world system, which cannot operate on a Carnot cycle, we call the heat pump efficiency a coefficient of performance (COP) as opposed to a Carnot efficiency.) $\begin{array}{l}Q\\ W\\ (Q+W)\\ Q/T_c=(Q+W)/T_h\\ ?=(Q+W)/W\end{array}$ heat energy withdrawn from a cold reservoir at temperature $T_c(K)$ mechanical energy supplied by the compressor heat energy supplied to the hot reservoir at temperature $T_h(K)$ true when a reversible process (the total entropy $S$ does not increase) definition of efficiency (also for the coefficient of performance) a. Express the work $W$ in terms of $Q,T_c$ and $T_h$ b. Express the Carnot heating efficiency in terms of $T_c$ and $T_h$ c. A refrigerator is the same as a heat pump except that the efficiency is defined as the heat energy withdrawn, $?=Q/W$. Express the Carnot cooling efficiency in terms of $T_c$ and $T_h$.