From the equation above (CH4 (g) + 2O2 (g) → 2H2O (l) + CO2), Δ H = -802.4 kJ, if we show water vapor rather than liquid water as a product.
Reversing the melting of ice and the combustion of methane, the thermochemical equation
(2H2O (l) + CO2 → CH4 (g) + 2O2 (g) ) Δ H = 890.4 kJ
and what was an endothermic process becomes exothermic, and vice versa.
2H2O (s) → 2H2O (l) Δ H = 2(6.01 kJ) = 12.0 kJ
the enthalpy change is -802.4 kJ rather than -890.4 kJ because 88.0 kJ are needed to convert 2 moles of liquid water to vapor
2H2O (l) → 2H2O (g) Δ H = 88.0 kJ
Remember this? Δ H = H (products) - H (reactants)
CH4 (g) + 2O2 (g) → 2H2O (l) + CO2, Δ H = -802.4 kJ
With these being said, where did you get 4.1798 J /g C?
I'm asking you this because specific heat of water is 4.184 J /g C.
You must find this from your Chemistry Textbook to do this
Δ Hf (kJ /mol) standard enthalpies of Formation of Inorganic Substance at 25 C.
By the way, molar mass of methane (CH4) is 16.01 g /mol. And there are 2 mol of H2O in this equation. And molar mass of water is 18.0 g /mol. You might need these so you can do the stoichiometry to solve this problem.
Good luck to you.