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