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Copper calorimeter final temperature

  • Thread starter Vuldoraq
  • Start date
  • #1
272
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Homework Statement



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)

Homework 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]


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?
 

Answers and Replies

  • #2
Chi Meson
Science Advisor
Homework Helper
1,789
10
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."
 
  • #3
272
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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]

Thanks for pointing that out, does this look right to you?
 
  • #4
272
0
Hi everyone,

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

Vuldoraq
 
  • #5
272
0
Hi,

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

Vuldoraq
 

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