Question

Explain in detail the working of a refrigerator.

Answer 1

Refrigerator: A refrigerator is a Carnot’s engine working in the reverse order.

Working Principle: 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 Q_H is rejected to the hot body (source) ie, the atmosphere at T_H. When you stand beneath of refrigerator, you can feel warmth air. From the first law of thermodynamics, we have

Q_L +W = Q_H ..............(1)

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

Schematic diagram of a refrigerator

COP = β = `"Q"_"L"/"W"` ............(2)

From the equation (1) β = `"Q"_"L"/("Q"_"H" - "Q"_"L")`

β = `1/("Q"_"H"/"Q"_"L" - 1)` .........(3)

But we know that `"Q"_"H"/"Q"_"L" = "T"_"H"/"T"_"L"`

Substituting this equation into equation (3), we get

β = `1/("T"_"H"/"T"_"L" - 1) = "T"_"L"/("T"_"H" - "T"_"L")`

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, the 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 objects to hot objects. It is not a violation of the second law of thermodynamics, because the heat is ejected into the surrounding air and the total entropy of (refrigerator + surrounding) is always increased.