Calculations of how much water evaporated and what is left

[apbuiii]

1. The problem statement, all variables and given/known data
Suppose you put a coffee pot containing 300 g of water at 20 C on the stove. You then add 350 kJ of heat to the water, which heats the water to 100 C and the water starts to boil. After all the heat is added to the water, then how much water do you still have left? The heat of fusion for water is 6.01 kJ/mol and the heat of vaporization is 40.7 kJ/mol.


2. Relevant equations

q=mass(specific heat)(Tf-Ti); q=moles(delta-enthalpy)

3. The attempt at a solution
I first found how much energy it would take to evaporate all of the 300 grams. (100-20)(300g)(4.18) + (40.7 KJ)(300)(1mol/18g)(1000J)= 778653 J. So I then set up a ratio that if 778653 evaporates 300 grams then how many grams will 350000 J evaporate. 778653/300 = 350000/X. That gave me 135 grams evaporated. So I then minused that from the 300 giving me 165 grams left. Seemed reasonable to me but I got it wrong! Please help, but showing the steps to find the answer would be even more helpful! Thanks

[Borek]

There are two heat sinks - one (water heating to 100 °C) doesn't depend on the amount of evaporated water, other (water evaporation) depends. When you use ratio it is equivalent to assumption that you have heated only part of water, while you have to heat it all.

Calculate how much heat was used to heat the water, excess heat was consumed by evaporation.

[apbuiii]

Okay, so I did (300)(4.18)(100-20)= 100320 J. Then I subtracted that from 350000J to get 249680 J. This is where I get a little confused.... Do I then do this: 249680=mols(40.7KJ)(1000J). That gave me 6.13 moles of water and then I change that to grams to give me 110.4 grams of water that it can evaporate. I then subtract 300-110.4 to give me the grams of water left 190 grams. Is that correct? Thank you so much...

[Borek]

I have not checked the numbers, but the logic is OK.