## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 6 - Exercises and Problems - Page 107: 77

#### Answer

Please see the work below.

#### Work Step by Step

We know that energy-efficient: $(0.225KW)(0.11t hour)(\$ 0.095 )=0.00235125t$standard:$(0.425KW)(0.20t\space hour)(\$0.095)=0.008075t$ Now the difference is $0.00235125t-0.008075t=1150-850$ We can solve it for time as: $t=\frac{1150-850}{0.008075-0.00235125}=52413h=6.0yr$

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