Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 20 - Entropy and the Second Law of Thermodynamics - Problems - Page 606: 40a

Answer

$|Q_H| = 49368 J$ or $49.37 kJ$

Work Step by Step

$K_C = \frac{|Q_L|}{|Q_H| - |Q_L|}$ Solve for $Q_H$ $K_C(|Q_H| - |Q_L|) = |Q_L|$ $|Q_H| - |Q_L|) = \frac{|Q_L|}{K_C}$ $|Q_H| = \frac{|Q_L|}{K_C} + |Q_L| $ $|Q_H| = |Q_L| [\frac{1 + K_C}{K_C}]$ Where $K_C = 5.7$ and $Q_L = 42000 J$ $|Q_H| = 42000 J [\frac{1 + 5.7}{5.7}]$ $|Q_H| = 49368 J$ or $49.37 kJ$
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.