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Homework Statement
How long would a 2.90 kW space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 2.58) does when it freezes 1.45 kg of water at 19.4°C into ice at 0°C?
Homework Equations
COP=Q_c/W
Q_h=Q_c + W
Q=cm(deltaT)
Q=mL_f
P=E/t
The Attempt at a Solution
I get something around 745s each time I do it, but that is wrong. The way I have been doing it is to find out the amount of energy it takes to change the water to ice which would be the work done by the refrigerator. Therefore Qc for the fridge would be coefficient of performance times work. Add Qc to work to get the amount of heat that exits the fridge and then divide by the number of watts used by the space heater to get the amount of time it would need to be run. Where did I go wrong in this reasoning? Thank you.
COP=Q_c/W therefore Q_c=COP*W
Q_total=mc(deltaT) + mL_f
Q_h=Q_c+W
Q_h=(COP+1)*(mc(deltaT) + mL_f)
Q_h/T=Power
[(COP+1)*(mc(deltaT) + mL_f)]/Power=T
[(2.58+1)*((1.45)(4186)(19.4) + (1.45)(33.5e4))]/2.9e3=745s
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