# Refridgerator Problem: Where Did I Screw Up?

1. Mar 11, 2009

### EstimatedEyes

1. The problem statement, all variables and given/known data

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?

2. Relevant equations
COP=Q_c/W
Q_h=Q_c + W
Q=cm(deltaT)
Q=mL_f
P=E/t

3. 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

Last edited: Mar 11, 2009
2. Mar 12, 2009

### Andrew Mason

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