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Specific Heat and aluminum

  1. Nov 25, 2008 #1
    This is a really simple problem, but I just can't remember how to combine the container and the water.

    1. The problem statement, all variables and given/known data
    An aluminum container weighing 200g contains 500g of water at 20*c. A 300g shot of the same aluminum at 100*c is put into the water. The specific heat for the aluminum is 0.215cal/(g*c). What is the final temperature for the entire system?

    2. Relevant equations
    dQ = c*m*dT

    3. The attempt at a solution
    (T - 20*c) x [(500g x 1cal/(g*c)) + (200 x 0.215cal/(g*c))] = (T - 100*c) x 300g x 0.215cal/(g*c)

    T = 9.1216*c

    However this answer seems really off.
  2. jcsd
  3. Nov 25, 2008 #2


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    Energy is conserved so, energy gained by:
    Econtainer = mct = 200 * 0.215 *(T-20)
    Ewater = mct = 500 * 1* (T-20)
    Equals energy lost by:
    Eshot = mct = 300 * 0.215* (100-T)

    (Final temperature T)
  4. Nov 25, 2008 #3
    Wait, so would you just make Eshot = Ewater and then solve for T and then Eshot = Ewater and solve for a different T?
  5. Nov 25, 2008 #4


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    You assume everything comes to the same final temperature. Eshot = Ewater+Econtainer
    Watch the signs, energy is lost by the shot and gained by the container and water.
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