Heat gain by ice to melt = mLf = 10 × 80 = 800 cal Heat gain by 10 g water to raise its temperature from 0°C to 100°C = 10 × 1 × 100 = 1000cal Total heat gain = 1800cal Mass of steam converted into water ∴ 1800 = m × 540 ⇒ m = 3.33 g ∴ Equilibrium temperature 100°C Amount of water = 10 + 3.33 = 13.33g Amount of water = 5 - 3.33 = 1.67g