Question

What are (a) the heat extracted from the cold reservoir and (b) the coefficient of performance for the refrigerator shown in Figure Ex-$21.20$?

Short answer

(a)The heat extracted from the cold reservoir, $64\mathrm{J},$

(b)The coefficient of performance for the refridgerator$1.56$

Step-by-step solution

Step: 1 Finding the value of workdone: (part a)

Net work done by the gas can be calculated as follows. This is the energy which is required to be served by external electrical power .

Electrical power used up = 119 J - 78 J = 41 J .

105 J is the total heat given out to the environment . This heat will be the sum of heat extracted and electrical energy used.

We can calculate the heat extracted as follows.

$Q_c=Q_h-W$

$$\begin{gathered} Q_c=105-41 \\ =64\mathrm{J} \end{gathered}$$

Step: 2 Finding the value of K: (part b)

The thermal efficiency is the ratio of the amount of heat extracted from a cold reservoir to the electrical energy required to extract it.. That is,

$K=\frac{Q_c}{W}$

$$\begin{gathered} K=\frac{64}{41} \\ =1.56 \end{gathered}$$