Chemistry. How much energy (in kJ) is required to boil 729 g of water? 1 mol water = 40.67 kJ

Question

1 mol water = 40.67 kJ

Answer 1

Hello, S F.,

There is no temperature given on the starting temperature of the water, so I'll assume it is already at 100^oC. At this point the information we need is the heat of vaporization for water. "1 mol water = 40.67kJ" is a little confusing, I'll rewrite is as the heat of vaporization:

H_vap = 40.67 kJ/mole (for water)

40.67 Joules of energy is required to vaporize 1 mole of water (at 100^oC, and at 1 atm standard pressure).

We have 729 grams of water, but need it expressed as moles in oder to use the H_vap constant.

Moles H₂O = (729 grams)/(18 grams/mole) = 40.5 moles

Now we can calculate the heat required:

(40.5 moles H₂O)*(40.67 kJ/mole H₂O) 1647 kJ

1647 kJ to boil 729 grams of water starting at 100^oC

[Nte that this does not include any heat associated with getting the water temperature to 1900^oF, and that the steam produced is at 100^oC, also.

Bob