Heat required by 1g ice at 0°C to melt into 1g water at 0°C,

Q 1 = mL (L = latent heat of fusion)

= 1 × 80 = 80 cal (L = 80 cal/g)

Heat required by 1g of water at 0°C to boil at 100°C,

Q 2 = ms (s = specific heat of water)

= 1 × 1(100 – 0) (s = 1 cal/g°C)

= 100 cal

Thus total heat required by 1g of ice to reach a temperature of 100°C,

Q = Q 1 + Q 2

= 80 + 100 = 180 cal

heat available with 1g of steam to condense into 1g of water at 100°C

Q' = mL' (L' = latent heat of vaporization)

= 1 × 536 cal (L' = 536 cal/g)

= 536 cal

Obviously, the whole steam will not be condensed and ice will attain temperature of 100°C. Thus, the mixture of temperature is 100°C