Question

Draw schematic diagram of a refrigerator. Define its coefficient of performance and mention the expression.

Answer 1

Hint: A device which takes heat from a cold body and transfers it to a hot body is called a refrigerator. It does the reverse of a heat engine. Work by the external agent must be done so that the process of refrigeration goes on. First we will draw the schematic diagram of a refrigerator and from that we will define the coefficient of performance of the refrigerator.

Complete answer:
The schematic diagram of a refrigerator is shown below.

 

As we can see from the diagram it takes $Q_2$ amount of heat from the cold body which is at a lower temperature of $T_2$. $W$ amount of work is done by the external agent and a total of
$Q_1=Q_2+W$ amount of heat is given to the hot body which is at a higher temperature of $T_1$.
The purpose of a refrigerator is to remove as much heat from the cold body at the expense of as little work required to run the refrigerator. Thus the coefficient of performance $K$ of a refrigerator is defined as the ratio of the heat taken from the cold body to the amount of work required to run the refrigerator, Thus
$K={\displaystyle \frac{Q_2}{W}}$.
Now
$\begin{array}{rl}& Q_1=Q_2+W\\ & or{\displaystyle \frac{W}{Q_2}}=({\displaystyle \frac{Q_1}{Q_2}}-1)\end{array}$
Now as in Carnot’s engine, if we use ideal gas as a working substance then it can be shown
${\displaystyle \frac{Q_1}{Q_2}}={\displaystyle \frac{T_1}{T_2}}$, thus
${\displaystyle \frac{W}{Q_2}}=({\displaystyle \frac{T_1}{T_2}}-1)=({\displaystyle \frac{T_1-T_2}{T_2}})$
Therefore the coefficient of performance
$K={\displaystyle \frac{Q_2}{W}}={\displaystyle \frac{T_2}{T_1-T_2}}$

Note:
A refrigerator can have a coefficient of performance greater than 1. In a heat engine, heat cannot be fully converted to work, similarly a refrigerator cannot work without some external work done on the system i.e. the coefficient of performance cannot be infinite. A good refrigerator should have its value around 5 or 6.