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Another problem dealing with heat transfer

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data
    How much Fe at 200.0C must be placed in 100.00g of water at 20C so that the temperature of both will be at 35.0?


    2. Relevant equations
    Q=MCdeltaT


    3. The attempt at a solution
    Energy required to raise water is 6300 J
    So I did by:
    (keep in mind I used sig figs)
    Q=100.00(4.18)15
    Q=6300

    And then..

    I assumed I must plug in 6300 as a negative to the FE to get the amount needed
    I did
    -6300=x(.449)165
    x=-6300/74.1
    x=85g

    I go to check my work by doing taking out the final temp and trying to solve for it by using the grams

    I did so by:
    (85).449(Tf-200.0)=(100.0)4.18(35.0-Tf)
    38Tf-7600=14600-418Tf
    456Tf=22200
    And got 48.7 Which is not so close to 35C


    Am I checking this wrong or am I doing the problem wrong?
     
  2. jcsd
  3. Sep 27, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I think you have done the problem correctly and the problem with the check is that the "35-Tf" should be "Tf - 20".

    You know, your work would be so much clearer if you put in a few words like this:
    Heat lost by iron = heat gained by water
    mC(delta T) = mC(delta T)
    m(.449)(165) = 100*4.18*15
    m = 100*4.18*15/(.449*165) = 84.6 g
     
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