How to calculate change in combustion calorific value based on change in temperature?

I have it on good authority (GPA-2145 09) that burning 1 M3 (ideal) of methane at 15 Deg. C and 101.325 kPaa, and then letting the combustion products cool back to 15 Deg. C will provide a net of 37.708 MJ of energy. How would a person recalculate the heating value if a different reference / combustion condition was used? I tried the following but didn't get what I expected:

CH4 + 2 O2 --> CO2 + 2 H2O

molar density at 15 C and 101.325 = P / R T (ideal). = 101.325 / (8.31451 * 288.15) * 1000 mol/kmol= 42.2923 mol

Molar HV = 37.708 MJ / 42.2923 mol = 891.604 kJ/mol at 15 C

Delta H (CH4) = Cp * moles * Delta T = 0.035 kJ/(mol.K) * 0.55556 K = 19.4 J / mol

Deltah H (O2) = Cp * moles * delta T = 0.55556 K * 0.029 kJ/(mol.K) * 2 (moles o2 per mole CH4) = 32.2 J / mol CH4

Delta H (CO2) = Cp * Moles * delta T = 0.55556 K * 0.037 kJ/(mol.K) = 20.6 J / mol

Delta H ( H2O) = Cp * moles * delta T = 0.55556 K * 0.075327 kJ/(mol-K) * 2 mol H2O per mol CH4 = 83.703 J / mol CH4

So, heating up the CH4 and O2 from 15 C to 15.555 C takes 51.6 J/mol.

Cooling down the CO2 and H2O to 15.555 C instead of 15 gives you 104.3 J/mol less.

You should have 155.9 J/mol less at 15.555 C.

The other reference conditions I know of are 60 F and 14.73 PSI which equate to 15.555 C and 101.56 kPaa. Methane is published to have a "BTU value" of 1010 BTU/Cu. Ft. Does this work out?

At 15.555 C and 101.56 kPaa, there are 101.56/8.31451*288.71 = 42.3082 mol per m3. So, the answer should be (891.6-0.1559) * 42.308 = 37.715 MJ/m3.

In fact, the result should be the equivalent of 1010 BTU/Cu. Ft which is 1010*1.055056/1000 / (0.3048^3) = 37.6315 MJ/m3. The molar heating value at 15.555 should work out to be 37.6315 / 42.3082 = 889.46 kJ/mol.

The difference per mole should be 2.142 kJ, but I can only account for a tenth of that. Clearly my methods are unsound! But I don't know why.

Can anyone clue me in?

Edited December 1, 201015 yr by Ban Yan