Question

Calculate the amount of energy in kilojoules needed to change 333 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

Solution :

Mass of water = 333 g

Moles of water(m) = 333 / 18 = 18.5 mol H₂O

We have to solve this question step by step ;

Q₁ = Amount of heat required to convert from ice at -10 ^oC to ice at 0 ^oC = mC_m(ice)T₁

Q₁ = Amount of heat required to convert from ice at 0 ^oC to water at 0 ^oC = mH_fus

Q₂ = Amount of heat required to convert from water at 0 ^oC to water at 100 ^oC = mC_m(water)T₂

Q₃ = Amount of heat required to convert from water at 100 ^oC to steam at 100 ^oC = mH_vap

Q₄ = Amount of heat required to convert from steam at 100 ^oC to steam at 125 ^oC = mC_m(steam)T₃

Amount of heat energy,

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

or, Q = mC_m(ice)T₁ + mH_fus + mC_m(water)T₂ + mH_vap + mC_m(steam)T₃

or, Q = [18.5 36.57 (0-(-10))] + [18.5 6.01 1000] + [18.5 75.40 (100-0)] + [18.5 40.67 1000] + [18.5 36.04 (125-100)]

or, Q = 6765.45 + 111185 + 139490 + 752395 + 16668.5

or, Q = 1026503.95 J

or, Q = 1026.5 kJ

Therefore, amount of energy = 1026.5 kJ