# Entropy Increase of Aluminum

1. Jul 17, 2014

### Art_Vandelay

1. The problem statement, all variables and given/known data

Imagine that the temperature of 255 g of aluminum sitting in the sum increases from 278 K to 294 K. By how much has its entropy increased?

2. Relevant equations

Q=mcΔT
ΔS=Q/T

3. The attempt at a solution

Q=(255 g)(.90 J/gK)(294 K -278 K)
Q=3672 J

ΔS=Q/T
ΔS=3672 J / 294 K
ΔS= 12.5 J/K

This answer is incorrect, but I don't know why.

2. Jul 17, 2014

### haruspex

T is not constant through the warming. Obtain a differential equation from ΔS=ΔQ/T and solve it.

3. Jul 18, 2014

### Art_Vandelay

Would a differential equation be subtracting ΔSCOLD from ΔSHOT?

Q = mcΔT = (.255 kg)*(910 J/kgK)*(294 K -278 K)
Q = 3713 J

ΔSCOLD = Q/T
ΔSCOLD = 3713 J / 278 K
ΔSCOLD = 13.36 J/K

ΔSHOT = 3713 J / 294 K
ΔSHOT = 12.63 J/K

ΔSnet = ΔSHOT - ΔSCOLD
ΔSnet = 12.63 J/K - 13.36 J/K
ΔSnet = -.73 J/K

4. Jul 18, 2014

### haruspex

No, I mean consider just a very small change.
If a little heat, dQ, enters the metal, there is a corresponding small change dT to the temperature and a small gain in entropy, dS = dQ/T.
You need a second equation to relate dT to dQ.

5. Jul 18, 2014

### Art_Vandelay

Hmm... Would I use 1/T = dS/dU? But how could I figure out dU without a given pressure?

6. Jul 18, 2014

### TSny

Following haruspex's lead, you need an expression for dQ to use in dS = dQ/T.

What does the equation Q = mcΔT look like for an infinitesimal amount of heat?

7. Jul 18, 2014

### Art_Vandelay

It would almost be equal to zero?

8. Jul 18, 2014

### haruspex

Well, yes, but TSny means using dQ, dT. The Q in Q = mcΔT is a change in heat content, not the total heat content. So really it should ΔQ = mcΔT. But, conventionally, Δ is used for (possibly) substantive differences, whereas d is used for infinitesimal changes.
So, rewrite Q = mcΔT using dQ, dT, and substitute in the entropy equation to eliminate dQ.

9. Jul 18, 2014

### Art_Vandelay

Oh! Ok, so I would end up with S*T=mc(dT). Would the temperatures then negate each other, or would I need to divide mc by T?

10. Jul 19, 2014

### haruspex

Again, that's not quite right. It's an infinitesimal change in S, so it's dS*T=mc*dT.
Can you solve that differential equation?

11. Jul 19, 2014

### Art_Vandelay

Unfortunately, I've never used a differential equation before. Is it calculus-based?

Here's my attempt:

dS*T=mc*dT
dS=((.255 kg)(910 J/kg K)(>0K?)/294 K
dS=.79 J/K

12. Jul 19, 2014

### TSny

Sorry, but that's not right at all.

Solving dS*T = mc*dT does involve calculus. Suppose you rearrange the equation as dS = mc*dT/T. In this form the variables S and T are "separated", with S on the left and all T's on the right. Can you see what to do to obtain ΔS from dS?

13. Jul 19, 2014

### Art_Vandelay

Since dS is infinitesimal, would ΔS be found by multiplying by ΔT?

14. Jul 19, 2014

### TSny

No.

Do you have some calculus background?

As a little exercise that is relevant to this problem, can you evaluate $$\int_{x_1}^{x_2} dx$$ in terms of $x_1$ and $x_2$?

How about this one? $$\int_{x_1}^{x_2} \frac{dx}{x}$$

15. Jul 19, 2014

### Art_Vandelay

I wish I did have some knowledge of calculus, but I do not. I really have no clue how to evaluate the practice problem.

16. Jul 19, 2014

### TSny

This problem requires calculus to solve it. If this is a problem from a course in physics that does not use calculus, then I don't see how you would be expected to solve it.

Could it be that the instructor or textbook has given you a formula (without derivation) that you can use for this particular type of problem? The formula would contain a logarithm function.

[Edit: You can get an approximate answer by using ΔS ≈ Q/Tavg, where Tavg is an average temperature of the aluminum during the heating process.]

Last edited: Jul 19, 2014
17. Jul 19, 2014

### TSny

For your particular problem, the temperature changes by only about 6%. In this case, the approximate formula ΔS = Q/Tavg gives an answer that is in agreement with the exact formula (from calculus) to 4 significant figures. Since the numbers given in the problem are only given to 3 significant figures, the approximate formula is accurate enough in this case.

For problems where the temperature change is greater, the approximate formula will be less accurate.

18. Jul 19, 2014

### Art_Vandelay

Yes, this is an algebra-based physics course :/
There have been a few others in the class which required calculus, so I was unable to answer them.

Thank you so much for your time, effort, and helpfulness! The approximate answer was close enough for the computer program. Thanks again!

19. Jul 19, 2014

OK. Good.