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Calorimeter with ice and water

  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)
    [tex]L_{f(water)}[/tex]=334*10^3(J/Kg)

    2. Relevant equations

    Q=m[tex]L_{f(water)}[/tex]

    Q=mc[tex]\Delta[/tex]T

    0=[tex]Q_{1}[/tex]+[tex]Q_{2}[/tex]+[tex]Q_{3}[/tex]


    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,

    [tex]-mc(lead)*(Tf-255)=mL_{f(water}+mc(copper)*(Tf-0)+mc(water)*(Tf-0)[/tex]

    Therefore,

    [tex]Tf=24790/877=28.3^{o}C[/tex]

    Which is apparently wrong. Can anyone please tell me why?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 13, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi Vuldoraq,

    What numbers did you use in your final equation to get 28.3 degrees?
     
  4. May 13, 2008 #3
    I used the following numbers,

    [tex]-97.5*T_{f}+24862.5J=72.45J+39T_{f}+745.82T_{f}[/tex]

    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]-97.5*T_{f}+24862.5J=6012J+39T_{f}+745.82T_{f}[/tex]

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

    Is that the right answer?
     
  5. May 13, 2008 #4
    Hi everyone,

    Please note I accidentally posted the same question twice. Sorry about that. I would delete one, but it seems I am unable.

    Vuldoraq
     
  6. May 13, 2008 #5
    Hey,

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

    Vuldoraq
     
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