91Ó°ÊÓ

The heat in the cold compartment is${\mathrm{Q}}_{\mathrm{L}}=560\mathrm{J}$.

The work done, w=150J

By using Equations 20-14, we can find the refrigerator’s coefficient of performance. By adding the heat to the cold compartment and work done, we can find the heat per cycle exhausted in the kitchen.

Formulae:

From Equation 20-14, the coefficient of performanceof a refrigerator,

$\mathrm{k}=\frac{\left|{\mathrm{Q}}_{\mathrm{L}}\right|}{\left|\mathrm{W}\right|}$ (1)

where$Q_L$is heat at cold compartment, W is work done.

The work was done using the first law of thermodynamics of a Carnot engine,

$\left|\mathrm{W}\right|=\left|{\mathrm{Q}}_{\mathrm{H}}\right|-\left|{\mathrm{Q}}_{\mathrm{L}}\right|$ (2)

Using the given data in equation (1), we can get the coefficient of performance of the refrigerator as given:

$$\begin{gathered} \mathrm{K}=\frac{560\mathrm{J}}{150\mathrm{J}} \\ =3.73 \end{gathered}$$

Hence, the value of the coefficient of performance is 3.73

The energy conservation requires the exhausting heat to be given using equation (2) as follows:

$$\begin{gathered} \left|{\mathrm{Q}}_{\mathrm{H}}\right|=\left|{\mathrm{Q}}_{\mathrm{L}}\right|+\left|\mathrm{W}\right| \\ =560\mathrm{J}+150\mathrm{J} \\ =710\mathrm{J} \end{gathered}$$

Hence, the value of the required exhausted energy is 710 J