• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! 4.10

Expert-verified Found in: Page 129 ### An Introduction to Thermal Physics

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.

See the step by step solution

## Step 1 : Given information

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

## Step 2 : Explanation

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:

$P=\frac{{P}_{\text{avg}}}{COP}\phantom{\rule{0ex}{0ex}}P=\frac{300\mathrm{W}}{5.2}\phantom{\rule{0ex}{0ex}}P=57.69\mathrm{W}$ ### Want to see more solutions like these? 