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 cal and Q2 = heat required to melt 6 gm of ice at 0°C into water at 0°C + heat required to raise the temperature 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 gmThus, the temperature of the mixture is 100°C and the mixture contains 8 gm of water and 3 gm of steam at a temperature of 100°C .