\(0.1 \mathrm{~m}^{3}\) of water at \(80^{\circ} \mathrm{C}\) is mixed with \(0.3 \mathrm{~m}^{3}\) of water at \(60^{\circ} \mathrm{C}\). The final temperature of the mixture is

Let the final temperature of the mixture be \(t\)
Heat lost by water at \(80^{\circ} \mathrm{C}\)
\(=\mathrm{ms} \Delta \mathrm{t} \)
\( =0.1 \times 10^{3} \times \mathrm{s}_{\text {water }} \times\left(80^{\circ}-\mathrm{t}\right) \)
\( \left(\because \mathrm{m}=\mathrm{V} \times \mathrm{d}=0.1 \times 10^{3} \mathrm{~kg}\right)\)
Heat gained by water at \(60^{\circ} \mathrm{C}\)
\(=0.3 \times 10^{3} \times \mathrm{s}_{\text {water }} \times\left(\mathrm{t}-60^{\circ}\right)\)
According to principle of calorimetry
Heat lost \(=\) Heat gained
\(\therefore 0.1 \times 10^{3} \times \mathrm{s}_{\text {water }} \times\left(80^{\circ}-\mathrm{t}\right)\)
\(=0.3 \times 10^{3} \times s_{\text {water }} \times\left(t-60^{\circ}\right)\)
or \(\quad\left(80^{\circ}-\mathrm{t}\right)=3 \times\left(\mathrm{t}-60^{\circ}\right)\)
or \(\quad 4 \mathrm{t}=260^{\circ}\)
or \(\quad \mathrm{t}=65^{\circ} \mathrm{C}\)