Heat gained by ice cubes would be equal to the - heat lost by warm water
The moles of ice is: 50.5 g / 18.0 g/mol = 2.81 mol
Heat required to melt all of the ice is equal to: 2.81 mol X 6.02 kJ/mol = 16.9 kJ = 16890 J
Now, know whether the warm water will still be above 0C when it loses this much heat:
-1690 J = 160 g (4.184 J/gC) (Delta T) Delta T = -25C
In order to solve for the final temperature, going back to include warming of the melted ice to a final temperature:
q(ice/water) = - q(warm water)
moles (Delta Hf) + m c (T2-T1) = - m c (T2-T1)
50.5 g / 18.0 g/mol = 2.81 mol
2.81 mol X 6.02 kJ/mol + 50.5g (4.184 J/gC) (T2-0) = -160g (4.184 J/gC) ( T2-80)
16916 + 211.3T2 = -669.4 T2 + 53555
36639 = 880.7 T2
T2 = 41.6 C
