As the quantities have been given in CGS units, therefore, answers have been worked out in same units.
Values used
Latent Heat of fusion of water = 79.5cal//gm =333 kJ//kg
Specific heat of water=1cal//gm=4.19kJ//kg
Latent heat exchanged for change of state Q_l=mL,
where m and L are mass and latent heat of the substance.
Amount of heat exchanged is given as DeltaQ=msDeltat,
where m is the mass, s is the specific heat and Deltat is change in temperature.
Case 1.
Let the final temperature of the mixture be water at t_1^o C.
Heat lost by 5gm of water at 90^oC to become Water at t_1^o C is given as
DeltaQ_(lost)=5 times 1times (90-t_1^o)
Simlarly heat gained by ice at 0^o to water at t_1^o C is given as
DeltaQ_(gai n ed)=heat gained by 5gm Ice at 0^oC for the change of state+Heat gained by 5gm of water at 0^oC to become Water at t_1^o C
DeltaQ_(gai n ed)=5 times 79.5+ 5 times 1 times (t_1^o-0)
Since, DeltaQ_(lost)= Delta Heat_(gai n ed)
5 times 1times (90-t_1^o)=5 times 79.5+ 5 times 1 times (t_1^o-0)
implies (90-t_1^o)= 79.5+ t_1^o
implies 2t_1^o= 10.5^oC
Case 2.
Let the final temperature of the mixture be water at t_2^o C.
Heat lost by 500gm of water at 90^oC to become Water at t_2^o C is given as
DeltaQ_(lost)=500 times 1times (90-t_2^o)
As in Case 1. heat gained by ice at 0^o to water at t_2^o C is given as
DeltaQ_(gai n ed)=heat gained by 5gm Ice at 0^oC for the change of state+Heat gained by 5gm of water at 0^oC to become Water at t_2^o C
DeltaQ_(gai n ed)=5 times 79.5+ 5 times 1 times (t_2^o-0)
Since, DeltaQ_(gai n ed)= Delta Heat_(lost)
500 times 1times (90-t_2^o)=5 times 79.5+ 5 times 1 times (t_2^o-0)
implies 100(90-t_2^o)= 79.5+ t_2^o
implies 101t_2^o= 8920.5^oC
