Answer
$T_f = 0^{\circ}C$
Work Step by Step
We can find the heat energy required to melt the ice:
$Q = L_F~m$
$Q = (79.5~cal/g)(500~g)$
$Q = 39,750~cal$
We can find the total heat energy removed from the tea:
$Q = cm\Delta T_f$
$Q = (1.00~cal/g\cdot K)(500~g)(70-T_f)$
$Q = 35,000~cal-(500~cal)~T_f$
Since the heat energy required to melt the ice is greater than the heat energy lost from the tea, the equilibrium temperature is $0^{\circ}C$ and some of the ice remains.
$T_f = 0^{\circ}C$
