Refridgerator Problem: Where Did I Screw Up?

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The discussion revolves around calculating the time a 2.90 kW space heater would need to run to generate the same heat as a refrigerator freezing 1.45 kg of water. The user attempts to find the total heat output from the refrigerator using the coefficient of performance (COP) and energy equations but consistently arrives at an incorrect time of 745 seconds. The key error identified is not properly accounting for the relationship between the work done by the refrigerator and the COP in the calculations. The correct approach involves dividing the total heat delivered by the refrigerator by the COP to find the work done. Clarifying these calculations is essential to arrive at the correct answer.
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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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EstimatedEyes said:

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
You have to divide your expression for Qh by COP:

The total heat delivered by the refrigerator is Qh = Qc + W = W(COP+1)

W = Qc/COP. That is what you have to work out by substituting:

Q_c = m(c\Delta T + L_f)

AM
 
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