Heat energy required for #5g# of water at #0^@C# to get converted to water at #100^@C# is latent heat required + heat required to change its temperature by #100^@C#=#(80*5)+(5*1*100)=900# calories.
Now,heat liberated by #5g# of steam at #100^@C# to get converted to water at #100^@C# is #5*537=2685# calories
So,heat energy is enough for #5g# of ice to get converted to #5g# of water at #100^@C#
So,only #900# calories of heat energy will be liberated by steam,so amount of steam that will be converted to water at the same temperature is #900/537=1.66g#
So,the final temperature of the mixture will be #100^@C# in which #5-1.66=3.34g# of steam and #5+1.66=6.66g# of water will coexist.
values used during solving this problem are,
Latent heat for melting of ice =#80#calorie #g^-1#,latent heat for vaporisation of water=#537# calorie #g^-1# and specific heat of water =#1# calorie #g^-1# #@C^-1#
