Heating Your Kitchen with Your Refrigerator

AI Thread Summary
To determine how long a 3.50 kW space heater must run to match the heat output of a refrigerator freezing 1.46 kg of water at 21.4°C, the total heat required was calculated as 618,427 J. Using the coefficient of performance (COP) of 3.13, the work done was found to be 197,581 J. Dividing this work by the heater's power output of 3,500 W resulted in a time of 56.45 seconds. However, there was confusion regarding the heat transfer process and where the heat went, indicating a potential error in the calculations. Clarification on the heat transfer mechanism is necessary to resolve the discrepancy.
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Homework Statement



How long would a 3.50 kW space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 3.13) does when it freezes 1.46 kg of water at 21.4°C into ice at 0°C?

m=1.46 kg
c_water=4186 J/kg
Delta T= 21.4°C
L_fusion= 3.34x10^5 J/kg
COP= 3.13
P= 3.50 kW

Homework Equations



Q_c = m x c_water x Delta T + m x L_f

COP = Q_c/W

P = W/t

The Attempt at a Solution



I plugged everything into the equation for the heat and got 618427 J. Then I divided that by the coefficient of performance to get that work = 197581 J. Then, I divided that by 3500 W for the power and found that t = 56.45 s. And apparently that's not the answer. I'm not sure where I went wrong!
 
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