Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 5 - Questions and Problems - Page 226: 5.86



Work Step by Step

Heat lost by water at 74.3$^{\circ}$C=heat gained by water at 24.8$^{\circ}$C $\implies -q_{1}=q_{2}$ As $q=mc\Delta T$, we can write $-88.6\,g\times c\times(T_{f}-74.3^{\circ}C)=57.9\,g\times c\times(T_{f}-24.8^{\circ}C)$ $\implies -88.6\,g\times T_{f}+88.6\,g\times74.3^{\circ}C=57.9\,g\times T_{f}-57.9\,g\times24.8^{\circ}C$ $88.6\,g\times74.3^{\circ}C+57.9\,g\times24.8^{\circ}C=(57.9\,g+88.6\,g)T_{f}$ $\implies T_{f}=\frac{88.6\,g\times74.3^{\circ}C+57.9\,g\times24.8^{\circ}C}{57.9\,g+88.6\,g}=54.7^{\circ}C$
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