Phyysics: Coefficient of Performance

Question

Consider a common refrigerator in a standard home, where the temperatures are 22.0 C in the home and the refrigerator is kept slightly above freezing. If a 0.750 kg ice cube initially at 0.00 C is allowed to melt into water at 0.00 C, and the refrigerator is running at a coefficient of performance of 17.5, find Q_H & Q_L for this process.

Answer 1

How much heat is required to melt ice at 0 °C?

heat removed from cold space: Q_L

Q_L = mL_f

Standard for melting ice at 0°C L_f = 3.34 x 10⁵ J/kg

Q_L = (0.750 kg) (3.34 x 10⁵ J/kg) = 2.505 x 10⁵ J OR 250 kJ

Use COP to get work W

for a refrigerator: W = (Q_L / COP)

W = (2.505 x 10⁵ J / 17.5) = 1.43 x 10⁴ J or 14.3 kJ

Heat added into the room: Q_H

Q_H = Q_L + W

QH = 2.505 x 10⁵ J + 1.43 x 10⁴ J = 2.65 x 10⁵ J or 265 kJ