Specific Heat and aluminum

  • Thread starter XenoPhex
  • Start date
  • #1
3
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This is a really simple problem, but I just can't remember how to combine the container and the water.

Homework Statement


An aluminum container weighing 200g contains 500g of water at 20*c. A 300g shot of the same aluminum at 100*c is put into the water. The specific heat for the aluminum is 0.215cal/(g*c). What is the final temperature for the entire system?


Homework Equations


dQ = c*m*dT


The Attempt at a Solution


(T - 20*c) x [(500g x 1cal/(g*c)) + (200 x 0.215cal/(g*c))] = (T - 100*c) x 300g x 0.215cal/(g*c)

T = 9.1216*c

However this answer seems really off.
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
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Energy is conserved so, energy gained by:
Econtainer = mct = 200 * 0.215 *(T-20)
Ewater = mct = 500 * 1* (T-20)
Equals energy lost by:
Eshot = mct = 300 * 0.215* (100-T)

(Final temperature T)
 
  • #3
3
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Wait, so would you just make Eshot = Ewater and then solve for T and then Eshot = Ewater and solve for a different T?
 
  • #4
mgb_phys
Science Advisor
Homework Helper
7,774
13
You assume everything comes to the same final temperature. Eshot = Ewater+Econtainer
Watch the signs, energy is lost by the shot and gained by the container and water.
 

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