Question

A refrigerator is used to remove 84 kj/min of heat from a tank. If the electric power consumed by the refrigerator is 1. 2 kw, what is the cop of the refrigerator?.

Answer 1

Answer:

84 kj/min = 1.4 kj/sec

Power Out / Power In = Heat Out / Heat In - Coefficient of Performance

1.4 kj/sec / 1.2 kj/sec = 1.17 = COP