Question

To make ice, a freezer that is a reverse Carnot engine extracts 42 kJ as heat at -13°C during each cycle, with coefficient of performance 5.9. The room temperature is 30.9°C. How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?

Answer 1

Answer:

(a) 49.09 kJ

(b) 7.09 kJ

Explanation:

Qc = 42 kJ

Tc = - 13 degree C = - 13 + 273 = 260 K

TH = 30.9 degree C = 30.9 + 273 = 303.9 K

β = 5.9

(a) Let the heat delivered to the room per cycle is QH.

TH / Tc = QH / Qc

303.9 / 260 = QH / 42

QH = 49.09 kJ

(b) Work done per cycle = QH - Qc = 49.09 - 42 = 7.09 kJ