JEE Advanced Model Paper 8 with solutions for online practice

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