Answer
$m_f = 60~g$
Work Step by Step
In part (c), we found that $T_f = 0^{\circ}C$
We can find the total heat energy removed from the tea:
$Q = cm\Delta T_f$
$Q = (1.00~cal/g\cdot K)(500~g)(70~C^{\circ})$
$Q = 35,000~cal$
We can find the amount of ice that melted:
$Q = L_F~m$
$m = \frac{Q}{L_F}$
$m = \frac{35,000~cal}{79.5~cal/g}$
$m = 440~g$
Therefore, since we started with $500~g$ of ice, $~~60~g~~$ of ice remains.
$m_f = 60~g$
