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 –3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50% of a perfect engine.

2 Views|Posted 7 months ago
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7 months ago

This is a short answer type question as classified in NCERT Exemplar

Temperature of the source is 270C

T1= 27+273= 300K

T2= -3+273= 270K

Efficiency of heat engine = 1-T2/T1= 1-270/300=1/10

Efficiency of refrigerator  is 50% of a perfect engine

= 0.5 * η = 1/20

Coefficient of performance of the refrigerator β = Q 2 W = 1 - η ' η

=

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physics ncert solutions class 11th 2023

physics ncert solutions class 11th 2023

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