In-class problem

I think a number of people were confused at how to approach the problem we did in class on Wednesday. The problems was:

"fuel" reacts with oxygen to produce carbon dioxide gas and water gas. If 10.00g of "fuel" is burned in excess oxygen and all of the energy is transferred to 5.00L of water initially at 11.24degC, what is the final temperature of the water?

I'll pick a fuel none of you had, benzene, C6H6(l). This is a coupled-systems problem, the benzene will burn to produce/liberate heat in an enthalpy process, then the 5.00L sample of water will absorb the heat in a heat capacity process. Start with a balanced equation:
2 C₆H₆(l) + 15 O₂(g) --> 12 CO₂(g) + 6 H₂O(g)
Now calculate the {delta}H for the reaction from the standard enthalpies of formation found in the table in the back of your book.
2(-49.03^kJ/_mol) + 15 (-0^kJ/_mol) + 12(-393.509^kJ/_mol) + 6(-241.818^kJ/_mol) = -6271.08^kJ/_{mol rxn}
NOTE: change sign on reactants, don't change sign on products.
10.00g of benzene does not represent a "mol of reaction", so we need to scale that number to the amount of fuel being used:
(10.00g C₆H₆) / (78.113 ^g/_mol) = 0.1280mols benzene
(0.1280 mols C₆H₆) (1 mol rxn / 2 mols C₆H₆) = 0.06401 mols rxn
(0.06401 mols rxn) (6271.08^kJ/_{mol rxn}) = 401.4kJ of energy released by the reaction

OK, now the heat capacity part of the problem. I'm putting 401400J of heat into this 5.00L sample of water and changing its temperature. Use the units on heat capacity to set up the problem correctly:
(401400J) (1 g.degC / 4.184J) (1 / 5000g) = 19.19degC
This is the change in temperature, so the final temperature of the water must be (11.24+19.19)degC = 30.43degC

Other questions, let me know...