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What is the entropy change of aluminum?

  • Thread starter KrazyX24
  • Start date
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1. Homework Statement
In an experiment, 240 g of aluminum (with a specific heat of 900 J/kgK) at 100°C is mixed with 40.0 g of water (4186 J/kgK) at 20°C, with the mixture thermally isolated.
a: What is the equilibrium temp in C?
b: What is the entropy change of aluminum?
c: What is the entropy change of the water?
d: What is the entropy change of the aluminum-water system?


2. Homework Equations
Q=mC(Tf - Ti)
MwCw(Tf - 20c) + MaCa(Tf - 100c)= 0
S(entropy)= Q/T
3. The Attempt at a Solution

Basically I'm kinda stuff trying to solve for the final temp, I get 11.9C but that cannot be right since that means the water-Al system lost heat. :grumpy:
any help is very welcome
 

Answers and Replies

2
0
Nevermind, solved it just now. T(Final) = CaMa + CwMwTi/CwMw + CaMa
Final temp came out to be 65.1*C.
B: S = (MaCa)ln(Tf/Ti) = -21.2 J/K
C: S = (MwCw)ln(Tf/Ti) = 24 J/K
D: 24-21.2= 2.8 J/K

Mod can close thread...
 

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