A refrigerator, whose coefficient of performance K is 5, extracts heat from the cooling compartment at the rate of 250 J per cycle.How much work per cycle is required to operate the refrigerator cycle ?

A refrigerator, whose coefficient of performance K is 5, extracts heat

1.	A refrigerator, whose coefficient of performance K is 5, extracts heat from the cooling compartment at the rate of 250 J per cycle.How much work per cycle is required to operate the refrigerator cycle ?
Answer» Solution :As coefficient of performance of a REFRIGERATOR is DEFINED as `K = Q_L//W`. so ` W= (Q_L)/(K )= ( 250)/(5) = 50J`