Fifty kg of water at 0oC must be frozen into ice in a refrigerator. The room temperature is 20oC. The latent heat of fusion of water is 3.33x105 J kg-1. What is the minimum power required if the freezing is to take place in one hour?
m = 50 kg
Lf = 3.33x105 J kg-1.
Twater = 0oC
Troom = 20oC
[itex]\Delta[/itex]t = 1 hour = 3600s
[itex]\Delta[/itex]U = Q1 + Q2 - W
W = Q1 + Q2
(Q1/Q2) = -(T1/T2)
The Attempt at a Solution
Since there are 50kg of water and Lf = 3.33x105 J kg-1, the total amount of energy that is required to turn the water into ice is 1.665x107 J. For the freezing to occur in one hour, at least 4625 J must be taken from the water per second. This led me to believe that the minimum power of the refrigerator must be 4625 watts. However, according to the selected answers in the back of the book, the minimum power should come out to be 340 watts.
After reading this, I assumed that 1.665x107 J (energy required to turn all water into ice) corresponds to Q1, or the amount of heat extracted from the system. I then assumed that T1=20oC (room temp), but without being given the value of T2, Q2, efficiency of the engine, etc, I am unsure how to proceed with the problem, perhaps I am simply overlooking something or do not have one of the necessary equations.
Any help is appreciated, Thanks!