Q. The reversible engine has thermal efficiency of 20%. What will be the COP if it is used as a refrigerator with other conditions unchanged:
Solution:
C.O.P is reciprocal of efficiency of heat engine.
So 1/0.2 = 5 = C.O.P of heat engine.
Now C.O.P of refrigerator = c.o.p of heat engine-1 = 5-1 = 4.
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