In a refrigerator, one removes heat from a lower temperature and deposits 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 -3oC to 27oC, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.
In a refrigerator, one removes heat from a lower temperature and deposits 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 -3oC to 27oC, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

Answer:

Expression for the efficiency of the Carnot engine is as follows;

\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}

Where T1 = 300 K and T2 = 270 K

We are given that the efficiency is 50% = 0.5

According to the question, the efficiency of refrigerator = 0.05

We know that,

{{\eta }_{ref}}=\frac{W}{{{Q}_{1}}}

\Rightarrow {{Q}_{1}}=\frac{W}{{{\eta }_{ref}}}=20kJ

As a result, the efficiency of the refrigerator is as follows:

{{Q}_{2}}={{Q}_{1}}-{{\eta }_{ref}}{{Q}_{1}}={{Q}_{1}}(1-{{\eta }_{ref}})

{{Q}_{2}}=19kJ