Question

In one cycle, a freezer uses 585 J of electrical energy in order to remove 1750 J of heat from its freezer compartment at 10.0 F.a. What is the coefficient of performance of this freezer?b. How much heat does it expel into the room during this cycle?

Answer 1

(a) 0.525

(b) 1750J

Step-by-step explanation:

Coefficient of Performance (COP) = Q2/(Q1 - Q2)

Q1 = heat rejected to the surrounding = 1750J

Q2 = heat absorbed in the environment = 585K

COP = 585/(1700 - 585) = 585/1115 = 0.525

(b). Heat expelled to the surrounding (room) = 1750J