Heat Capacity and Entropy Textbook Definition - Quick Question


by weeksy
Tags: capacity, definition, entropy, heat, textbook
weeksy
weeksy is offline
#1
Jun7-12, 12:48 AM
P: 14
Just a quick question of something I found in my textbook but cant get how they produced it.

C_p =(∂Q/∂T)_p

that is the definition of heat capacity at a constant pressure p. Q is heat and T is temperature. This equation is fine and I know how to derive it. Now it is the next line which worries me.

C_p=(∂Q/∂T)_p = T(∂S/∂T)_p

where S is the entropy

why this bothers me is that this equation as I understand should only hold for C_V (heat capacity with constant volume)
thats because dQ=TdS for a REVERSE-ABLE ONLY expansion (i.e dQ=0 , dS=0) ie adiabatic all of which occur as V, the volume is held constant for C_V. Hence dU=dQ and dQ(rev)=TdS = dU , which can then be simply substituted into the definition. OK

Sorry for the spiel, but my question is how can this same line of reasoning be true for C_p where dV is not constant and dU is not equal to dQ and expansion isn't reversible?

Thanks
Phys.Org News Partner Physics news on Phys.org
Better thermal-imaging lens from waste sulfur
Scientists observe quantum superconductor-metal transition and superconducting glass
New technique detects microscopic diabetes-related eye damage
Studiot
Studiot is offline
#2
Jun7-12, 10:14 AM
P: 5,462
I think you are confusing U and H, have another look at your textbook.

The partial differntial expressions you give for Cp are correct, but they are obtained by substituting Tds into the definition of H not U.

Similar partials can be derived for Cv by substituting Tds into the definition of U.
Darwin123
Darwin123 is offline
#3
Jun7-12, 12:30 PM
P: 741
Quote Quote by weeksy View Post
Just a quick question of something I found in my textbook but cant get how they produced it.

C_p =(∂Q/∂T)_p

that is the definition of heat capacity at a constant pressure p. Q is heat and T is temperature. This equation is fine and I know how to derive it. Now it is the next line which worries me.

C_p=(∂Q/∂T)_p = T(∂S/∂T)_p

where S is the entropy
The definition of specific heat, C, under any condition is:
C= ∂Q/∂T
The definition of S, under any condition is:
dQ=TdS
where Q is the heat energy being transferred.
So regardless of whether the process is isobaric or isovolumeteric,
C= T(∂S/∂T)

Quote Quote by weeksy View Post
why this bothers me is that this equation as I understand should only hold for C_V (heat capacity with constant volume)
thats because dQ=TdS for a REVERSE-ABLE ONLY expansion (i.e dQ=0 , dS=0) ie adiabatic all of which occur as V, the volume is held constant for C_V. Hence dU=dQ and dQ(rev)=TdS = dU , which can then be simply substituted into the definition. OK

Sorry for the spiel, but my question is how can this same line of reasoning be true for C_p where dV is not constant and dU is not equal to dQ and expansion isn't reversible?
Thanks
You have made an incorrect hypothesis. The definition of specific heat, C, is not:
C= (∂U/∂T)
where U is the internal energy. The definition of specific heat is:
C= (∂Q/∂T)

One can easily prove that:
C_V= (∂U/∂T)_V
C_P= (∂H/∂T)_P
where H is the enthalpy, not the internal energy. Your logic worked
for C_V, but it didn't work for C_P because of the incorrect hypothesis.

Andrew Mason
Andrew Mason is online now
#4
Jun7-12, 12:37 PM
Sci Advisor
HW Helper
P: 6,562

Heat Capacity and Entropy Textbook Definition - Quick Question


Quote Quote by weeksy View Post
Just a quick question of something I found in my textbook but cant get how they produced it.

C_p =(∂Q/∂T)_p

that is the definition of heat capacity at a constant pressure p. Q is heat and T is temperature. This equation is fine and I know how to derive it. Now it is the next line which worries me.

C_p=(∂Q/∂T)_p = T(∂S/∂T)_p

where S is the entropy

why this bothers me is that this equation as I understand should only hold for C_V (heat capacity with constant volume)
thats because dQ=TdS for a REVERSE-ABLE ONLY expansion (i.e dQ=0 , dS=0) ie adiabatic all of which occur as V, the volume is held constant for C_V. Hence dU=dQ and dQ(rev)=TdS = dU , which can then be simply substituted into the definition. OK

Sorry for the spiel, but my question is how can this same line of reasoning be true for C_p where dV is not constant and dU is not equal to dQ and expansion isn't reversible?
Your question is why:

[tex]C_p=\left(\frac{\partial{Q}}{\partial{T}}\right)_p = \left(\frac{\partial{Q_{rev}}}{\partial{T}}\right)_p[/tex]

The answer is that it is implicit in the left side that Q is Qrev. Cp is determined by a reversible process:

Cp = dQ/dT = dU/dT + dW/dT

Now if dW = PdV and if P is constant,

Cp = dU/dT + PdV/dT = Cv + PdV/dT = Cv + P(RdT/P)dT = Cv+R

But dW = PdV ONLY in a quasi-static process.

AM
weeksy
weeksy is offline
#5
Jun7-12, 10:00 PM
P: 14
Ok very good Andrew, makes sense though seems a rather large implication that C_p is determined by a reversible process only. Oh and Darwin as I understand it entropy S is not defined by dQ=TdS for all conditions but dQ(rev)=TdS where (res) is for a reversible process only. Huge difference.

Thanks for the fast responses, I have an exam next week where I have to use that substitution and am much happier now I know where It comes from. Thanks again
weeksy
weeksy is offline
#6
Jun7-12, 10:09 PM
P: 14
I'm new to this forum, is there some sort of thanks button / points system to show appreciation ?
weeksy
weeksy is offline
#7
Jun7-12, 11:00 PM
P: 14
Quote Quote by Studiot View Post
I think you are confusing U and H, have another look at your textbook.

The partial differntial expressions you give for Cp are correct, but they are obtained by substituting Tds into the definition of H not U.

Similar partials can be derived for Cv by substituting Tds into the definition of U.
your 100% right.
dH=dU+pdV
=TdS-pdV+pdV+Vdp
=TdS+Vdp
since dQ(rev)=TdS
dH=dQ(rev) at at constant pressure
since C_p=∂Q/∂T we have

C_p=T(∂S/∂T)
awesome
thanks a bunch
Studiot
Studiot is offline
#8
Jun8-12, 02:23 AM
P: 5,462
The fact that you have come back to the thread to let us know how you got on is great, keep it up.

It is quite dispiriting if you put in a deal of effort to write something and you never hear further. Others have put in more the I have here.

go well

Andrew Mason
Andrew Mason is online now
#9
Jun8-12, 02:27 PM
Sci Advisor
HW Helper
P: 6,562
Quote Quote by weeksy View Post
your 100% right.
dH=dU+pdV
=TdS-pdV+pdV+Vdp
=TdS+Vdp
You seem to be missing a step at the beginning. Start with:

H = U + PV

dH = dU + PdV + VdP

For constant pressure dP = 0 so:

dH = dU + PdV = dQrev (first law, substituting PdV for ∂W).

Since dQrev = TdS by definition (ie dS = dQrev/T), then

[tex]\left(\frac{\partial H}{\partial T}\right)_P = \left(\frac{T\partial S}{\partial T}\right)_P = \left(\frac{\partial Q_{rev}}{\partial T}\right)_P = C_p[/tex]

Similarly, since

dU = TdS - PdV , for a constant volume process (dV = 0):

[tex]\left(\frac{\partial U}{\partial T}\right)_V = \left(\frac{T\partial S}{\partial T}\right)_V = \left(\frac{\partial Q_{rev}}{\partial T}\right)_V = C_V[/tex]


AM
Darwin123
Darwin123 is offline
#10
Jun8-12, 04:20 PM
P: 741
Quote Quote by weeksy View Post
Ok very good Andrew, makes sense though seems a rather large implication that C_p is determined by a reversible process only. Oh and Darwin as I understand it entropy S is not defined by dQ=TdS for all conditions but dQ(rev)=TdS where (res) is for a reversible process only. Huge difference.
This hypothesis is not valid for conditions under which the process is irreversible. Once example would be if a Carnot engine where modified by having kinetic friction between the movable piston and the cylinder. I would call this the "the rusty Carnot problem".
This probably won't appear on any test. However, I am now interested for its
own sake. I am curious how one would take into account kinetic friction. I think I have an idea, but I am not sure. I also wonder if a solution to "the rusty Carnot problem"
could be published anywhere.
Andrew Mason
Andrew Mason is online now
#11
Jun9-12, 10:37 AM
Sci Advisor
HW Helper
P: 6,562
Quote Quote by Darwin123 View Post
This hypothesis is not valid for conditions under which the process is irreversible. Once example would be if a Carnot engine where modified by having kinetic friction between the movable piston and the cylinder. I would call this the "the rusty Carnot problem".
This probably won't appear on any test. However, I am now interested for its
own sake. I am curious how one would take into account kinetic friction. I think I have an idea, but I am not sure. I also wonder if a solution to "the rusty Carnot problem"
could be published anywhere.
The work done by the rusty Carnot would include the friction. There would be no change the efficiency of the engine if you include that friction as part of the output.

The Carnot cycle is reversible in the sense that if you were able to store the output mechanical work you could reverse the heat flow and restore the original state of the reservoirs using that stored energy. So in the forward Carnot engine cycle if the output work was used to lift a weight, you could then lower that weight to reverse the direction of the cycle so that it acts as a Carnot refrigerator and flow all that heat back to the hot reservoir.

AM


Register to reply

Related Discussions
Definition of specific heat by via entropy General Physics 0
Calculating Entropy of a Star using Heat Capacity Introductory Physics Homework 2
Heat capacity & Latent heat Simple question Introductory Physics Homework 1
the difference between the entropy and the heat capacity? they are very similar!! Mechanical Engineering 19
Quick question: Grade 12 stuff (Heat capacity) Biology, Chemistry & Other Homework 7