Posted by Melissa on Tuesday, November 9, 2010 at 7:34pm.
I would do this although it may be the long way around. I would convert all of the "moles" to grams.
1300 kJ/mol = 1300000/46.069 = ??J/g for the heat of combustion.
65.6 J/mol = 65.6/46.00 = xx J/g for the heat capacity (specific heat) of ethanol.
1.00 L ethanol = 1000 mL and that has a mass of 1000 x 0.789 g/mL = xx grams.
Let X = mass ethanol removed.
Xg*heat combustion = (mass ethanol-X)*specific heat ethanol* (Tfinal-Tinitial).
Solve for X and 789-X.
I obtain something a little less than 3 grams to be removed and that will heat the remainder from 25 to 90.
Post your work if you get stuck.
Again Melissa?
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