How Much Does Entropy Increase for Aluminum When Heated?

[Art_Vandelay]

Homework Statement

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?

Homework Equations

Q=mcΔT
ΔS=Q/T

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.

 

[haruspex]

Homework Statement

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?

Homework Equations

Q=mcΔT
ΔS=Q/T

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.

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

 

[Art_Vandelay]

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

Would a differential equation be subtracting ΔS_COLD from ΔS_HOT?

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

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

ΔS_HOT = 3713 J / 294 K
ΔS_HOT = 12.63 J/K

ΔS_net = ΔS_HOT - ΔS_COLD
ΔS_net = 12.63 J/K - 13.36 J/K
ΔS_net = -.73 J/K

 

[haruspex]

Would a differential equation be subtracting ΔS_COLD from ΔS_HOT?

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.

 

[Art_Vandelay]

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.

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

 

[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?

 

[Art_Vandelay]

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?

It would almost be equal to zero?

 

[haruspex]

It would almost be equal to zero?

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.

 

[Art_Vandelay]

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.

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?

 

[haruspex]

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?

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?

 

[Art_Vandelay]

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?

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

 

[TSny]

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

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?

 

[Art_Vandelay]

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?

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

 

[TSny]

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

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? $$\int_{x_1}^{x_2} \frac{dx}{x}$$

 

[Art_Vandelay]

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? $$\int_{x_1}^{x_2} \frac{dx}{x}$$

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

 

[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/T_avg, where T_avg is an average temperature of the aluminum during the heating process.]

 

[TSny]

For your particular problem, the temperature changes by only about 6%. In this case, the approximate formula ΔS = Q/T_avg 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.

 

[Art_Vandelay]

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/T_avg, where T_avg is an average temperature of the aluminum during the heating process.]

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!

 

[TSny]

OK. Good.

You might ask your instructor about how you were expected to answer this question.