1. Show that for a reversible heat pump the energy required per unit of heat delivered inside the building is given by the Carnot efficiency: \begin{align} \frac{W}{Q_H} &= \eta_C = \frac{T_H-T_C}{T_H} \end{align} What happens if the heat pump is not reversible?

2. Assume that the electricity consumed by a reversible heat pump must itself be generated by a Carnot engine operating between the even hotter temperature \(T_{HH}\) and the cold (outdoors) temperature \(T_C\). What is the ratio \(\frac{Q_{HH}}{Q_H}\) of the heat consumed at \(T_{HH}\) (i.e. fuel burned) to the heat delivered at \(T_H\) (in the house we want to heat)? Give numerical values for \(T_{HH}=600\text{K}\); \(T_{H}=300\text{K}\); \(T_{C}=270\text{K}\).

3. Draw an energy-entropy flow diagram for the combination heat engine-heat pump, similar to Figures 8.1, 8.2 and 8.4 in the text (or the equivalent but sloppier) figures in the course notes. However, in this case we will involve no external work at all, only energy and entropy flows at three temperatures, since the work done is all generated from heat.

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These lecture notes covering week 8 of https://paradigms.oregonstate.edu/courses/ph441 include a small group activity in which students derive the Carnot efficiency.