Question

A refrigerator whose coefficient of performance $K=5$, extracts heat from the cooling compartment at the rate of $250J/cycle$. What is the work done per cycle to operate the refrigerator? How much heat is discharged per cycle to the room which acts as the high-temperature reservoir?

Answer 1

Step 1: Given data

1. The coefficient of performance of the refrigerator is $\eta =5.$
2. The cooling rate of the cold reservoir is $Q_c=250J/cycle.$

Step 2: Formula Used

1. We know that the performance of a refrigerator is the ratio of the amount of heat extracted from the cold body to the work done on it.
2. The performance or efficiency of a refrigerator is defined by the form, $\eta =\frac{Q_C}{W}=\frac{heatextractedfromthecoldbody}{workdoneonthemachine}.$
3. Heat discharged by a refrigerator is the sum of total heat taken by the machine and the work done by the machine, i.e, $Q_d=Q_c+W$

Step 3: Diagram

Step 4: Calculation

As we know, the work done by a refrigerator is $W=\frac{Q_C}{\eta }$

So,

$$\begin{gathered} W=\frac{250}{5}=50 \\ orW=50joule. \end{gathered}$$

Therefore, the work done per cycle to operate the refrigerator is $50joule.$

Now heat discharged by the refrigerator is $Q_d=Q_c+W$.

So,

$$\begin{gathered} Q_d=250+50=300 \\ or{Q}_d=300joules \end{gathered}$$

Therefore, heat discharged per cycle to the room is $300joules$.