Question:-
A heat engine with 30% efficiency drives a refrigerator of COP. 5.0. What would be the net heat input to the engine for each MW of heat removed in the refrigerator?
Option (A)
66.67 kJ
Option (B)
600 kJ
Option(C)
666.67 kJ
Option(D)
6600 kJ
Correct Option:
(C)
Question Solution:
Power required for refrigerator
= (106J/5) = 200KJ
Heat input for heat pump = (200/0.3) = 666.67KJ
question-answer-faq-9463