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Calorimetry Question ;(

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data
    To find cx, the specific heat of material X, I place 75g of X in a 30g copper
    calorimeter that contains 65g of water, all initially at 20°C. When I add 100g of
    water at 80°C, the final temperature is 49°C. What is cx?
    Data: cCU = 386 J kg
    -1
    K
    -


    2. Relevant equations

    Q=mcDt


    3. The attempt at a solution

    I tried finding out the amount of energy around in the system of water and cupper and then added the energy provided by the addition of X, with its variable Cx.
    I then added the energy provided by the 100g of water at 80. and equalled that to the ( Sum of masses x Cw x Cc x Cx x 49. and then tried to calculate the variable Cx by rearranging. The answer is 2.18 Kj. I can't find it :(
     
  2. jcsd
  3. Feb 1, 2012 #2

    tiny-tim

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    welcome to pf!

    hi davekardle! welcome to pf! :wink:

    show us what you've tried, and then we'll know how to help! :smile:
     
  4. Feb 1, 2012 #3
    Energy Available in the system ( H2O + Cu)

    Q= 0.03 x 0.386 x 20 = 0.2316 KJ (Cu)
    Q= 0.065 x 4.2 x 20 = 5.5

    total Qw + Qcu = 5.7 Kj KG- C-

    ENERGY offered by X:

    Q= 0.075x Cx x 20= 1.5Cx

    Energy offered by water:

    Q= 0.1 x 4.2 x 80 = 33.6 Kj Kg- c-

    ADDING energies=

    5.5 + 1.5Cx + 33.6 = 0.170kg(TOTAL mass of mix) x (0.27 x 4.2) x Cx x 49 ( final temp)


    rearranging

    39.3 = 19.9Cx
    Cx= 1.97Kj

    Which is wrong :(
     
  5. Feb 1, 2012 #4

    tiny-tim

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    i don't understand this bit …
    (what is the 0.27 ? and …)

    don't you need to continue to treat each of the materials separately?

    (btw, you've used a "zero-energy-level" of 0° …

    it would be easier and quicker to use 20°)​
     
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