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Heat of Combustion

  1. Jan 12, 2009 #1
    What's the formula for calculating this, given: grams, heat capacity of substance, and temperature change?

    How about heat combustion per kJ/Mol?

    Thanks
     
  2. jcsd
  3. Jan 12, 2009 #2

    mgb_phys

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    You would write a balanced equation for
    Fuel + oxygen -> product (typically water + CO2)

    Then account for all the energy needed to break the bonds in the original fuel and oxygen plus the energy released when bonds form in the product.
    For real world energy values the water formed is usually steam so you have to account for the heating and vapourisation energy needed.

    Or you can lookup the heat of combustion/gram for lots of different fuels on the web or in databooks
     
    Last edited: Jan 12, 2009
  4. Jan 12, 2009 #3
    Thanks for the reply.
    Don't need an equation, according to solution; just a plug and chug formula. It seems like it is heat capacity * temp diff ALL divided by the grams converted to mols -- wondering why?
     
  5. Jan 12, 2009 #4

    mgb_phys

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    For simply burning the temperature difference shouldn't come into it - if you burn something in liquid oxygen (-200degC) you get just the same energy out!

    There is a temperature effect if you have to take into account the density of the fuel (as in liquified gas, or jet fuel at high altitudes) or if you have a non-condensing boiler where some energy is lost as hot steam.
     
  6. Jan 12, 2009 #5
    A .922 sample of napthalene (C10H8) is burned in a calorimeter that has a heat capacity of .944 kJ/K. The temperature of the calorimeter rose from 15.73 C to 19.75 C. Calculate the heat of combustion for this chemical in kJ/mol.

    I'm sure there's an equation, wondering how?
     
  7. Jan 12, 2009 #6

    mgb_phys

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    The heat of combustion doesn't depend on the actual temperature - the temperature change of the calorimeter is telling you how much energy was given off.
    You get the point - if the fuel burned and heated the calorimeter from 115.73 C to 119.75 C you would get the same result.

    You need to work out how many moles of fuel you used, and how much energy the calorimter received that gives you the joules/mole for the fuel.
     
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