Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 6 - Sections 6.1-6.10 - Exercises - Problems by Topic - Page 288: 49a

Answer

$7.6\times10^{2}{^{\circ}C}$

Work Step by Step

To find the temperature we are going to use the equation $q= m\times C_{s}\times\Delta T$ where $q$ is the heat in joules, $m$ is the mass in grams, $C_{s}$ is the specific heat (found on table on pg. 257), and $\Delta T$ is the change in temperature in $^{\circ}C$. Now we plug in the information from the question into our equation. $2.35\ kJ= 2,350J$ as $1\ kJ= 1000J$. The heat capacity of gold is $0.128$. $2,350J= 25g\times 0.128\frac{J}{g \times^{\circ}C}\times\Delta T$ Rearranging we get: $\frac{2,350}{25\times 0.128}^{\circ}C=\Delta T$ So the change in temperature is $734^{\circ}C$. The temperature is then $730+27.0=757^{\circ}C$ with the correct number of significant figures is $7.6\times10^{2}{^{\circ}C}$
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