SSC JE Mechanical Engineering 2018 Set 2 Paper

Question : 133 of 200

Marks: +1, -0

If the COP of Carnot refrigerator is 4, then thethermal efficiency of the Carnot engine would be________.

Solution:

COP of Carnot refrigerator = 4
COP of heat pump = 1 + (Cop)

_\Ref.

= 5
Efficiency of Carnot =

1/(Cop) _{\H.P.}=1/5=0.2

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