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Is entropy the same as heat capacity?

by CraigH
Tags: capacity, difference, entropy, heat, specific
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CraigH
#1
Jan30-13, 01:28 AM
P: 202
A formula for entropy:
dS=dQ/dT

This is a formula that represents the rate of change of heat with respect to temperature.

Q=mcdT

This is another formula that states how the temperature changes as heat is added.

from this can we say that entropy is just specific heat capacity * Mass, i.e entropy is just heat capacity?
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DrDu
#2
Jan30-13, 01:34 AM
Sci Advisor
P: 3,595
You got the first formula wrong, it is dS= dQ /T. Furthermore it only holds for reversible changes, but ok.
Hence
dS=mc/T dT, so you can calculate the entropy change from heat capacity although they are not identical.
CraigH
#3
Jan30-13, 02:54 AM
P: 202
Quote Quote by DrDu View Post
You got the first formula wrong, it is dS= dQ /T
This confuses me, surely as the heat added increases ( dQ), then the temperature will also increase, so it should be (dT)

Unless T is a different constant for each example?

In the first few seconds of this video
http://www.youtube.com/watch?v=xJf6pHqLzs0
it is said that "if the temperature is changing while we add the heat, which is normally the case, then we'l have to do some calculus"

khanacademy is usually a pretty reliable source, so what did he mean by this if the formula is actually dS= dQ /T

Studiot
#4
Jan30-13, 03:17 AM
P: 5,462
Is entropy the same as heat capacity?

Trust Dr Du, he knows a great deal about thermodynamics.

You got the first formula wrong, it is dS= dQ /T.
This was the original definition of entropy in classical thermodynamics.

Now as to what happens when we add heat to something, note that Dr Du also said

Furthermore it only holds for reversible changes
.

This is very important. You should always note the conditions that apply to any statement in Physics.

Of course this presents a quandrary because there are very few natural processes that take place at constant temperature.

One such is change of state ie melting or boiling when the heat added/removed is the latent heat.

In this case the entropy of the change can be calculated from the formula.

More generally when the heat input goes to changing the temperature the formula cannot be used directly and other methods must be used to calculate the entropy. Changes that include temperature change are always irreversible.

Entropy cannot be measured directly in the laboratory.
CraigH
#5
Jan30-13, 03:42 AM
P: 202
Oh okay thank you. I think my confusion was with what a reversible change meant, I've always assumed that if the temperature of something changes, this is a reversible process because if something heats up, it can cool down. I think I just took the name too literally.

So if adding and removing latent heat is reversible, why is adding and removing temperature not?
chill_factor
#6
Jan30-13, 03:47 AM
P: 899
Quote Quote by CraigH View Post
This confuses me, surely as the heat added increases ( dQ), then the temperature will also increase, so it should be (dT)

Unless T is a different constant for each example?
It is dS = dQ/T.

Why am I so confident I am right?

Well lets note some important facts. As temperature approaches infinity, entropy S is at a maximum. The way to increase temperature is to add heat Q. So maybe entropy S is a function of Q.

lets differentiate S. dS/dQ. To maximize S, dS/dQ = 0. This is exactly true when temperature is infinite for the expression dS/dQ = 1/T. Makes sense?

But wait, you say. Maybe entropy is actually a minimum at T -> ∞. No that's silly, because you can check the 2nd derivative. d/dQ (dS/dQ) = d/dQ (1/T) = -1/T^2. When the first derivative is zero and the second derivative is negative at a point, the maximum is described.

Therefore dS/dQ = 1/T is a correct formula, give or take some coefficients, perhaps like the Boltzmann constant. Redistribute it. dS = dQ/T.
DrDu
#7
Jan30-13, 04:23 AM
Sci Advisor
P: 3,595
Of course also in a reversible process T may be changing. I think your problem is more a mathematical one than a physical one.
e.g. you can calculate the change in energy of a particle as dE=1/2 mvdv where v is velocity. Although v will change this does not mean that you have to write dv dv.


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