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The difference between the entropy and the heat capacity? they are very similar

by su214
Tags: capacity, difference, entropy, heat, similar
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su214
#1
Apr21-11, 10:45 AM
P: 3
hello ...
as we all know that
specific heat capacity = joule/ k
entropy = joule/k
they are same in units

Q= m Cv dT

Q/dT=m Cv................1
dS= dQ/T .....2
from 1 & 2

dS= m Cv ..............??????????????????????????
i'm now confused ,,entropy can't equal mCv i know i'm wrong ,,but don't know why? and in the same time what does it mean that they have the same units ?

and if entropy measuers the chaos in the system and the specific heat capacity is the amount of heat to raise one Kg of the matter one degree of tempreture
aren't the entropy and the heat capacity related somehow ?
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Mapes
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Apr22-11, 06:09 PM
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Quote Quote by su214 View Post
aren't the entropy and the heat capacity related somehow ?
Sure, the heat capacity for a given constraint X (like constant volume or constant pressure) is defined as [itex]C_X=T(\partial S/\partial T)_X[/itex]. In your derivation, dQ/T doesn't equal Q/dT, so this can't be used to show that entropy is the same as heat capacity. And it doesn't mean anything that they have the same units; work, heat, and torque are fundamentally different parameters that also share the same units (N-m), for example.
SteamKing
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Apr22-11, 08:17 PM
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You all have confused entropy with enthalpy. Enthalpy is a measure of the total heat content of a substance. Entropy is a much more subtle concept, and its description and use form the basis for the Second Law of Thermodynamics.

Mapes
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Apr22-11, 08:33 PM
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The difference between the entropy and the heat capacity? they are very similar

Quote Quote by SteamKing View Post
You all have confused entropy with enthalpy. Enthalpy is a measure of the total heat content of a substance. Entropy is a much more subtle concept, and its description and use form the basis for the Second Law of Thermodynamics.
Who is the "all" you're addressing? Nobody is discussing enthalpy.
Curl
#5
Apr22-11, 08:51 PM
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Quote Quote by SteamKing View Post
You all have confused entropy with enthalpy. Enthalpy is a measure of the total heat content of a substance. Entropy is a much more subtle concept, and its description and use form the basis for the Second Law of Thermodynamics.
Son, please. Do not give wrong information on here.


and in the same time what does it mean that they have the same units ?
It means nothing.

Also, manipulating some symbols doesn't really give any insight into what entropy is. Actually, the "engineering thermo" education will never explain what entropy or temperature are. For that you have to study on your own

http://en.wikipedia.org/wiki/Entropy...rmodynamics%29
SteamKing
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Apr22-11, 09:37 PM
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If you still think you have not confused enthalpy with entropy, check out the following link:
http://en.wikipedia.org/wiki/Enthalpy

And, Curl, I'm not your son.
Mapes
#7
Apr23-11, 09:48 AM
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Quote Quote by SteamKing View Post
If you still think you have not confused enthalpy with entropy, check out the following link:
http://en.wikipedia.org/wiki/Enthalpy
It's not clear what you're talking about. What statement do you have a problem with, and how would you revise it?
jonsayago
#8
Apr29-11, 07:40 AM
P: 7
Quote Quote by su214 View Post
hello ...
as we all know that
specific heat capacity = joule/ k
entropy = joule/k
they are same in units

Q= m Cv dT

Q/dT=m Cv................1
dS= dQ/T .....2
from 1 & 2

dS= m Cv ..............??????????????????????????
i'm now confused ,,entropy can't equal mCv i know i'm wrong ,,but don't know why? and in the same time what does it mean that they have the same units ?

and if entropy measuers the chaos in the system and the specific heat capacity is the amount of heat to raise one Kg of the matter one degree of tempreture
aren't the entropy and the heat capacity related somehow ?
I think is a good question. The Fourier law you started: Q= m Cv dT, I think is missing a dot over the Q which means is the heat flux (derivate with respect to time). The heat capacity has a subindex V which means it is a process at constant volume. However, it would still be valid to say:

Q = mCT (1)

Notice Q (without point) refers just to the internal energy in the system, C would be the heat capacity -without specifying if is at constant pressure or volume- and of course, m stands for mass and T for temperature (the units of temperature should be consistent with those of C and should be in kelvin units to make the relationship with entropy concept).

From Eq. (1) it can find out Q/T= mC = entropy

In my opinion is correct, entropy can be deducted from mC, considering C is not a constant anymore, but a parameter that changes with respect to volume and pressure. However, remember that entropy itself is not useful, instead we always find out the change of entropy (dS) from one state to another. This is the reason why I mentioned before that in order to make a comparison with entropy we should use kelvin temperature units.

