What is the entropy change of aluminum?

[KrazyX24]

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

 

[KrazyX24]

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...

 

[m2003]

I would respond by saying that the change in entropy of aluminum can be calculated using the equation S = Q/T, where S is the change in entropy, Q is the heat transferred, and T is the temperature. In this experiment, the aluminum and water are thermally isolated, so no heat is transferred between them. Therefore, the entropy change of aluminum would be zero.

To calculate the equilibrium temperature, you can use the equation Q = mC(Tf - Ti), where Q is the heat transferred, m is the mass, C is the specific heat, Tf is the final temperature, and Ti is the initial temperature. In this case, the heat lost by the aluminum (Q = -Q) must equal the heat gained by the water (Q = +Q). So, you can set up the equation -MaCa(Tf - 100°C) = MwCw(Tf - 20°C) and solve for Tf.

Once you have the final temperature, you can calculate the entropy change of the water using the same equation S = Q/T. Finally, to calculate the entropy change of the aluminum-water system, you can simply add the individual entropy changes of aluminum and water together.