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Copper calorimeter problem

  1. May 12, 2008 #1
    1. The problem statement, all variables and given/known data

    A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure.

    If 0.750 kg of lead at a temperature of 255[tex]^{o}[/tex]C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

    c(copper)=390 (J/k*kg)
    c(water)=4190 (J/k*kg)
    c(lead)=1930 (J/k*kg)

    2. Relevant equations




    3. The attempt at a solution

    I said that the heat lost by the lead is equal to the heat gained by the calorometer. So,

    [tex]-mc(lead)\Delta T(lead)=mL_{f(water)}+mc(copper)\Delta T(cal)+mc(water)\Delta T(cal)[/tex]

    Than I noted that [tex]\Delta T(lead)\ and \Delta T(cal)[/tex] both share the same final temperature, say Tf. Letting [tex]\Delta T=Tf-Ti[/tex] and subbing in the Ti (initial temp) values and then rearranging to get Tf on it's own,




    Which is apparently wrong. Can anyone please tell me why?
  2. jcsd
  3. May 12, 2008 #2

    Chi Meson

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    Science Advisor
    Homework Helper

    Your equation is correct. Try crunching the numbers again. I haven't got my calculator, but if the numbers given are correct you should get the correct answer.

    Did you put the correct masses in the correct places? I noticed you wrote "mLf(water)" where I would have written "ice."
  4. May 13, 2008 #3
    I used the following numbers,


    Where the 72.45J is the latent heat of fusion of the ice, which I now can see is wrong. It should be 6012J. I foolishly used the waters specific heat incorrectly.

    Redoing the above gives,


    [tex]Therefore T_{f}=18850/882.32=21.4^{o}C[/tex]

    Thanks for pointing that out, does this look right to you?
  5. May 13, 2008 #4
    Hi everyone,

    Please note i accidentally posted the same question twice. Sorry.

  6. May 13, 2008 #5

    21 degrees C is the right answer. Thanks for all the help.

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