Question

Calculate the amount of energy in kilojoules needed to change 207 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

Answer 1

no of moles of water = weight of water / molar mass of water

= 207 / 18.015

= 11.49 moles

energy to warm ice from –10 to 0

Q1 =mc delta t = 36.57 J/(mol⋅∘C) x 11.49 g/mol (0-(-10)) = 4201.893 J

2. energy to melt ice

Q2 = 6010 J/mol x 11.49 g/mol = 69054.9 J

3, energy to warm water from 0 to 100

Q3 = 75.40 J/(mol⋅∘C) x 11.49 g/mol (100-0) = 86634.6 J

4. energy to boil water

Q4 = 40670 J/mol x 11.49 g/mol = 467298.3 J

5. energy to heat steam from 100 to 125

Q5 = 36.04 J/(mol⋅∘C) x 11.49 g/mol (125-100 ºC) = 10352.49 J

now Q = Q1 + Q2 + Q3 + Q4 + Q5

Q = 4201.89 + 69054.9 + 86634.6 + 467298.9 + 10352.49

Q = 637542.78 J

Q = 637.5427 kJoules