5 gm of steam at 100°C is passed into 6 gm of ice at 0°C. If the latent heat of steam and ice in cal per gm are 540 and 80 respectively, then the final temperature is

Let us assume here that resulting temperature=100°C∴Q1 heat supplied by 5 gm of steam if it was to condense totally into water at 100°C= 5 X 540= 2700 caland Q2 = heat required to melt 6 gm of ice at 0°Cinto water at 0oC + heat required to raise thetemperature of 6 gm of water from 0°C to 100°C= 480 + 600 = 1080 calAs Q1 > Q2, hence whole of steam will not condense. If m' gm be the mass of steam which condenses into water at 100°C , then m' 540= 1080∴m'= 2 gm