Question

The motor in a refrigerator has a power output of 250 watts. The freezing compartment is at 270 K and outside air at 300 K. Assuming ideal efficiency, what is the amount of heat that can be extracted from the freezing compartment in 10 minutes? What is the shortest time in which 10 kg of water at 273 K can be converted into ice? J = 4.2 × 10-3 J K ca-1 .​

Answer 1

Answer:

1.08 × 10^6 J

Explanation:

The work done by the motor in t=10.0min is  

W = Pt = (200 W)(10 min)(60 s/min)

W = 1.2 × 10^5 J  

The heat extracted is then

 Q(L) = K W = T(L) W / T(H) - T(L)

                   = (270 K) (1.2 × 10^5 J) / 300 K - 270 K

                   = 1.08 × 10^6 J