Chemistry: Molecular Approach (4th Edition)

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

Chapter 6 - Exercises - Page 289: 47

Answer

$4.7\times 10^{5}\,J$

Work Step by Step

We find: $m=(1.0\,g/mL)(1.50\times10^{3}\,mL)=1500\,g$ $c= 4.18\,J/g\cdot\,^{\circ}C$ $\Delta T= 100.0^{\circ}C-25.0^{\circ}C=75.0^{\circ}C$ $q=mc\Delta T$ $=(1500\,g)(4.18\,J/g\cdot\,^{\circ}C)(75.0^{\circ}C)$ $= 4.7\times 10^{5}\,J$
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