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Homework Help: Quick Thermo Question

  1. Sep 28, 2010 #1
    Question: The heat capacity of a bomb calorimeter was determined by burning 6.79 g of methane (energy of combustion = -802 kJ/mol CH4) in the bomb. The temperature changed by 10.8 degrees C.
    a. What is the heat capacity of the bomb?
    b. A 12.6 g sample of acetylene, C2H2, produced a temperature increase of 16.9 degrees C in the same calorimeter. What is the energy of combustion of acetylene (in kJ/mol) ?

    2. Relevant equations
    I wasn't even sure if there were specific equations to use...I did figure out the problem, but I just did so by logic...kinda.

    3. The attempt at a solution
    So for Part A, I first converted the 6.79 g of CH4 to mols, and I got about 0.423 mols. Since I didn't remember if there was a formula for this or not, I kind of just logically figured out to multiply by the -802 kJ/mol, so that mols would cancel out. I then divided by the temperature change, 10.08 degrees C, so that my units would be in kJ/C...which is what heat capacity is measured in.

    I ended up getting -31.4 kJ/mol. When I compare this to the answer in my book, it's not supposed to be negative. I can't figure out why.

    For Part B, I basically used the same process. I looked at all of the data I had, and saw that I needed to get to kJ/mol. I converted the 12.6 g C2H2 to mols; I got about 0.483 mols. I took my answer from Part A, -31.4 kJ/C, and multiplied by the 16.9 degree temp change to cancel out celcius. I then divided by the mols to get -1098.67 kJ/mol.

    So...why are my signs wrong? I believe all of the actual math is right. And is there a specific formula or easy way to do this?

    Also, when it says "energy of combustion"...is that E, for energy, or H, for heat? Thanks!
  2. jcsd
  3. Sep 29, 2010 #2


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    Staff: Mentor

    You are on the right track.

    Energy of combustion is just an enthalpy of the reaction.

    [tex]Q = m\: c\: \Delta T[/tex]

    Sign is a matter of convention, you probably forgot that minus means energy is produced in the reaction.

    And this

    is a little bit off.
    Last edited by a moderator: Aug 13, 2013
  4. Sep 29, 2010 #3
    So if energy of combustion is just enthalpy, it's delta H? I think my problem is just in identifying what information the problems give me. And I would use that formula...?

    And by a little bit off, did you mean the math? When I actually did it, I got 33 point something, but my book said -31.4, so I just put that. I rounded at one point or another.
  5. Sep 29, 2010 #4


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    Staff: Mentor


    That's the most basic thing in all heat balance questions. Amount of heat is mass times specific heat times delta T. Heat capacity of calorimeter is mc.

    Watch your units.
    Last edited by a moderator: Aug 13, 2013
  6. Sep 29, 2010 #5
    Okay, thank you so much! We went over it in class, and it helped me understand it better, too.

    And just to double check signs...the energy flowing out of the methane (the "system") would be -802, but the energy is also flowing into the calorimeter (the "surroundings") so it's positive 802 with respect to this question?
  7. Sep 30, 2010 #6


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    Staff: Mentor

    Yes, that's the way it is done.
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