The motor in a refrigerator has a power output of 210 W. The freezing compartment is at -3.0ºC and the outside air is at 26ºC. Assuming that the efficiency is 85%of the ideal, calculate the amount of heat that can be extracted from the freezing compartment in 15 min.
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Answer:
Solution:-
First we have to find out the coefficient of performance K and work done W.
To obtain coefficient of performance K, substitute 270 K for TL and 299 K for TH in the equation K = 0.85 (TL / TH - TL),
K = 0.85 (TL / TH - TL)
= 0.85 (270 K / 299 K - 270 K)
= 7.91
To obtain work done W, substitute 210 W for power P and 15 min for time t in the equation W = (P) (t),
W = (P) (t)
= (210 W) (15 min)
= (210 W) (15 min) (60 s/1 min)
= (210 W) (900 s)
= (1.89×10^5 Ws) (1 J/1 Ws)
= 1.89×10^5 J
To obtain the amount of heat QL that can be extracted from the freezing, substitute 7.91 for coefficient performance K and 1.89×105 J for work done W in the equation
QL = (K) (W),
QL = (K) (W)
= (7.91) (1.89×10^5 J)
= 1.50×10^6 J
From the above observation we conclude that, the amount of heat QL that can be extracted from the freezing compartment in 15 min would be 1.50×10^6 J.
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