Final Equilibrium Temperature of Lead Shot Problem

[Confused_07]

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)]
= ?

 

[Confused_07]

Help... anyone?

 

[mgb_phys]

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.

 

[Confused_07]

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...

 

[Doc Al]

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.

 

[mgb_phys]

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 )