1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Enthalpy, heat of neutralization, calorimetry

  1. Jun 29, 2010 #1
    1. The problem statement, all variables and given/known data
    A quantity of 2.00x10^2 mL of .862 M HCl is mixed with 2.00x10^2 mL of .431 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 20.48 C.

    For the process
    H+(aq) + OH-(aq) --> H2O (l)
    the heat of neutralization is -56.2 kJ/mol. What is the final temperature of the mixed solution?

    2. Relevant equations
    I'm thinking all I need is q=ms delta t

    3. The attempt at a solution
    q=(heat of neutralization per mol)(number of moles H2O)
    q= -56.2 kJ/mol x (200 mL*.862mol/1000mL) = -9.69 kJ

    If -9.69 = q = ms(T(f)-T(i))
    Then T(f) = q/ms + T(i)
    = -9690 J/(400g)(4.184g/J*C) + 20.48*C
    = 121.8 degrees C

    The answer should be 26.3 degrees C. Any help?
  2. jcsd
  3. Jun 29, 2010 #2


    User Avatar

    Staff: Mentor

    Do you know how to use calculator?

    [tex]\frac{9690}{400\times4.184} + 20.48[/tex]

    and not

    [tex]\frac{9690}{400} 4.184 + 20.48[/tex]

    (not to mention fact that you have lost minus sign, in a way luckily for you).
    Last edited by a moderator: Aug 13, 2013
  4. Jun 29, 2010 #3
    well that's embarrassing. thanks.
  5. Jun 30, 2010 #4


    User Avatar

    Staff: Mentor

    Happens :devil:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook