Question

The motor in a refrigerator has a power output of 294 W. If the freezing compartment is at 271 K and the outside air is at 310 K, assuming ideal efficiency, what is the maximum amount of heat (in joules) that can be extracted from the freezing compartment in 81.6 minutes

Answer 1

Answer:

$Q_{L} = 11.442\times 10^{6}\,J$

Explanation:

The ideal efficiency of a refrigerator is given by the model of a Carnot refrigerator, whose indicator is known as Coefficient of Performance:

$COP_{R} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{R} = \frac{310\,K}{310\,K-271\,K}$

$COP_{R}=7.949$

Heat rate extracted from the freezing compartment is:

$\dot Q_{L} = COP_{R}\cdot \dot W$

$\dot Q_{L} = 7.949\cdot 294\,W$

$\dot Q_{L} = 2337.01\,W$

The heat that can be extracted in abovementioned period of time is:

$Q_{L} = \dot Q_{L}\cdot \Delta t$

$Q_{L} = (2337.01\,W)\cdot (81.6\,min)\cdot (\frac{60\,s}{1\,min} )$

$Q_{L} = 11.442\times 10^{6}\,J$

Answer 2

Answer:

Maximum amount of heat = 10002151.38J

Explanation:

Workdone by motor in 86.1 minutes I given by:

W = power × time

W= 294 × 86.1×60

W= 1439424 Joules

W= 1.4 ×10^6Joules

The amount of heat extracted is given by:

/QL /= K/W/ = TL/W/ /(TH - TL)

Where TL= freezing compartment temperature

TH = Outside air temperature

/QL /= 271 × 1439424 / (310 - 271)

/QL/ = 390083904/39

/QL/ = 10002151.38 Joules