Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Urgent -- Bomb Calorimetry Combustion Reaction

  1. Nov 2, 2017 #1
    1. The problem statement, all variables and given/known data
    The combustion of 0.1577 g benzoic acid increases the temperature of a bomb calorimeter by 2.51°C. a) Calculate the heat capacity of this calorimeter. (The energy released by combustion of benzoic acid is 26.42 kJ/g.)

    A 0.2123-g sample of vanillin is then burned in the same calorimeter, and the temperature increases by 3.25°C.
    b) What is the energy of combustion per gram of vanillin?
    C)How about per mol?

    2. Relevant equations

    qv=mCv delta T

    3. The attempt at a solution
    a)26.42 kj/g x 0.1557g= 4.113594 kj=q
    4.113594 kj= (0.1577g)(Cv)(2.51 Celsius) -> Cv=10.39240375 kj/celcius

    b) (0.2123g)(10.39240375 kj/celsius)(3.25 celsius) = 7.17... kj/g ->7.17.../0.2123g = 33.78...

    c) (33.78 kj/1g) x (152.15g/1 mol)= 5139.627 kj/mol

    All those are wrong apparently... Help?
  2. jcsd
  3. Nov 3, 2017 #2


    Staff: Mentor

    I don't understand what you calculate here, but the result is wrong and the units in the calculation are inconsistent as well. If your calorimeter would need 10.4 kJ to get heated by 1 K (don't use Celsius for differences), then 4 kJ couldn't heat it by more than 1 K, and certainly not by 2.51.
    Here the units are inconsistent as well.

    (c) looks fine, it just starts with a wrong value from (b).
  4. Nov 3, 2017 #3


    User Avatar

    Staff: Mentor

    You are confusing specific heat capacity (which requires some scaling factor, like mass or number of moles, to calculate the heat capacity) with a heat capacity, which is already a capacity of the whole calorimeter.

    Do you know the difference between intensive and extensive properties?
  5. Nov 3, 2017 #4
    In (b), you should have divided by 0.2123 gm, not multiplied by 0.2123 gm.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?