- #1
XenoPhex
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This is a really simple problem, but I just can't remember how to combine the container and the water.
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?
dQ = c*m*dT
(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.
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.