Question

Calculate the amount of energy in kilojoules needed to change 405 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

Q Calculate the amount of energy in kilojoules needed to change 405 g of water ice at −10 ∘C to steam at 125 ∘C. The following constants may be useful:

Given data :- 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 = 6001 J /mol

ΔHvap=+40.67 kJ/mol = 40670 J/mol

Solution :-

405 g water

Moles of water = 405 g / 18.0148 g per mol = 22.4815 mol

Q total = q ice + delta H fus + q water + delta H vap + q steam

= (n*c*delta T)+(n*delta Hfus)+(n*c*delta T)+(n*delta Hv)+(n*c*delta T)

= (22.4815*36.57*10)+(22.4815*6001)+(22.4815*75.40*100 )+(22.4815*40670 ) + (22.4815* 36.04*25 )

= 1.25*10^6 J

Now lets convert joules to kJ

1.25*10^6 J * 1 kJ / 1000 J = 1.25*10^3 kJ

So the amount of the heat needed is 1.25*10^3 kJ