SSC JE Mechanical Engineering 27 Oct 2020 Shift 2 Papers

Question : 114 of 200

Marks: +1, -0

A Carnot engine receiving heat at 400 K has an efficiency of 50 %. What is the COP of a Carnot refrigerator working between the same temperature limits?

Validate

Solution:

Given:

\η _{\text "Carnot "}=50 \%=0.5, {~T}_{{H}}=400 {K}

\η _{\text "carnot "}={{T}_{{H}}-{T}_{{L}}}/{{T}_{{H}}}=1-{{T}_{{L}}}/{{T}_{{H}}}

\⇒{{T}_{{L}}}/{{T}_{{H}}}=0.5={1}/{2}

Now,

({COP})_{\text "carnot "}={{T}_{{L}}}/{{T}_{{H}}-{T}_{{L}}}

={1}/{{{T}_{{H}}}/{{T}_{{L}}}-1}={1}/{2-1}=1

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