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posted by  ere on 10/13/2007 8:51:57 PM  |  status: Live  

Thermochemistry, pleaseeee

Course Textbook Chapter Problem
General Chemistry N/A N/A N/A
Question Details:
A 26.0 - aluminum block is warmed to 65.6^\circ C and plunged into an insulated beaker containing 55.2 g water initially at 22.0^\circ C. The aluminum and the water are allowed to come to thermal equilibrium.
Assuming that no heat is lost, what is the final temperature of the water and aluminum?

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posted by Sara2029 on 10/13/2007 10:07:23 PM  |  status: Live
Asker's Rating: Lifesaver   
ere's comment:
"thank you"
Response Details:
use the conservation of energy concept...
thus, we know that the heat of the block is lost to the water, because of the temperature difference...
 
so we get
 
-mc ΔT = mc ΔT
 
where the negative sign means that heat is lost...
 
i do not have the c value for aluminium....
so ull have to do the calculations urself...
 
take note that ΔT is (Tfinal - Tinitial)...
 
so, u get the formula
 
- mAl cAl (Tfinal - Tinitial) = mwater cwater (Tfinal - Tinitial)
 
solve for Tfinal....
and u have ur answers
....
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