SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

Question : 105 of 200

Marks: +1, -0

The efficiency of a Carnot engine is given by 0.80. If the cycle direction is reversed, then will be the COP for the Carnot refrigerator?

Solution:

Given:

\η =0.80

{COP}_{{HP}}={{T}_{{H}}}/{{T}_{{H}}-{T}_{{L}}}

={1}/{\η }={1}/{0.8}=1.25

Now,
COP of Refrigerator:

{COP}_{{\Ref}}={{T}_{{L}}}/{{T}_{{H}}-{T}_{{L}}}

={COP}_{{HP}}-1=1.25-1

1.25-1=0.25

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