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

  

A)

Coefficient of performance = Q_c/ W = T_c / (T_h - T_c )

here T_c = 260.93 K or 10 F and T_h = 298K

Coefficient of performance = 260.93 / 37.07 = 7.03 or approximately 7

B)

According to law of thermodynemics

Q_h = Q_c + W

Q_h = 1765+792 = 2557 J