So entropy can be seen as a disorder parameter, or the capacity to store energy of each component of a system of mass m.

My answer is just the way I understand the physics related to this issue therefore I cannot cite further works. I would appreciate reading more comments or citations about this topic.
Mapes
#9
Apr29-11, 07:54 AM
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Hi jonsayago, welcome to PF, but please note that personal theories are not appropriate here; your posts should be based on consensus physics.

Quote Quote by jonsayago View Post
The Fourier law you started: Q= m Cv dT, I think is missing a dot over the Q which means is the heat flux (derivate with respect to time).
No, this would make the units inconsistent. dQ is measured in Joules, so that dQ = mcVdT has units [J] = [kg][J/kg/C][C]. Note that in this equation, dQ must be an infinitesimal quantity to match dT.

Quote Quote by jonsayago View Post
However, it would still be valid to say: Q = mCT (1)
No. If you integrate dQ = mcdT, you get Q = mcΔT, which is different.

Quote Quote by jonsayago View Post
From Eq. (1) it can find out Q/T= mC = entropy
Eq. (1) is not correct, so this does not hold.
jonsayago
#10
Apr29-11, 12:06 PM
P: 7
First comment I accept Q shouldn't be a derivative with respect to time. It is simply the change with respect to one state and another.
However, Ec. (1) stands under equilibrium conditions. I mean in just one state. The rest of my derivation should be fine.
jonsayago
#11
Apr29-11, 12:08 PM
P: 7
Quote Quote by Mapes View Post
Hi jonsayago, welcome to PF, but please note that personal theories are not appropriate here; your posts should be based on consensus physics.



No, this would make the units inconsistent. dQ is measured in Joules, so that dQ = mcVdT has units [J] = [kg][J/kg/C][C]. Note that in this equation, dQ must be an infinitesimal quantity to match dT.



No. If you integrate dQ = mcdT, you get Q = mcΔT, which is different.



Eq. (1) is not correct, so this does not hold.
"No. If you integrate dQ = mcdT, you get Q = mcΔT, which is different"

How did you integrate this??? I am sure you did a mistake here.
Mapes
#12
Apr29-11, 12:24 PM
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Quote Quote by jonsayago View Post
"No. If you integrate dQ = mcdT, you get Q = mcΔT, which is different"

How did you integrate this??? I am sure you did a mistake here.
[tex]\int dQ=\int_{U_1}^{U_2}dU=\int_{T_1}^{T_2}mc_VdT[/tex]

[tex]\Delta U = Q=mc_V\Delta T[/tex]

where ΔU=U2-U1 is the difference in energy due to the addition of thermal energy Q at constant volume, and where ΔT=T2-T1 is the temperature difference. See here for the differential version and here for the integrated version, for example. What other way is there to integrate it?
jonsayago
#13
Apr29-11, 02:04 PM
P: 7
You evaluated the right part of the integral (from T1 to T2) and not left part... You have to evaluate left part too from Q1 to Q2. Someone else out there to comment???
jonsayago
#14
Apr29-11, 02:07 PM
P: 7
By the way which software you use to write equations?
Mapes
#15
Apr29-11, 02:15 PM
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Quote Quote by jonsayago View Post
You evaluated the right part of the integral (from T1 to T2) and not left part... You have to evaluate left part too from Q1 to Q2.
dQ is not an exact differential, and there is no such thing as Q1 or Q2. (A state can have an associated energy U or an associated temperature T, but not an associated heat Q; heat describes a path-dependent transfer of energy between two states.) See equation 15.11 here or equation 5.6(a) here, for example, where in this case the work W is zero because of the constant-volume constraint.

If you think I'm wrong, please show a reference from the literature.
Mapes
#16
Apr29-11, 02:17 PM
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Quote Quote by jonsayago View Post
By the way which software you use to write equations?
You can enclose your equations between {tex} and {/tex}, using straight brackets instead of curly brackets. Or use {itex} and {/itex} for inline equations (within a paragraph). Click on anyone's equation to get a popup window showing their code.
jonsayago
#17
Apr29-11, 03:40 PM
P: 7
Here is the importance of using the absolute scale (kelvin) for temperature. When using it Q = U. Eq 33.4 of

http://books.google.com/books?id=8se...ntropy&f=false

Though Q is not a state variable, the entropy S is a state variable (Dunod, Physique statistique). So we could have divided both sides of the equation by T first and then integrate from one state to another.
Check the entropy section of the book you quoted.
jonsayago
#18
Apr29-11, 03:41 PM
P: 7
By the way don't forget to tell me which program you use to set up equations in the internet!!!


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