Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 6 - Exercises - Page 289: 33

Answer

(a) $1.92\times10^{9}\,J$ (b) $51.4\,Cal$ (c) $2.37\times 10^{6}\,J$ (d) $681\,cal$

Work Step by Step

Using energy conversion factors, we have (a) $534\,kWh= 534\,kWh\times\frac{3.60\times10^{6}\,J}{1\,kWh}$ $=1.92\times10^{9}\,J$ (b) $215\,kJ=215\,kJ\times\frac{1000\,J}{1\,kJ}\times\frac{1\,Cal}{4184\,J}$ $=51.4\,Cal$ (c) $567\,Cal= 567\,Cal\times\frac{4184\,J}{1\,Cal}$ $=2.37\times 10^{6}\,J$ (d) $2.85\times10^{3}\,J=2.85\times10^{3}\,J\times\frac{1\,cal}{4.184\,J}$ $=681\,cal$
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