Answer
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 $T_{1}=24.8^{\circ}C$
$\Delta T= T_{2}-T_{1}\implies T_{2}=\Delta T+ T_{1}$
$=63.47^{\circ}C+ 24.8^{\circ} C=88.3^{\circ}C$
