Question 8 of 14 presanted by Map do -open Physics sapling leaming What is the b

Question

Question 8 of 14 presanted by Map do -open Physics sapling leaming What is the best possible coefficient of performance for a refrigerator that cools an environment at-33.0 and exhausts heat to another environment at 45.0 ? Number How much work in joules must be done for a heat transfer of 4.725 x 10% J from the cold environment? Number What is the cost of doing this, if the work costs 10.5 cents per 3.60 x 10% J(a kilowatt-hour)? Number cents How many joules of heat transfer occurs into the warm environment? Number Previous Give Up &View; Solution Check Answer Next Exit

Explanation / Answer

coefficient of performance = Th/(Th-Tc) = (273+45)/((273+45)-(273-33)) = 4.1

COP = Q/W

W = Qc/COP = (4.725*10^6)/4.1 = 1152439.02 J

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cost = (1152439.02/(3.6*10^6))*10.5 = 3.36 cents

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W = Qh - Qc

Qh = W + Qc = 1152439.02+(4.725*10^6) = 5877439.02 J

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