If \(T\) be the resultant temperature, then heat gained by \(5 \mathrm{~kg}\) (at \(20^{\circ} \mathrm{C}\) ) water
\(\begin{aligned}
& H_{\text {gain }} =m c(T-20) \quad(\because \text { given, } m=5 \mathrm{~kg}) \\
\Rightarrow & H_{\text {gain }} =5 c(T-20) \quad \ldots (i)
\end{aligned}\)
Heat lost by \(10 \mathrm{~kg}\) water at \(60^{\circ} \mathrm{C}\) is given as
\(H_{\text {loss }}=10 c(60-T)\) ...(ii)
By the principle of calorimetry
\(\begin{array}{ll}
& H_{\text {gain }}=H_{\text {loss }} \\
\Rightarrow & 5 c(T-20)=10 c(60-T) \\
\Rightarrow & T-20=120-2 T \\
\Rightarrow & 3 T=140 \\
& T=46.67^{\circ} \mathrm{C} \simeq 47^{\circ} \mathrm{C}
\end{array}\)