University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 20 - The Second Law of Thermodynamics - Problems - Exercises - Page 677: 20.6

Answer

(a) $𝜖 = 59.4 \%$ (b) $|Q_C|= 4060 J $ $W = 5940 J $

Work Step by Step

(a) $𝜖 = 1 - \frac{1}{r^{\gamma -1 }} $ $𝜖 = 1 - \frac{1}{9.50^{1.40 -1 }} $ $𝜖 = 1 - \frac{1}{9.50^{0.40}} $ $𝜖 = 59.4 \%$ (b) $ 𝜖 = 1 - \frac{|Q_C|}{|Q_H|} $ $|Q_C|= (1 - 𝜖 )Q_H$ $|Q_C|= (1 - 0.594 )(10000 J) $ $|Q_C|= 4060 J $ The work output is $W = |Q_H| - |Q_C| $ $W = 10000J - 4060 J $ $W = 5940 J $
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.