Question

We have of ice at -11.9°C and it is ultimately completely converted to liquid at 9.9°C. For ice, use a specific heat (Cs) of 2.01 J/gK, for liquid water use a specific heat of 4.18 J/gK, and ΔHfusion = 6.01kJ/mole, Molar mass H20 = 18.01g/mol. If 1 mole of ice at -11.9°C is heated by the combustion of 0.010 mole of CH4(g), what would the final temperature be? ΔHf°(CH4) = -74.6kJ/mol; ΔHf°(CO2)=-393.5kJ/mol; ΔHf°(H2O(g))=-241.8kJ/mol.
Balanced equation:
CH4 + 2O2 = CO2 + 2H2O

Answer 1

dHcomb = (n*dHformation products) - (n*dHformation products)
I ran through this and estimate about 802.5 kJ/mol so the heat emitted by 0.01 mol CH4 will be 8.025 kJ or 8,025 J.

I would do the rest of it in pieces.
It will take
q1 = heat to raise T of solid ice from -11.9 to solid ice at zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) = ?

Then it will take 6.01 kJ/mol to melt the ice at zero to liquid H2O at zero. Since you had 1 mol that will take 6.01 kJ or 6010 J and that is q2.

Now you are faced with how much will the remaining heat raise the temperature
so the question is how much heat is remaining. That is
8025 J - q1 - q2 = ? = q3

Finally, plug in q3 and see how much Tf is.
q3 = mass H2O x specific heat H2O x (Tf-Ti). Solve for Tf; you know Ti is zero.

I think the answer is about 20C but that's just a close answer.

Answer 2

What's the Tfinal - Tinitial? i tried (0 - (-11.9)) but I don't think its the right thing.