To make ice, a freezer that is a reverse Carnot engine extracts 42 kJas heat at $\mathbf{-}\mathbf{15}\mathbf{°}\mathbf{C}$ during each cycle, with coefficient of performance 5.7. The room temperature is$\mathbf{30}\mathbf{.}\mathbf{3}\mathbf{°}\mathbf{C}$. (a) How much energy per cycle is delivered as heat to the room and (b) how much work per cycle is required to run the freezer?

Temperature of heat extraction,${\mathrm{T}}_{\mathrm{L}}=-15°\mathrm{C}=258\mathrm{K}$

Coefficient of performance, K=5.7

Energy extracted by the reverse Carnot Cycle,${\mathrm{Q}}_{\mathrm{L}}=42\mathrm{kJ}$

Room temperature,${\mathrm{T}}_{\mathrm{H}}=30.3°\mathrm{C}=303.3\hspace{0.33em}\mathrm{K}$

We use the formula of coefficient of performance to find the amount of energy per cycle delivered as heat to the room. We use the conservation of energy formula to find the amount of work per cycle required to run the freezer.

Formula:

The formula for coefficient of performance of a Carnot Engine,

$\mathrm{K}=\frac{{\mathrm{Q}}_{\mathrm{L}}}{{\mathrm{Q}}_{\mathrm{H}}-{\mathrm{Q}}_{\mathrm{L}}}$ (1)

The work done per cycle of a Carnot Engine,

${\mathrm{W}}_{\mathrm{in}}={\mathrm{Q}}_{\mathrm{H}}-{\mathrm{Q}}_{\mathrm{L}}$ (2)

Using equation (1) and the given values, the required energy per cycle to be delivered to the room can be given as:

$$\begin{gathered} \frac{42\mathrm{kJ}}{{\mathrm{Q}}_{\mathrm{H}}}=\frac{258\mathrm{K}}{\left(303.3\mathrm{K}-258\mathrm{K}\right)} \\ \frac{42\mathrm{kJ}}{{\mathrm{Q}}_{\mathrm{H}}-42\mathrm{kJ}}=5.69 \\ 42\mathrm{kJ}=\left({\mathrm{Q}}_{\mathrm{H}}-42\mathrm{kJ}\right)5.69 \\ 42\mathrm{kJ}=5.69{\mathrm{Q}}_{\mathrm{H}}-239.20\mathrm{kJ} \\ 5.69{\mathrm{Q}}_{\mathrm{H}}=281.20\mathrm{kJ} \\ {\mathrm{Q}}_{\mathrm{H}}=\frac{281.20\mathrm{kJ}}{5.69} \\ {\mathrm{Q}}_{\mathrm{H}}=49.42\mathrm{kJ} \end{gathered}$$

Therefore, the amount of energy per cycle delivered as heat to the room is 49.42 kJ

Using above values in equation (2), the work done per cycle to run the freezer can be given as:

$$\begin{gathered} \mathrm{W}=49.42\mathrm{kJ}-42.0\mathrm{kJ} \\ =7.42\mathrm{kJ} \end{gathered}$$

Therefore, the amount of work per cycle required to run the freezer is 7.42 kJ