An ideal refrigerator does 150 Jof work to remove 560 Jas heat from its cold compartment. (a) What is the refrigerator’s coefficient of performance? (b) How much heat per cycle is exhausted to the kitchen?

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