What is the entropy change of aluminum?

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SUMMARY

The discussion centers on calculating the entropy change of aluminum when mixed with water in a thermally isolated system. The final equilibrium temperature of the mixture is determined to be 65.1°C. The entropy change for aluminum is calculated as -21.2 J/K, while the entropy change for water is 24 J/K, resulting in a total entropy change of 2.8 J/K for the aluminum-water system. The calculations utilize the specific heat capacities of aluminum and water, along with the formula Q=mC(Tf - Ti).

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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?


Homework Equations


Q=mC(Tf - Ti)
MwCw(Tf - 20c) + MaCa(Tf - 100c)= 0
S(entropy)= Q/T

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.
any help is very welcome
 
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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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