Final Equilibrium Temperature of Lead Shot Problem

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The problem involves calculating the final equilibrium temperature when 0.20 kg of lead shot at 90°C is mixed with 0.50 kg of water at 20°C in an ideal calorimeter. The correct formula to use is (c*m*delta T)_lead = (c*m*delta T)_water, where delta T for both substances is expressed in terms of the final temperature T_f. To find T_f, set up the energy balance equations for both lead and water, ensuring to account for the specific heat capacities and initial temperatures. The final temperature can be solved by equating the total energy before and after mixing, using absolute temperatures in Kelvin. This approach will yield the final equilibrium temperature of the system.
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Homework Statement


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).


Homework 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?

The Attempt at a Solution



delta T_lead = (c*m*delta T)_water / (c*m)_lead
= [(4186)(.50)(?)] / [(128)(.20)]
= ?
 
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Help... anyone?
 
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.
 
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 textbook 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...
 
Confused_07 said:
Am I using the right formula, (c*m*delta T)_lead = (c*m*delta T)_water
That's correct. (Assuming you get the signs right.)

What is the delta T of water in this equation?
Call the final temperature T_f. Express those "delta T"s in terms of the given initial temperatures and T_f. Then solve for T_f.
 
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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