Explanation

(a) Let the final temperature of the mixture be $t$. Heat lost by water
at 80$℃\ \ \ \ \ \ \ = ms\mathrm{\Delta}t$
=$\ 0.1 \times 10^{3} \times s_{water} \times (80{^\circ} - t)$
$(\because m = V \times d = 0.1 \times 10^{3}kg$)
Heat against by water at 60$℃$
=$\ 0.3 \times 10^{3} \times s_{water} \times (t - 60{^\circ})$
According to principle of Calorimetry,
Heat lost = Heat against
$0.1 \times 10^{3} \times S_{water} \times (80{^\circ} - t) = 0.3 \times 10^{3} \times S_{water} \times (t - 60{^\circ})$
or $(80{^\circ} - t) = 3 \times (t - 60{^\circ})$
or $4t = 260℃$
or $t = 65℃$