Working Principle:

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Fig 8.49 (a) Schematic diagram of a refrigerator (b) Actual refrigerator

The working substance (gas) absorbs a quantity of heat _Q_L from the cold body (sink) at a lower temperature _T_L. A certain amount of work W is done on the working substance by the compressor and a quantity of heat QH is rejected to the hot body (source) ie, the atmosphere at TH. When you stand beneath of refrigerator, you can feel warmth air. From the first law of thermodynamics , we have

\[ Q_L + W = Q_H \]

As a result the cold reservoir (refrigerator) further cools down and the surroundings (kitchen or atmosphere) gets hotter.

Coefficient of performance (COP) (β): COP is a measure of the efficiency of a refrigerator. It is defined as the ratio of heat extracted from the cold body (sink) to the external work done by the compressor W.

\[\text{COP} = \beta = \frac{Q}{W_L} \]

From the equation (8.68)

\[ \beta = \frac{Q_t}{Q_n- Q_t} \] \[ \beta = \frac{1}{\frac{Q_n}{ Q_t}-1} \]

But we know that \[ \beta = \frac{Q_t}{ Q_t}=\frac{T_n}{ T_t} \]

Substituting this equation into equation (8.70) we get \[ \beta = \frac{1}{\frac{T_n}{ T_l}-1} =\frac{T_t}{T_n- T_t} \]

Inferences: 1. The greater the COP, the better is the

condition of the refrigerator. A typical refrigerator has COP around 5 to 6.

2. Lesser the difference in the temperatures of the cooling chamber and the atmosphere, higher is the COP of a refrigerator.

3. In the refrigerator the heat is taken from cold object to hot object by doing external work. Without external work heat cannot flow from cold object to hot object. It is not a violation of second law of thermodynamics, because the heat is ejected to surrounding air and total entropy of (refrigerator + surrounding) is always increased.

EXAMPLE 8.27

A refrigerator has COP of 3. How much work must be supplied to the refrigerator in order to remove 200 J of heat from its interion?

\[\text{COP} = \beta = \frac{Q}{W_L} \] \[\frac{W}{Q} = \text{COP} = \frac{L}{J} = \frac{200}{366} \]

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