Question

250mL of water at 15.0 ?C is placed in the freezer compartment of a refrigerator with a coefficient of performance of 4.00. How much heat energy is exhausted into the room as the water is changed to ice at -20.0 ?C ?

Answer 1

First, develop a formula for the heat that the refrigeration device sucks out of the freezing water. Then, use the coefficient of performance, and its meaning, to understand what heat is implied to be rejected to the environment.

Heat sucked in to the refrigeration piping = Heat required to cool water to freezing + Heat required to freeze water + heat required to cool ice to -20 C.

Q_in = m*(c_w*(T1 - Tmelt) + h_f + c_i*(Tmelt- T2))

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

COP = Q_in/W_in

No work is captured in the throttling valve expansion process. Conservation of energy will tell us that the work used to operate the compressor plus the heat absorbed from the freezing space both become the rejected heat.

Q_out = Q_in + W_in

Replace W_in with Q_in/COP:

Q_out = Q_in*(1 + 1/COP)

Thus:
Q_out = m*(c_w*(T1 - Tmelt) + h_f + c_i*(Tmelt- T2))*(1 + 1/COP)

Data:
T1:=15 C; T2:= -20 C; m:=0.1 kg;
Tmelt:=0 C; h_f:=334.944 kJ/kg; c_w:=4.184 kJ/kg-C; c_i:=2.0934 kJ/kg-C;
COP:=4;

Result:
Q_out = 53.38 kJ

alternatively expressed as
Q_out = 53,380 J