# Thermodynamics I have been working on this for 3 hours

Thermodynamics...I have been working on this for 3 hours :(

Please help anyone :), I have been trying this problem for the past 3 hours and I am not getting the right answer. I would appreciate any hint or guideance.

Thanks all
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" A copper calorimeter can with a mass of 0.100 kg contains 0.160 kg of water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure. If 0.755 kg of lead at a temperature of 280°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings."

Mass of copper= .1kg
Mass of water=.16kg
Mass of ice= .018kg
Mass of lead= .755kg @ 280ºC

I assume that since the ice is in thermal equilibrium with the copper cup and water then the temperature is 0ºC.

The target variable is the common final temperature(T).

Using 17.13(Q = mc∆T) I set up all of the above elements including the heat transfer of the phase change of ice(Q=mL).

(.018kg)ּ(2100J/kgּK)ּ(T-0ºC)+(.1kg)ּ(390J/kgּK)(T-0ºC)+(.16kg)ּ(4190J/kgּK)ּ(T-0ºC)+(.018kg)ּ(334000J/kg)+(.755kg)ּ(130J/kg)ּ(T-280ºC)=0

Solving for final temperature, I get close to 25.3977ºC which doesn't seem to be the correct answer.

Thanks

andrevdh
Homework Helper
(.018kg)ּ(2100J/kgּK)ּ(T-0ºC)+(.1kg)ּ(390J/kgּK)(T-0ºC)+(.16kg)ּ(4190J/kgּK)ּ(T-0ºC)+(.018kg)ּ(334000J/kg)+(.755kg)ּ(130J/kg)ּ(T-280ºC)=0

The first term for ice water should have the specific heat of water (4190). You can add the mass of the ice water to the water for one less term in the equation. The heat of the lead should be subtracted in this equation from: heat gained = heat lost.

andrevdh said:
(.018kg)ּ(2100J/kgּK)ּ(T-0ºC)+(.1kg)ּ(390J/kgּK)(T-0ºC)+(.16kg)ּ(4190J/kgּK)ּ(T-0ºC)+(.018kg)ּ(334000J/kg)+(.755kg)ּ(130J/kg)ּ(T-280ºC)=0

The first term for ice water should have the specific heat of water (4190). You can add the mass of the ice water to the water for one less term in the equation. The heat of the lead should be subtracted in this equation from: heat gained = heat lost.

Cool thanks! I got the right answer now,

I did what you said and changed the ice to 4190 and Subtracted lead and switched the terms for heat, -.755g*130j/kg*(280-T)

Thanks again!!! This one was killing me.