What is the Entropy Change in Cooling Aluminum from 373K to 333.45K?

[jmcgraw]

I know the final and initial temperatures of .259kg of aluminum. Specific heat of almuninum is 900 J/(kg*K).

The aluminum is cooled from 373K to 333.45K. I need to find the entropy change. This is what I did:

let c = specific heat, m = mass, T = final temperature

Q = cm(T - 373) <=> Q = Tcm - 373cm <=> T = Q/cm + 373

Now that I had T in terms of Q, I subsituted it into the entropy integral:

delta S = [integral from initial heat to final heat] (Q/cm + 373) dQ

Initial heat lost is cm(373K-373K) = 0, final amount of heat lost is cm(333.45K-373K) = -9219 J

My integration yielded -3.256 x 10^6 J/K. The right answer should be only around -20 j/k or so. I am WAY off, but my approach seems like it should work.

What did I do wrong?

 

[lightgrav]

if you integrate dQ/T , instead of T dQ , you might get closer to 25 J/K

 

[jmcgraw]

 

[jmcgraw]

The only problem is that it gives a horrible integral (partial fraction decomposition? )

This is much better:

dQ =cmdT

so integrate

cm [integral from initial temp to final temp] dT/T

Ahhhhhhh... How much nicer.