Answer
54.7$^{\circ}$C
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
$-88.6\,g\times c\times(T_{f}-74.3^{\circ}C)=57.9\,g\times c\times(T_{f}-24.8^{\circ}C)$
$\implies -88.6\,g\times T_{f}+88.6\,g\times74.3^{\circ}C=57.9\,g\times T_{f}-57.9\,g\times24.8^{\circ}C$
$88.6\,g\times74.3^{\circ}C+57.9\,g\times24.8^{\circ}C=(57.9\,g+88.6\,g)T_{f}$
$\implies T_{f}=\frac{88.6\,g\times74.3^{\circ}C+57.9\,g\times24.8^{\circ}C}{57.9\,g+88.6\,g}=54.7^{\circ}C$
