We know latent heat of melting of ice is 80 calories/g
So,to convert 10g of ice at 0^@C to same amount of water at the same temperature, heat energy required would be 80*10=800 calories.
now,to take this water at 0^@C to 100^@C heat energy required will be 10*1*(100-0)=1000 calories (using,H=ms d theta where,m is the mass of water,s is specific heat,for water it is 1 C.G.S unit,and d theta is the change in temperature)
Now,we know,latent heat of vaporization of water is 537 calories/g
So,to convert water at 100^@C to steam at 100^@C heat energy required will be 537*10=5370 calories.
Now,to convert steam at 100^@C to 110^@C,heat energy required will be 10*0.47*(110-100)=47 calories (specific heat for steam is 0.47 C.G.S units)
So,for this entire process heat energy required will be (800+1000+5370+47)=7217 calories