Physics: Principles with Applications (7th Edition)

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

Chapter 6 - Work and Energy - General Problems - Page 168: 87

Answer

(a) $1~kWh = 3.6\times 10^6~J$ (b) 420 kWh (c) $1.5 \times 10^9~J$ (d) The cost is $50 The monthly bill does not depend on the rate at which the energy is used. The bill depends on the total amount of energy that is used each month.

Work Step by Step

(a) $1~kWh = (1\times 10^3~W)(3600~s) = 3.6\times 10^6~J$ (b) Let's assume there are 30 days in one month. So, $(0.580~kW)(24~h/day)(30~days) = 420~kWh$ (c) $(420~kWh)(3.6\times 10^6~J/kWh) = 1.5 \times 10^9~J$ (d) $Cost = (420~kWh)(\$0.12/kWh) = \$50.40$ The monthly bill does not depend on the rate at which the energy is used. The bill depends on the total amount of energy that is used each month.
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