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Temp Equilibrium 
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#1
Jul2307, 02:16 PM

P: 8

1. The problem statement, all variables and given/known data
A .20kg lead shot is heated to 90 degrees C and dropped into an ideal calorimeter containing .50kg of water initially at 20 degrees C. What is the final equilibrium temperance of the lead shot? Specific heat capacity of lead is 128 J/(kg*C), and water is 4186 J/(kg*C). 2. Relevant equations Am I using the right formula, (c*m*delta T)_lead = (c*m*delta T)_water What is the delta T of water in this equation? 3. The attempt at a solution delta T_lead = (c*m*delta T)_water / (c*m)_lead = [(4186)(.50)(?)] / [(128)(.20)] = ? 


#2
Jul2307, 05:35 PM

P: 8

Help... anyone?



#3
Jul2307, 05:45 PM

Sci Advisor
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P: 8,953

You have to assume that the final temperature of the lead and water are the same. Write your equations in terms of this rather than delta T.
Once you have found this you can obviously work out each dT. 


#4
Jul2307, 06:00 PM

P: 8

Temp Equilibrium
The equation doesn't give me the final temp of the water, so how do you set up the equations for the final temps? I'm confuse because my text book only gives me one sample problem in it and that one gives you the final temp of the water, so the equation is simple.
Sorry if I am not seeing it.... 


#5
Jul2307, 06:51 PM

Mentor
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#6
Jul2307, 06:54 PM

Sci Advisor
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P: 8,953

Calculate the total energy at the start.
=mass lead * c lead * T lead + mass water * c water * T water ( use the absolute temperature in kelvin ie 273 + C) At the end you have the same energy and the lead / water temperature are the same. = mass lead * c lead * T + mass water * c water * T Simply solve for T ( remember this is in kelvin ) 


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