Question

A refrigerator with a coefficient of performance of 1.70 absorbs 3.60×10⁴ J of heat from the low-temperature reservoir during each cycle.

How much mechanical work is required to operate the refrigerator for a cycle?

How much heat does the refrigerator discard to the high-temperature reservoir during each cycle?

Answer 1

Recall the definition of COP:
COP = what we want/what we pay for

In a refrigerator, we want to "suck" heat out of the cold chamber and put it elsewhere. We pay for mechanical work in the form of electricity to operate the compressor. We reject it out the condenser in to the kitchen, in order to provide fresh refrigerant to absorb more heat.

Part A:

As an analogy, we need to buy garbage bags to (safely) reject the garbage we want to dispose of.

COP = Q_in/W_in

Part A: Solve for W_in
W_in = Q_in/COP
W_in = 3.60*10^4 / 1.70

W_in = 2.11 * 10^4 Joules

Part B:
Heat is added, work is added, net sum must be rejected. Just like the garbage + garbage bags end up in the landfill.

Q_out = W_in + Q_in

Q_out = Q_in*(1 + 1/COP)

Q_out = 3.60*10^4 (1 + 1/1.70)

Q_out is 57176.47 Joules