Question

Calculate the amount of energy in kilojoules needed to change 369 g of water ice at −10 ∘C to steam at 125 ∘C. The following constants may be useful:

Cm (ice)=36.57 J/(mol⋅∘C)

Cm (water)=75.40 J/(mol⋅∘C)

Cm (steam)=36.04 J/(mol⋅∘C)

ΔHfus=+6.01 kJ/mol

ΔHvap=+40.67 kJ/mol

Express your answer with the appropriate units.

Answer 1

(369 g H2O) / (18.01532 g H2O/mol) = 20.48 mol H2O

(36.57 J/(mol⋅∘C)) x (20.48 mol) x (0∘C − (−10 ∘C)) =7490.4 J to warm the ice to 0 ∘C

(6.01 kJ/mol) x (20.48 mol) = 123.0848 kJ to melt the ice

(75.40 J/(mol⋅∘C)) x (20.48 mol) x (100 - 0) ∘C = 154419.2 J to warm the melted ice to 100 ∘C

(40.67 kJ/mol) x (20.48 mol) = 832.9216 kJ to vaporize the water

(36.04 J/(mol⋅∘C)) x (20.48 mol) x (125 - 100) ∘C = 18452.48 J to warm the steam to 125∘C

7.490 kJ + 123.0848 kJ + 154.419 kJ + 832.9216 kJ + 18.452 kJ = 1136.36 kJ total