Question 7.47: A household refrigerator that has a power input of 450 W and......
A household refrigerator that has a power input of 450 \mathrm{~W} and a COP of 2.5 is to cool five large watermelons, 10 \mathrm{kg} each, to 8^{\circ} \mathrm{C}. If the watermelons are initially at 20^{\circ} \mathrm{C}, determine how long it will take for the refrigerator to cool them. The watermelons can be treated as water whose specific heat is 4.2 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}. Is your answer realistic or optimistic? Explain.
The COP and the power consumption of a refrigerator are given. The time it will take to cool 5 watermelons is to be determined.
Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls, door, etc. is negligible. 3 The watermelons are the only items in the refrigerator to be cooled.
Properties The specific heat of watermelons is given to be c=4.2 \mathrm{~kJ} / \mathrm{kg} .{ }^{\circ} \mathrm{C}.
Analysis The total amount of heat that needs to be removed from the watermelons is
Q_{L}=(m c \Delta T)_{\text {watermelons }}=5 \times(10 \mathrm{~kg})\left(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(20-8)^{\circ} \mathrm{C}=2520 \mathrm{~kJ}
The rate at which this refrigerator removes heat is
\dot{Q}_{L}=\left(\mathrm{COP}_{\mathrm{R}}\right)\left(\dot{W}_{\text {net, in }}\right)=(2.5)(0.45 \mathrm{~kW})=1.125 \mathrm{~kW}
That is, this refrigerator can remove 1.125 \mathrm{~kJ} of heat per second.
Thus the time required to remove 2520 \mathrm{~kJ} of heat is
\Delta t=\frac{Q_{L}}{\dot{Q}_{L}}=\frac{2520 \mathrm{~kJ}}{1.125 \mathrm{~kJ} / \mathrm{s}}=2240 \mathrm{~s}=37.3 \mathrm{~min}
This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air, which will increase the work load. Thus, in reality, it will take longer to cool the watermelons.
