Calorimeter with ice and water

[Vuldoraq]

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

Homework Equations

Q=mL_{f(water)}

Q=mc\DeltaT

0=Q_{1}+Q_{2}+Q_{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,

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


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

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

Therefore,

Tf=24790/877=28.3^{o}C

Which is apparently wrong. Can anyone please tell me why?

 

[alphysicist]

Hi Vuldoraq,

What numbers did you use in your final equation to get 28.3 degrees?

 

[Vuldoraq]

I used the following numbers,

-97.5*T_{f}+24862.5J=72.45J+39T_{f}+745.82T_{f}

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,

-97.5*T_{f}+24862.5J=6012J+39T_{f}+745.82T_{f}

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

Is that the right answer?

 

[Vuldoraq]

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

 

[Vuldoraq]

Hey,

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

Vuldoraq