A heat engine operates between a hot reservoir at 1500 K and a cold reservoir at 500 K.During each cycle,1.0 × 10⁵ J of heat is removed from the hot reservoir and 5.0 × 10⁴ J of work is performed.
-If the coefficient of performance for a refrigerator is 5.0 and 72 J of work are done on the system,how much heat is rejected to the room?

Question

Quizplus · Accepted Answer

The coefficient of performance for a refrigerator is given by:$$COP = \frac{Q_c}{W}$$where:* $$Q_c$$ is the heat rejected to the room* $$W$$ is the work done on the systemWe are given that $$COP = 5.0$$ and $$W = 72 J$$, so we can solve for $$Q_c$$:$$Q_c = COP 	imes W = 5.0 	imes 72 J = 360 J$$Therefore, the heat rejected to the room is 360 J.

A Heat Engine Operates Between a Hot Reservoir at 1500