Optimizing Refrigerator Efficiency: Solving for Necessary Power Draw

AI Thread Summary
To optimize refrigerator efficiency, the average heat leak of 300W requires a power draw from the wall calculated using the coefficient of performance (COP). With a typical operating temperature range of 298 K to 250 K, the maximum COP is determined to be 5.2. Using the formula W = Q/COP, the necessary power draw is approximately 57.69W. The discussion emphasizes the benefits of maintaining algebraic expressions until the final calculation for improved accuracy. This approach leads to a more precise result of 57.6W.
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


\eta = \dfrac{T_C}{T_H - T_C}

W= \dfrac{Q}{COP}

The Attempt at a Solution



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

COP = \dfrac{250}{298-250}=5.2

W = \dfrac{Q}{COP}= \dfrac{300 \, W}{5.2} = 57.69 \, W
 
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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


\eta = \dfrac{T_C}{T_H - T_C}

W= \dfrac{Q}{COP}

The Attempt at a Solution



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

COP = \dfrac{250}{298-250}=5.2

W = \dfrac{Q}{COP}= \dfrac{300 \, W}{5.2} = 57.69 \, W
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.
 
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