Consider a refrigerator as reversible heat engine working backwards. It works between 00C and 270C. The power required to remove heat from the refrigerated space at the rate of 50 kilocalorie per minute is (J = 4.2 Joule/calorie)
0.346 kilowatt Here, we have to find out the power required to remove heat from refrigerated space. We know that, a refrigerator is a reversible Carnot’s engine working backwards. For a carnot engine we have
W
Q1
=
T1−T2
T1
where W = work done Q1 = heat absorbed T1 = temperature of source T2 = temperature of sink Also, From a refrigerator
W
Q2
=
T2−T3
T3
where T1 temp of room (higher temperature) T2 - temperature of refrigerated space (food material) Q2 - heat absorbed from the substance W - work done for absorbing the heat in this process it is given Q2 = 50 Kcal = 50 × 103 cal = 50 × 103 × 4.2 J T1 = 27°C = 300 K T2 = 0° C = 273 That is
W
50×103×4.2
=
300−273
273
=
27
273
W = 50 × 103 × 4.2 ×
27
273
J We have, Power P =
work
time
=
W
t
and time , t = 1 minute = 60 seconds P =
50×103×4.2×27
60×273
watt = 0.346 watt Therefore, the power required to remove heat from the refrigerated space at the rate of 50 kilocalorie per minute = 0.346 kW.