Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 19 - Exercises and Problems - Page 350: 41


a) 17.4% b) 83.3%

Work Step by Step

a) Using our knowledge of the different processes, we find: $e = \frac{W}{Q} \times 100$ $e = \frac{[(1.5)(12-6)]+[(2.5)(-18+6)]+[1.5(18-36)]+[2.5(36-12)]}{60+9} \times 100$ $e = \frac{[(1.5)(12-6)]+[(2.5)(-18+6)]+[1.5(18-36)]+[2.5(36-12)]}{69} \times 100$ $e = \frac{12}{69} \times 100=17.39$% b) We use the equation for the efficiency of the Carnot engine (equation 19.3) to find: $e = 1-\frac{(2)(3)}{(6)(6)}=83.33$% Note, Carnot's engine is more efficient because, as explained in the book, it is the most efficient when the high temperature is as high as possible, so it gets more efficient the warmer it gets.
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