One gram of ice is mixed with one gram of steam. After thermal equilibrium is reached, the temperature of the mixture is
(Latent heat of fusion of ice; Lice=80 cal/g, latent heat of condensation of vapour; LV=540 cal/g)
The correct option is A 100∘C
Latent heat of fusion of ice; Lice=80 cal/g
Latent heat of condensation; LV=540 cal/g of vapour.
Specific heat capacity of water; c=1 cal/g∘C
Heat supplied by steam if it is to condense totally and convert into water at 100∘C.
Heat required by ice to convert totally into water at 0∘C
Heat required by ice to convert totally into water at 100∘C
Since Q1>Q3, entire steam will not condense and the final temperature of mixture of water and steam will be 100∘C.