Heat energy liberated for water at #35# degree celcius to get converted to zero degree celcius is #(0.04*1000*1(35-0)) =1400 C# (using #H=ms d theta#)
Heat energy required for ice to attain a temperature of zero degrees celcius is #(0.05*100*0.5 (0-(-30))= 750 C# (specific heat for ice is #0.5# CGS units)
So,ice will increase its temperature to zero degrees celcius.
Now,remaining heat energy i.e #(1400-750)=650 J# will try to convert the ice to water at zero degrees,
So,let #m# gm of ice will get converted to water,
So, #m*80 = 650# (latent heat for melting of ice is #80C/g#)
or, #m=8.125 g#
So, the final temperature of the mixture will be #0# degree celcuis in which #(0.04*1000+8.125)=48.125 g# of water and #(0.05*1000-8.125)=41.875 g# of ice will coexist.
