Optimizing Refrigerator Efficiency: Solving for Necessary Power Draw

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patrickmoloney
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Homework Statement


Suppose that heat leaks into your kitchen refrigerator at an average rate of 300W. Assuming ideal operation, how much power must it draw from the wall.

Homework Equations


[tex]\eta = \dfrac{T_C}{T_H - T_C}[/tex]

[tex]W= \dfrac{Q}{COP}[/tex]

The Attempt at a Solution



A typical refrigerator works between [itex]298 \, K[/itex] [itex]250 \, K[/itex]. The maximum possible coefficient of performance is

[tex]COP = \dfrac{250}{298-250}=5.2[/tex]

[tex]W = \dfrac{Q}{COP}= \dfrac{300 \, W}{5.2} = 57.69 \, W[/tex]
 
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patrickmoloney said:

Homework Statement


Suppose that heat leaks into your kitchen refrigerator at an average rate of 300W. Assuming ideal operation, how much power must it draw from the wall.

Homework Equations


[tex]\eta = \dfrac{T_C}{T_H - T_C}[/tex]

[tex]W= \dfrac{Q}{COP}[/tex]

The Attempt at a Solution



A typical refrigerator works between [itex]298 \, K[/itex] [itex]250 \, K[/itex]. The maximum possible coefficient of performance is

[tex]COP = \dfrac{250}{298-250}=5.2[/tex]

[tex]W = \dfrac{Q}{COP}= \dfrac{300 \, W}{5.2} = 57.69 \, W[/tex]
Looks right.
The are many benefits in working algebraically, only plugging in numbers right at the end. One is improved accuracy. In the present case you would have got 57.6W exactly.
 
Yeah I think I'll do it algebraically so until the final line. Thanks very much.