0.1 ~m^3 of water at 80^ C is mixed with 0.3 ~m^3 of water at 60^ C...

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