Answer
The final temperature is $~~0^{\circ}C$
Work Step by Step
Let $T_f$ be the equilibrium temperature.
We can find the heat energy required to increase the temperature of the ice:
$Q = cm\Delta T$
$Q = (0.530~cal/g~K)(100~g)(15~C^{\circ})$
$Q = 795~cal$
We can find the heat energy required to melt the ice:
$Q = L_F~m$
$Q = (79.5~cal/g)(100~g)$
$Q = 7950~cal$
We can find the heat energy removed from the water:
$Q = cm\Delta T_f$
$Q = (1.00~cal/g\cdot K)(200~g)(25-T_f)$
$Q = 5,000~cal-(200~cal)~T_f$
Since the heat energy required to raise the temperature of the ice and then to melt the ice is greater than the heat energy lost from the water, the equilibrium temperature is $0^{\circ}C$ and some of the ice remains.
The final temperature is $~~0^{\circ}C$
