Answer
The coffee has cooled $13.5~C^{\circ}$
Work Step by Step
We can find the heat energy required to melt the ice:
$Q = L_F~m$
$Q = (333~kJ/kg)(0.0120~kg)$
$Q = 4000~J$
Let $T$ be the equilibrium temperature.
We can find the total heat energy required to melt the ice and raise the temperature:
$Q = (4000~J)+cmT$
$Q = (4000~J)+(4187~J/kg\cdot K)(0.0120~kg)T$
We can find the total heat energy removed from the coffee:
$Q = cm\Delta T$
$Q = (4187~J/kg\cdot K)(0.130~kg)(80.0-T)$
We can equate the two expression for the heat energy to find the equilibrium temperature $T$:
$(4000~J)+(4187~J/kg\cdot K)(0.0120~kg)T = (4187~J/kg\cdot K)(0.130~kg)(80.0-T)$
$(544.31~J/C^{\circ})~T+(50.244~J/C^{\circ})~T = 43,544.8~J-4000~J$
$(594.554~J/C^{\circ})~T = 39,544.8~J$
$T = \frac{39,544.8~J}{594.554~J/C^{\circ}}$
$T = 66.5^{\circ}C$
We can find the amount the coffee has cooled:
$\Delta T = 80.0^{\circ}C-66.5^{\circ}C = 13.5~C^{\circ}$
The coffee has cooled $13.5~C^{\circ}$
