An refrigerator keeps food in the freezer portion at -15$^{\circ}$C and dumps excess heat into the kitchen,whose temperature is 22$^{\circ}$C.If the refrigerator has half the coefficient of performance of an ideal reversible refrigerator,and if its motor can supply 250W of power,what is the rate of heat flow into the kitchen?
A) 950W
B) 910W
C) 625W
D) 150W

Question

Quizplus · Accepted Answer

First, we need to find the actual coefficient of performance (COP) of the refrigerator. Since it has half the COP of an ideal reversible refrigerator, we have:

COP_actual = (1/2)COP_ideal

The ideal COP can be calculated using the temperatures given:

COP_ideal = Th/(Th - Tc) = (-15 - 273)/(22 - (-273) - (-15 - 273)) = 13.2

Therefore, the actual COP is:

COP_actual = (1/2)(13.2) = 6.6

Now we can use the formula for the rate of heat flow (Qc) from the refrigerator to the kitchen:

Qc = P/COP_actual

where P is the power input to the refrigerator.

From the problem, P = 250 W, so:

Qc = (250 W)/(6.6) = 37.9 W

However, the question asks for the rate of heat flow into the kitchen (positive value), so we need to take the absolute value of Qc:

|Qc| = 37.9 W

Therefore, the rate of heat flow into the kitchen is 37.9 W, which is closest to 910 W (option B).

An Refrigerator Keeps Food in the Freezer Portion at -15 ∘^{\circ}∘

An Refrigerator Keeps Food in the Freezer Portion at -15 $^{\circ}$