Ncert Exercises & Solutions

In a refrigerator one removes heat from a lower temperature and deposits it to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1kW power and heat is transferred from —3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

Hint: The ratio of the heat delivered to the surroundings to the work done by the motor will be equal to the coefficient of performance of the refrigerator.

Step 1: Find the coefficient of performance of the refrigerator.

Given, the temperature of the source,

T_{1} = (27 + 273) K = 300 K

The temperature of the sink,

T_{2} = (- 3 + 273) K = 270 K

Efficiency of a perfect heat engine is given by,

η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{270}{300} = \frac{1}{10}

The efficiency of the refrigerator is 50% of a perfect engine.

∴

η' = 0.5 \times η = \frac{1}{2} η = \frac{1}{20}

Step 2: Find the heat delivered to the surroundings.

∴

Coefficient of performance of the refrigerator,

β = \frac{Q_{2}}{W} = \frac{1 - η'}{η'}

= \frac{1 - (1 / 20)}{(1 / 20)} = \frac{19 / 20}{1 / 20} = 19

\Rightarrow Q_{2} = β W = 19 W (∵ β = \frac{Q_{2}}{W})

= 19 \times (1 k W) = 19 k W = 19 k J / s

Therefore, the heat is taken out of the refrigerator at a rate of 19 kJ per second.

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