Question

In one cycle, a freezer uses 792 J of electrical energy in order to remove 1765 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? (J)

Answer 1

coefficient of performance is 2.22

heat expel into the room during this cycle is 2557 J

Step-by-step explanation:

Given data

electrical energy w = 792 J

heat Qc = 1765 J

temperature = 10.0°F

to find out

coefficient of performance of this freezer and How much heat does it expel

solution

we know that coefficient of performance calculate by formula that is

coefficient of performance = Qc / W .................1

put both value we get

coefficient of performance = 1765 / 792

coefficient of performance is 2.22

and

we know heat expel that is calculated by

by heat pump thta is

heat expel = w + Qc ................2

put here both value we get

heat expel = 1765 + 792

heat expel = 2557

so heat expel into the room during this cycle is 2557 J