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 349: 26


a) 287 J b) 38.5% c) 363.94 K d) 5,338.2 W

Work Step by Step

a) We find that the amount of work done is: $ W = 745-458=287 \ J$ b) We know that the efficiency is equal to: $=100(1-\frac{458}{745})=38.5 $% c) We multiply the temperature of the warmer reservoir by the efficiency to find: $T=(592)(1-.385)=363.94 K$ d) The power output is equal to the work done each cycle times the number of cycles per second, for Power is Joules per second. Thus, it follows: $P = 287 \times 18.6 = 5,338.2 \ W$
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