The temperature inside a refrigerator is t2°C and the room temperature is t1°C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:
(1) t 2 + 273 t 1 - t 2                (2) t 1 + t 2 t 1 + 273           (3) t 1 t 1 - t 2              (4) t 1 + 273 t 1 - t 2

Dear Student ,

Coefficient of performance (COP) of a refrigerator ,β=Q2Q1-Q2Therefore ,COPβ=Q2Q1-Q2=T2T1-T2=t1+273t1+273-t2-273=t1+273t1-t2
Regards

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