Calculate the volume of natural gas (methane) required to bring 3.9 L of water to a boil. Assume the initial temperature of the water is 39 C, and the methane is at 1 bar and 25 C. Assume that methane is an ideal gas, the specific heat capacity of water to be constant and equal to 4.184 J/g/K over this temperature range, and the density of water to be 1.0 g/cm3. The combustion of methane is given by

CH4 + 2 O2 → CO2 + 2 H2O ΔHcomb = -890.36 kJ/mol.

1 answer

What amount of heat do you need to raise the temperature of H2O from 39 C to 100 C.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial).
q = 3,900g x 4.184 x (100-39) = ? J energy needed.

Next, you know you obtain 890.36 kJ/mol CH4. How many moles CH4 do we need to supply ? J energy? That is
890,360 J/mol x y mol = ? J (from the water part above.) Solve for y mol

Last step, you have y mol CH4, which occupies 22.4L at STP. Convert y mol to L and convert to conditions in your problem if those conditions listed are not STP. I get confused between UK conditions and US conditions and new recommendations of IUPAC.