1. Not finding help here? Sign up for a free 30min 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!

Basic thermodynamic chemistry, heat transfer

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data

    How many grams of steam at 100 deg. Celcius would be required to raise the temperature of 35.8 g solid benzene from 5.5 deg Celcius to 45.0 deg Celcius?
    Assume that heat is only transferred from the steam (not the liquid water) and that the steam/water and benzene are separated by a glass wall and do not mix.


    2. Relevant equations
    Boiling point of benzene 5.5 C
    Δfus of benzen is 9.87 kJ/mol
    Specific heat of benzene is 1.63 J/g.C and 4.18 for water
    ΔHvap for steam at 100 C is 40.7 kJ/mol

    im not sure about this part, since i cant find it in my book.

    so the total heat of the system Qtotal =0= Qgained+Qlost
    or Qgained = -Qlost

    Qgained = molesbenzen ×ΔH+mbenzen×Cbenzen×(45-5.5)

    Qlost = moleswater×ΔHvap + mwater×C×water×(45-100)

    3. The attempt at a solution
    So it seem like my equation above are wrong
    because m comes out as a negative value, which made no sense.
    I am pretty sure I got the heat gained correct.
    I just uncertain about the heat lost of the steam.

    Please give me some help.
    Thanks for your time
     
    Last edited: Nov 17, 2012
  2. jcsd
  3. Nov 17, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    Wording is confusing. I guess you have to assume water vapor cooled down to 100 deg C and condensed, but the final temperature of water was 100 deg C.
     
  4. Nov 17, 2012 #3
    it is the exact wording from the book, and this is just general chemistry, so I dont think there is any complexity in the problem.

    so for the heat lost part, if the final temperature is 100 then the initial is 45?
    and still, if it is (100-45) based on the above equation I cannot get m as a positive value???
     
  5. Nov 17, 2012 #4

    Borek

    User Avatar

    Staff: Mentor

    Oops, sorry, there is no cooling of the vapor, only condensation.
     
  6. Nov 17, 2012 #5
    So what should I do with this problem now?
    Is my equation correct?
    or should it be Qgained = Qlost instead?
     
  7. Nov 17, 2012 #6

    Borek

    User Avatar

    Staff: Mentor

    Just leave the mcΔT part for water.
     
  8. Nov 17, 2012 #7
    What about the negative sign in the equation Qgained = -Qlost
    Should it be there?
    even though we take out the mcΔT of water, with the negative sign, the answer still comes out to be negative =(
     
  9. Nov 18, 2012 #8

    Borek

    User Avatar

    Staff: Mentor

    This sign thing can be confusing, as it can be done in many ways. What is important is to not mix conventions and stick to one. Sorry, no time to give more elaborate explanation now.

    Good thing is usually when you mix conventions you get a nonsense result, which makes it easy to spot something is wrong. Negative mass is a sure sign something went wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook