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The difference between the entropy and the heat capacity? they are very similar 
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#1
Apr2111, 10:45 AM

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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 ? 


#2
Apr2211, 06:09 PM

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#3
Apr2211, 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.



#4
Apr2211, 08:33 PM

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



#5
Apr2211, 08:51 PM

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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 


#6
Apr2211, 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. 


#7
Apr2311, 09:48 AM

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#8
Apr2911, 07:40 AM

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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. 


#9
Apr2911, 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.



#10
Apr2911, 12:06 PM

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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. 


#11
Apr2911, 12:08 PM

P: 7

How did you integrate this??? I am sure you did a mistake here. 


#12
Apr2911, 12:24 PM

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[tex]\Delta U = Q=mc_V\Delta T[/tex] where ΔU=U_{2}U_{1} is the difference in energy due to the addition of thermal energy Q at constant volume, and where ΔT=T_{2}T_{1} 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? 


#13
Apr2911, 02:04 PM

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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???



#14
Apr2911, 02:07 PM

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By the way which software you use to write equations?



#15
Apr2911, 02:15 PM

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If you think I'm wrong, please show a reference from the literature. 


#16
Apr2911, 02:17 PM

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#17
Apr2911, 03:40 PM

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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. 


#18
Apr2911, 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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