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A reversed Carnot cycle is operating between temperature limits of 272 K and (+) 49°C. If it acts as a heat engine gives an efficiency of 15.52%. What is the value of C.O.P. of a refrigerator operating under the same conditions?

A reversed Carnot cycle is operating between temperature limits of 272 K and (+) 49°C. If it acts as a heat engine gives an efficiency of 15.52%. What is the value of C.O.P. of a refrigerator operating under the same conditions? Correct Answer 5.44

T1 = 272 K, T2 = 49°C = 322 K Temperature limits are given in the question so; we can calculate C.O.P. using the formula C.O.P. = T1 / (T2 − T1) = 272 / (322 − 272) = 5.44 Alternative approach: But as the efficiency of the heat engine is given so directly by the relation, we can find out the C.O.P. C.O.P. of heat pump = 1 / ηE = 1 / (0.1552) = 6.44 C.O.P. of refrigerator = C.O.P. of heat pump − 1 = 6.44 − 1 = 5.44.

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