Explanation

(b) Heat required to melt 1 g of ice at 0$℃$ to water at 0$℃$
= 1$\times 80\ $cal.
Heat required to raise temperature of 1 g of water from 0$℃\ $to 100$℃ = 1 \times 1 \times 100$ cal
Total heat required for maximum temperature of 100$℃ = 80 + 100 = 180\ $cal
As one gram of steam gives 540 cal of heat when it is converted to water at 100$℃,$ therefore, temperature of the mixture would be 100$℃$