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Specific Heat of CO2 and Methane Combustion

  1. Aug 5, 2008 #1
    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.
     
  2. jcsd
  3. Aug 7, 2008 #2
    I did a mistake... the Energy of Combustion of CH4 is 50.1 MJ/Kg instead "KJ/Kg".

    Thanks.
     
  4. Aug 21, 2008 #3
    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.
     
  5. Aug 22, 2008 #4

    GCT

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    One slight complication is due to the fact that carbon dioxide is produced with the combustion of the methane.
     
  6. Aug 23, 2008 #5
    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.
     
  7. Aug 24, 2008 #6

    GCT

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    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.
     
  8. Aug 26, 2008 #7
    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.
     
  9. Aug 29, 2008 #8

    chemisttree

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    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.
     
  10. Aug 30, 2008 #9
    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!
     
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