Answer
$T_f = 5.3^{\circ}C$
Work Step by Step
We can find the heat energy required to melt the ice:
$Q = L_F~m$
$Q = (79.5~cal/g)(500~g)$
$Q = 39,750~cal$
Let $T_f$ be the equilibrium temperature.
We can find the total heat energy required to melt the ice and raise the temperature:
$Q = (39,750~cal)+cmT_f$
$Q = (39,750~cal)+(1.00~cal/g\cdot K)(500~g)T_f$
We can find the total heat energy removed from the tea:
$Q = cm\Delta T_f$
$Q = (1.00~cal/g\cdot K)(500~g)(90-T_f)$
We can equate the two expressions for the heat energy to find the equilibrium temperature $T_f$:
$(39,750~cal)+(1.00~cal/g\cdot K)(500~g)T_f = (1.00~cal/g\cdot K)(500~g)(90-T_f)$
$(500~cal)~T_f+(500~cal)~T_f = 45,000~cal-39,750~cal$
$(1000~cal)~T_f = 5,250~cal$
$T_f = 5.3^{\circ}C$
