Answer
$ T_f = 66.5 ^oC$
Work Step by Step
Mass of water, $m_{water}$ at 80 celcius = 130 g
Mass of ice, $m_{ice} $ at 0 celcius = 12 g
$T_f$ is the equilibrium temperature of both ice and water
To find the $T_f$, use the equation
$Q_1 = Q_2$
$m_{ice} C (T_f-T_i) + L_Fm_{ice}= m_{water} C (T_f -T_i)$
Plug in the necessary information and rearrange the equation to solve for $T_f$
$L_Fm_{ice} + cm_{ice} (T_f -0^o) + cm_{water} (T_f - 80^o) = 0$
$(4190J/Kg.^oC)(0.012kg)T_f +(4190J/Kg.^oC)(0.130kg)(T_f) = -(333000 J/kg)(0.012 kg) + (4190J/Kg.^oC)(0.130kg)(80^o C)$
$ 50.28 T_f + 544.7 T_f = 39580 J$
$ 594.98J/^oC \space T_f = 39580 J$
$ T_f = \frac{39580 J }{594.98J/^oC}$
$ T_f = 66.5 ^oC$
