Chemistry (4th Edition)

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

Chapter 5 - Section 5.4 - Calorimetry - Checkpoint - Page 209: 5.4.3


a) 88.3°C

Work Step by Step

$C_{cal}=\frac{q_{cal}}{\Delta T}=\frac{-q_{rxn}}{\Delta T}$ $\implies \Delta T= \frac{-q_{rxn}}{C_{cal}}=\frac{-(-318\,kJ)}{5.01\,kJ/^{\circ}C}=63.47^{\circ}C$ Given $T_{1}=24.8^{\circ}C$ $\Delta T= T_{2}-T_{1}\implies T_{2}=\Delta T+ T_{1}$ $=63.47^{\circ}C+ 24.8^{\circ} C=88.3^{\circ}C$
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