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Specific Heat of Metals/Homework Help?

  1. Apr 20, 2006 #1
    Here is a homework problem I am working on:

    1. Suppose 50 cm^3 of water at 85 degrees C is addedto 100 cm^3 of water at 10 degrees C. What is the equilibrium temperature of the mixture, assuming there is no heat lost to the environment?

    * = multiplied by
    T = Temperature of water
    m = mass of water
    c = specific heat of water = 4184 J

    T= {(m * c * T) + (m * c * T)} all that divided by {(m * c + m * c)}

    which gives me the following:

    {(50 * 4184 * 85) + (100 * 4184 * 10)}
    ------------divided by---------------
    {(50 * 4184) + (100 * 4184)}

    = 35 degrees C for equilibrium temperature

    Is this correct? Did I set it up and do it the right way??

    Appreciate your help on this :) THANKS!!
  2. jcsd
  3. Apr 20, 2006 #2


    User Avatar

    Staff: Mentor

    Is the temperature supposed to be in C or K? (I honestly don't know, since I don't work with this type of problem, but it seems like absolute temperature might be needed for the calculation of the combined average, and then convert back to C.)
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