The key idea here is that a refrigerator acts as a heat pump. It removes heat from the cold reservoir (the inside of the refrigerator) and releases that heat plus the work input into the surrounding room.
Given:
m = 0.750 kg
Latent heat of fusion of ice: Lf = 3.34 × 10^5 J/kg
Coefficient of performance: COP = 17.5
Step 1: Find the heat removed from the refrigerator (QL)
Since the ice is melting at 0.00°C, the heat removed is the latent heat required to melt the ice:
QL = mLf
QL = (0.750 kg)(3.34 × 10^5 J/kg)
QL = 2.505 × 10^5 J
QL = 250.5 kJ
Therefore:
QL = 250.5 kJ
Step 2: Find the work input (W)
The coefficient of performance for a refrigerator is:
COP = QL / W
Solving for W:
W = QL / COP
W = (2.505 × 10^5 J) / 17.5
W = 1.43 × 10^4 J
W = 14.3 kJ
Step 3: Find the heat released into the home (QH)
Using conservation of energy:
QH = QL + W
QH = 250.5 kJ + 14.3 kJ
QH = 264.8 kJ
Therefore:
QH = 264.8 kJ
Final Answers:
QL = 250.5 kJ
QH = 264.8 kJ
The refrigerator removes 250.5 kJ of heat from the ice as it melts and releases 264.8 kJ of heat into the home. The difference (14.3 kJ) comes from the electrical work required to operate the refrigerator. A common mistake is assuming QH = QL, but the refrigerator must also release the energy supplied by the compressor motor.
