Register to reply 
Heat Capacity and Entropy Textbook Definition  Quick Question 
Share this thread: 
#1
Jun712, 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 REVERSEABLE 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 


#2
Jun712, 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 C_{p} are correct, but they are obtained by substituting Tds into the definition of H not U. Similar partials can be derived for C_{v} by substituting Tds into the definition of U. 


#3
Jun712, 12:30 PM

P: 741

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) 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. 


#4
Jun712, 12:37 PM

Sci Advisor
HW Helper
P: 6,679

Heat Capacity and Entropy Textbook Definition  Quick Question
[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 quasistatic process. AM 


#5
Jun712, 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 


#6
Jun712, 10:09 PM

P: 14

I'm new to this forum, is there some sort of thanks button / points system to show appreciation ?



#7
Jun712, 11:00 PM

P: 14

dH=dU+pdV =TdSpdV+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 


#8
Jun812, 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 


#9
Jun812, 02:27 PM

Sci Advisor
HW Helper
P: 6,679

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 


#10
Jun812, 04:20 PM

P: 741

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. 


#11
Jun912, 10:37 AM

Sci Advisor
HW Helper
P: 6,679

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 