Specific Heat of CO2 and Methane Combustion

[Kasper_NYC]

Hi,

Could you please confirm these estimations for me:

Knowing that the Energy of Combustion of CH4 is 50.1 KJ/Kg, and that the specific heat for the CO2 at 273.15 K is approximately 0.81 KJ/Kg*K; please, could you confirm that in order to increase the temperature of 1 liter of gas of CO2 at 273.15K and 100 KPa (density = 1.98 Kg/m^3, so 0.00198 Kg), from 273.15 K to 573.15 K, it is only needed to combust 0,0000096 Kg of CH4. It is supposed an ideal / isolated system, without heat looses.

Q = m * c * AT

Q = 0.00198 * 0.81 * 300 = 0.481 KJ

Then: 0.481 / 50.1 * 10^3 = 9.6 * 10^-6 Kg, approximately just 0.015 liter.

It seems to be a very little portion of CH4 to me.

Thanks for your help.

PS – I apologize for my English, I am learning it.

 

[Kasper_NYC]

I did a mistake... the Energy of Combustion of CH4 is 50.1 MJ/Kg instead "KJ/Kg".

Thanks.

 

[Kasper_NYC]

Hi, I got the answer from another forum, and yes, it is right what I estimated:

http://es.answers.yahoo.com/questio...dFdX4woazKIX;_ylv=3?qid=20080807085536AAK6vg8

This wasn’t a coursework or homework question, because I am not a student in any school or university. It was just a personal question about one project that I am thinking it. I am sorry I posted it in the wrong forum.

 

[GCT]

Hi,

Could you please confirm these estimations for me:

Knowing that the Energy of Combustion of CH4 is 50.1 KJ/Kg, and that the specific heat for the CO2 at 273.15 K is approximately 0.81 KJ/Kg*K; please, could you confirm that in order to increase the temperature of 1 liter of gas of CO2 at 273.15K and 100 KPa (density = 1.98 Kg/m^3, so 0.00198 Kg), from 273.15 K to 573.15 K, it is only needed to combust 0,0000096 Kg of CH4. It is supposed an ideal / isolated system, without heat looses.

Q = m * c * AT

Q = 0.00198 * 0.81 * 300 = 0.481 KJ

Then: 0.481 / 50.1 * 10^3 = 9.6 * 10^-6 Kg, approximately just 0.015 liter.

It seems to be a very little portion of CH4 to me.

Thanks for your help.

PS – I apologize for my English, I am learning it.

One slight complication is due to the fact that carbon dioxide is produced with the combustion of the methane.

 

[Kasper_NYC]

Thanks GCT for your answer, but CH4 can’t combust with CO2, so it will need for example Oxygen as well in the system; and in the problem I supposed a hypothetical isolated system, where the heat from the CH4 combustion would be transferred to the CO2 without losses; because that, the only one mass indicate in the formula it was the relative to the liter of CO2.

Thanks again for your answer.

 

[GCT]

Thanks GCT for your answer, but CH4 can’t combust with CO2, so it will need for example Oxygen as well in the system; and in the problem I supposed a hypothetical isolated system, where the heat from the CH4 combustion would be transferred to the CO2 without losses; because that, the only one mass indicate in the formula it was the relative to the liter of CO2.

Thanks again for your answer.

Methane requires oxygen for combustion and carbon dioxide is a direct byproduct of this combustion process ; in order for a more accurate reading you need the specific heat of a calorimetric system.

 

[Kasper_NYC]

GCT,

Also the combustion of CH4 in a poor oxygen atmosphere could give Carbon Monoxide and Water… and the reading of the combustion could be a lot more complicate with so many other index if needed, but that wasn’t the case. Just I wanted to confirm the estimation that I posted.

Thanks again.

 

[chemisttree]

I think that GCT was referring to the fact that the calorimeter will have a slightly greater amount of CO2 than 1L. The energy of combustion you used only accounts for the energy released during the combustion process of methane + oxygen (both at standard conditions,T=273.15K) and producing some amount of CO2 (also at standard state, T=273.15K). You need to account for the energy required to heat the additional CO2 generated from the combustion process from standard conditions to 573.15 K as well.

It is probably just a minor point.

 

[Kasper_NYC]

Chemisttree and GCT,

Thanks again for your answers, but I think there is a misunderstanding in all this, because I didn’t explain completely my proposal. The system that I suggested was supposed to be “ideal and isolated, without heat losses”. So, what I wanted to suggest was a system where the CO2 would receive the heat directly from the Methane combustion, without considering any other variable.

It was like the CO2 was in a hypothetical / special container, that would receive directly the heat from the Methane combustion; but being the methane and the CO2 separated, without any contact between them… so the atmospherics conditions, the sub-products of the methane combustion, as any other index or variable would not account in the problem.

What I wanted to confirm it was just that the combustion of such a little amount of Methane could increase 300 degrees 1 liter of CO2, in an ideal system.

Thanks!