Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 19 - Exercises and Problems - Page 361: 16


Please see the work below.

Work Step by Step

We know that the heat absorbed from the water is $Q_c=mL_f$ We plug in the known values to obtain: $Q_c=(0.670)(334000)$ $Q_c=223780J$ For a refrigerator $COP=\frac{Q_c}{W}$ This can be rearranged as: $W=\frac{Q_c}{COP}$ We plug in the known values to obtain: $W=\frac{223780}{4.2}$ $W=53000J$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.