Refridgerator Problem: Where Did I Screw Up?

[EstimatedEyes]

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

 

[Andrew Mason]

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

 

[nlightner]

Dear student,

First of all, it is important to note that the equation COP=Q_c/W is only valid for a heat pump, not for a refrigerator. A refrigerator is a type of heat pump, but it is specifically designed to remove heat from a space (e.g. a kitchen) rather than transfer heat from a lower temperature source to a higher temperature sink.

To solve this problem, we can use the equation Q_h=Q_c+W, where Q_h is the amount of heat removed from the kitchen by the refrigerator, Q_c is the amount of heat rejected by the refrigerator, and W is the work done by the refrigerator. We know that Q_c=mL_f, where m is the mass of water frozen and L_f is the latent heat of fusion for water. Therefore, Q_h=mL_f+W.

Next, we can use the equation Q=mC(deltaT) to calculate the amount of heat required to freeze the water. We know that the initial temperature of the water is 19.4°C and the final temperature is 0°C, so deltaT=19.4°C. We also know that the mass of water frozen is 1.45 kg. Plugging these values into the equation, we get Q=mC(deltaT)=(1.45 kg)(4186 J/kg°C)(19.4°C)=120,280 J.

Now, we can set Q_h=Q and solve for W, the work done by the refrigerator. This gives us W=Q_h-Q=mL_f-W. Plugging in the values we know, we get W=(1.45 kg)(33.5e4 J/kg)+(120,280 J)=487,420 J.

Finally, we can use the equation P=E/t to calculate the time (t) it would take for a 2.90 kW space heater to do the same amount of work (E) as the refrigerator. Rearranging the equation, we get t=E/P=487,420 J/2.90 kW=168.07 seconds.

In conclusion, a 2.90 kW space heater would have to run for approximately 168 seconds to put into a kitchen the same amount of heat as a refrigerator does when it freezes 1.45 kg of water at 19.4°C into ice at 0°C. The error in your reasoning was using the equation COP=Q_c/W, which is not applicable for a refrigerator. It