Question

In one cycle, a freezer uses 800 J of electrical energy in order to remove 1735 J of heat from its freezer compartment at 10.0°F. Part A)

What is the coefficient of performance of this freezer?
Part B)
How much heat does it expel into the room during this cycle?

Answer 1

Step-by-step explanation:

A)

W = work done by the freezer = 800 J

Q = heat removed from the freezer = 1735 J

Q' = Heat expelled into the room

Coefficient of performance is given as

`\beta = (Q)/(W)`

inserting the values

`\beta = (1735)/(800)`

`\beta = 2.2`

B)

Heat expelled is given as

Q' = W + Q

Q' = 800 + 1735

Q' = 2535 J