Four ice cubes at exactly 0 ∘c with a total mass of 52.5 g are combined with 160 g of water at 90 ∘c in an insulated container. (δh∘fus=6.02 kj/mol, cwater=4.18j/g⋅∘c). if no heat is lost to the surroundings, what is the final temperature of the mixture?

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