Answer
$T_f = 2.5^{\circ}C$
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:
$Q = cm\Delta T$
$Q = (0.530~cal/g~K)(50~g)(15~C^{\circ})$
$Q = 397.5~cal$
We can find the heat energy required to melt the ice:
$Q = L_F~m$
$Q = (79.5~cal/g)(50~g)$
$Q = 3975~cal$
We can find the total heat energy required to raise the temperature of the ice, melt the ice, and increase the temperature to $T_f$:
$Q = cm\Delta T_f$
$Q = (397.5~cal)+(3975~cal)+(1.00~cal/g\cdot K)(50~g)~T_f$
We can find the heat energy removed from the water:
$Q = cm\Delta T_f$
$Q = (1.00~cal/g\cdot K)(200~g)(25-T_f)$
$Q = 5,000~cal-(200~cal/K)~T_f$
We can equate the two expressions to find $T_f$:
$(397.5~cal)+(3975~cal)+(1.00~cal/g\cdot K)(50~g)~T_f = 5,000~cal-(200~cal/K)~T_f$
$(200~cal/K)~T_f+(50~cal/K)~T_f = 5,000~cal-4372.5~cal$
$(250~cal/K)~T_f = 627.5~cal$
$T_f = \frac{627.5~cal}{250~cal/K}$
$T_f = 2.5^{\circ}C$
