A Carnot refrigerator extracts 35.0 kJ as heat during each cycle, operating with a coefficient of performance of 4.60 . What are (a) the energy per cycle transferred as heat to the room and (b) the work done per cycle?

1. The value of energy extracted as heat,${\mathrm{Q}}_{\mathrm{L}}=35.0\mathrm{kJ}$
2. The coefficient of performance, ${\mathrm{K}}_{\mathrm{c}}=4.60$

A refrigerator is a device that uses work to transfer energy from a low-temperature reservoir to a high-temperature reservoir. For the Carnot refrigerator, the coefficient of performance from equation 20-15 is: ${\mathbf{K}}_{\mathbf{c}}\mathbf{=}\left|Q_L\right|\mathbf{/}\left|Q_H\right|\mathbf{-}\left|Q_L\right|$where, $\mathbf{\left|}{\mathbf{Q}}_{\mathbf{L}}\mathbf{\right|}$is the magnitudeof the energy transformed as heat to the low-temperature reservoir and $\mathbf{\left|}{\mathbf{Q}}_{\mathbf{H}}\mathbf{\right|}$ is the magnitudeof the energy transformed as heat to the high-temperature reservoir.

Formula:

The coefficient of performance of Carnot refrigerator,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|{\mathrm{Q}}_{\mathrm{H}}\right|-\left|{\mathrm{Q}}_{\mathrm{L}}\right|} \\ {\mathrm{K}}_{\mathrm{c}}=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|\mathrm{W}\right|} \end{gathered}$$ (1)

The calculation of i.e. the energy per cycle transformed as heat to the room is given using equation (1) and the given values are as follows:

$$\begin{gathered} \left|{\mathrm{Q}}_{\mathrm{H}}\right|=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{{\mathrm{K}}_{\mathrm{c}}}{+}_{\mathrm{c}}-\left|{\mathrm{Q}}_{\mathrm{L}}\right| \\ =\left(\frac{1}{{\mathrm{K}}_{\mathrm{c}}}+1\right)\left|{\mathrm{Q}}_{\mathrm{L}}\right| \\ =\left(\frac{1}{4.6}+1\right)35\mathrm{kJ} \\ =42.60\mathrm{kJ} \end{gathered}$$

Hence, the value of the energy transferred as heat to the room is 42.60 kJ

We have to calculate the work done per cycle. So we rearrange the formula for work done from equation (1) and the required values as given:

$$\begin{gathered} \left|\mathrm{W}\right|=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\mathrm{K}} \\ =\frac{35\mathrm{kJ}}{4.6} \\ =7.61\mathrm{kJ} \end{gathered}$$

Hence, the value of work done per cycle by the refrigerator is 7.61 kJ