Question

To an insulated container with 100.0 g H2O (l) at 20.0 ºC, 175 g steam at 100.0 ºC and 1.65 kg of ice at 0.0 ºC are added. What mass of ice remains un-melted after equilibrium is established?

Answer 1

The problem clearly suggest that ice and water are in equilibrium at 0 deg.c

So heat given by 100 gm of water (from 20 deg.c to 0 deg.c )= mass of water* specific heat* temperatre difference

100*4.18*20=8360 joules

Heat given by steam in condensing at 100 deg.c = mass of steam* latent heat of vaporization = 175g*2260 j/g=395500 j/g

from 100 deg.c to 0 deg.c steam give sensible heat and this is = 175*4.18*(100-0)=73150 joules

total heat energy = 73150+395500+8360=477010 joules

Latent heat of fusion of water= 334 J/g

ice that can be melted = 477010/334=1428 gm =1.428 kg

Rest of the water remains as ice only and this is = 1.65-1.428=0.222kg