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4.10

Expert-verifiedFound in: Page 129

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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

The power drawn from wall is 57.69 W.

Assume temperatures T1 = 250 K and T2 = 298 KPower leak = 300 W

COP can be found by using $\mathrm{COP}=\frac{{T}_{1}}{{T}_{2}-{T}_{1}}$

Substitute the values

$\mathrm{COP}=\frac{250}{298-250}\phantom{\rule{0ex}{0ex}}COP=5.2$

The amount of power drawn from the wall is:

$P=\frac{{P}_{\text{avg}}}{COP}\phantom{\rule{0ex}{0ex}}P=\frac{300\mathrm{W}}{5.2}\phantom{\rule{0ex}{0ex}}P=57.69\mathrm{W}$

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