Thermodynamic Question (refrigerator)

[gotrilum]

I'm having a little trouble in completing this problem. I would appreciate your answer if anyone knows how to solve it. Thanks

An ideal refrigerator removes heat at a rate of 0.15kW from an interior temperature of 1 degree celsius. The exhaust from the refrigerator is 45 degrees celsius. How much electrical power is used.

*Note that this is a refrigerator problem (reverse process from a carnot engine)

I have found the efficiency using Th/Tc - 1 = .29 I do not know how to proceed after this.

 

[xxChrisxx]

Remember efficieny is always

what you get / what you pay for.

so in this case its
Ql/Win

and for the reversible carnot this comes to:

COP = Ql/(Qh-Ql)

You must remember the for refrigiration cycles its not technically efficiecny, its coefficient of performance we use. This avoids the confision of having efficienys over 100%.

Where you were going wrong was you were taking the efficieny for a carnot cycle and applying it directly.

 

[Andrew Mason]

I'm having a little trouble in completing this problem. I would appreciate your answer if anyone knows how to solve it. Thanks

An ideal refrigerator removes heat at a rate of 0.15kW from an interior temperature of 1 degree celsius. The exhaust from the refrigerator is 45 degrees celsius. How much electrical power is used.

*Note that this is a refrigerator problem (reverse process from a carnot engine)

I have found the efficiency using Th/Tc - 1 = .29 I do not know how to proceed after this.

Coefficient of Performance is Q_c/W. W = Q_h - Q_c.

So:

COP = (Q_c /(Q_h - Q_c)

1/COP = Q_h/Q_c - 1

For a Carnot cycle, this becomes:

1/COP = T_h/T_c - 1

You have to determine the COP from the temperatures. Using the value for Q_c you can determine W.

AM