Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

Chapter 19 - Exercises and Problems - Page 349: 30

Answer

a) The efficiency is about 2/3 of the maximum possible efficiency. b) $2154\ MW$ c) $93,673 \ houses$

Work Step by Step

a) We find the maximum efficiency, being sure to use the temperatures in Kelvin: $= 1-\frac{303K}{553K}\times 100=45$% b) We first find the power that the plant is producing: $=\frac{880}{.29}=3034\ MW$ We subtract the amount that it actually gets to find that the amount wasted is: $= 3034-880=2154\ MW$ c) We find: $=\frac{2154\times10^6\ W }{23\times10^3}=93,673 \ houses$

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