Explanation

(b) Heat released by 5 kg of water when its temperature falls from $20℃\ to\ 0℃\ is$,
$\mathcal{Q}_{1} = m_{1}c_{1}\ \mathrm{\Delta}\theta_{1} = (5)\left( 10^{3} \right)(20 - 0) = 10^{5}\ cal$
When 2 kg ice at $-$20$℃$ comes to a temperature of 0$℃$, it takes an energy
$\mathcal{Q}_{2} = m_{2}c_{2}\ \mathrm{\Delta}\theta_{2} = (2)(500)(20) = 0.2 \times 10^{5}\ cal$
The remaining heat
$\mathcal{\ \ \ \ \ \ \ \ \ \ \ \ \ \ Q =}\mathcal{Q}_{1} - \mathcal{Q}_{2} = 0.8 \times 10^{5}\ cal\ $will melt a mass $m\ $of the ice, thus
$m = \frac{\mathcal{Q}}{L} = \frac{0.8 \times 10^{5}}{80 \times 10^{3}} = 1$ kg
So, the temperature of the mixture will be 0$℃$, mass of water in it is 5$+$1=6 kg and mass of ice is 2$-$1=1 kg