- #1

Vuldoraq

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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?