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Q=mcT heat problem, were did i go wrong?

  1. Dec 4, 2007 #1
    [SOLVED] Q=mcT heat problem, were did i go wrong?

    1. The problem statement, all variables and given/known data
    If a 45g sample of aluminum at 22 degrees C is given 6.0 x 10^3 J of heat, what will its final temperature be?


    2. Relevant equations
    [tex]Q = mc \Delta T[/tex]


    3. The attempt at a solution
    i found in my textbook that aluminum has a specific heat capacity of 900
    so c = 900.

    [tex]Q = mc \Delta T [/tex]
    [tex]Q = mc( T' - T ) [/tex]
    [tex]T' - T = \frac{Q}{mc} [/tex]
    [tex]T' = \frac{Q + T}{mc}[/tex]
    [tex]T' = \frac{(6.0 x 10^3 J) + 22\deg}{(0.045kg)(900)} [/tex]
     
  2. jcsd
  3. Dec 4, 2007 #2
    i got 148 Deg C which is wrong... i need 170 Deg C
     
  4. Dec 4, 2007 #3

    Kurdt

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    You've just rearranged wrongly. Its not (Q+T)/mc.
     
  5. Dec 4, 2007 #4
    so is it [tex] T' = \frac{Q}{mc} + T [/tex] ??
     
  6. Dec 4, 2007 #5

    Kurdt

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    That should work.
     
  7. Apr 8, 2009 #6
    Re: [SOLVED] Q=mcT heat problem, were did i go wrong?

    kurdt ,
    what would the worked out solution look like for the last step?
     
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