Question is attached as picture as it is from Quiz 6.3

It is a very interesting question ...in this case we have two systems :
1) 20 g steam at 100 degree C 
2)100 g of ICE at -10 degree C  in calorimeter of 10 gm water equivalent
Now you added both the systems in a calorimeter.
System 1 going to lose heat which will be gained by system 2.
Now let us calculate total amount of heat lost by system 1 in converting 20 g of  steam into water at the same temperature.
H2O(g) ----> H2O (l)
q = mass * ∆ H(condensation) in cal/g = 20*540 cal/g =-10800cal (lost)
Now we can say that system one has 20 g of water at 100 deg C (initial temperature)
Now system 2 will gain heat released when steam is condensed to liquid water.
This heat gained by system 2 will do the following changes in it
First it will raise the temperature to zero deg C(∆ t1) and then it will melt the ice(m*∆ H(fusion) and finally it will further raise the temp from zero to some temperature x deg C (∆ t2)
It means
10800 cal (gained by system 2) = ms∆ t1 (ice)+ m*∆ H(fusion) + ms∆ t 2(liquid water)
10800 = 100(0.5) (10)+100*80+(100)*1(x-0)
Get the value of x
x= 23 deg C
Now this has become the initial temperature of system 2 which is now in liquid water form.
When these systems together achieve the equilibrium in calorimeter ,final temperature of both systems will be same.
-Heat lost (system1) = heat gained(system2)
- ms∆t(system 1) = ms∆t(system 2)
-20*1*(x-100) = 100 *1(x – 23.0)
Solve it for x
X=35.83 deg C