Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 18 - Electric Currents - General Problems - Page 523: 66

Answer

a) $40.7 b) 1400kg

Work Step by Step

a) $P=4.4kWh+2.4kWh+3.0kWh+2.0kWh=11.8kWh$ $11.8kWh\frac{$0.115}{kWh}=\$1.4$ $\$1.4\times30=\$40.7$ b) $\frac{11.8kWh}{1day}\times\frac{365days}{1year}=4307kWh/year$ $\frac{7500kcal}{1kg}=\frac{8.7kWh}{1kg}$ At 100% efficiency, $495kg$ of coal. At 35% efficiency, $1400kg$
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