# When is path-independent heat a useful idea?

1. Dec 23, 2015

### trekkiee

For Fourier's law $\vec{J}=-k\nabla{T},$ where
$\vec{J}=$ conductive heat flux in $\frac{W}{m^2}$
k=thermal conductivity in W/m-K
T=temperature,
$\frac{1}{k}\vec{J}$ certainly appears to be a conservative field with temperature as potential and for any path from position a to position b, $\int\limits_{path}\vec{J}\cdot d\vec{r}=-k(T_b-T_a)$, i.e, conductive heat flows from an equipotential surface at $T_a$ to an equipotential surface at [lower] $T_b$.
This is analogous to:
$\vec{F}_{grav.}=-\nabla{\mbox{[potential energy]}}$ where $\int\limits_{path}\vec{F}_{grav}\cdot d\vec{r}=-(P.E._b-P.E._a)$
and mass flows (falls) from a to b.

Note: The path from position a to position b is a path in three dimensional space with coords x,y,z, not a path in a thermodyn. phase space, such as p-V diagram, in which heat is always path dependent.

But this path-independent heat seems to only be useful when heat is considered to be an idealized conserved quantity, i.e.,
1. heat is a perfect fluid - no meaningful internal or thermal energies, no chemical interaction, nuclear interations, partical collisions, particle velocity distributions, etc.
2. heat is conserved - no unwanted heat losses.
Which is a clearly limited model.

On the other hand, isn't it true that, whenever we do any relatively-straightforward engineering analysis involving conductive heat flow, such as the very-common one dimensional application of Fourier's law, that we are treating heat as a ideal conserved quantity?

But on the other hand, heat transfer requires $\Delta T\Rightarrow$ heat transfer never occurs in thermodyn. equilibrium $\Rightarrow$ heat transfer is always irreversible $\Rightarrow$ heat transfer always involves unwanted heat losses $\Rightarrow$ heat transfer can never involve a perfectly idealized conserved quantity.

So I think the conclusion is:
We do pretend that heat is an idealized conserved quantity when we do relatively-straightforward conductive heat transport analysis, even though the 2nd law $\Rightarrow$ there'r always unwanted heat losses.

Does this seem reasonable, or am I missing something important?

2. Dec 23, 2015

### Staff: Mentor

This applies only if the conductive heat transfer is taking place at steady state. For transient conductive heat transfer, it is not correct.

Steady state heat conduction in a solid is definitely not a limited model. Certainly inside a conductive solid, there are no heat losses. Just consider steady state heat flow along an insulted rod, with one end at a higher temperature than the other. All the heat that enters one end flows out the other end. In 3D steady state situations, the same "heat conservation" applies; the only locations where heat is gained or lost by the solid is at the boundaries.

So what. How does that differ conceptually from what we do when we calculate the area of a circle, even though we know that there is no such thing as a perfect circle. And, as I said above, inside the solid, there are no heat losses.
Irreversibility does not imply unwanted heat losses. Both steady state and unsteady state (transient) heat transfer calculations have provided accurate quantitative predictions of actual experimentally observed behavior for hundreds of years. That's why we study it.
No. The 2nd law does not relate to unwanted heat losses at all. And, conservation of energy is addressed exclusively by the 1st law.
No, this doesn't seem reasonable. If you want to learn more about the first and second laws of thermodynamics, please see the following Physics Forums Insights link: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/. If you want to learn more about conductive heat transfer, see the classic work by Carslaw and Jaeger, Conduction of Heat in Solids. If you still feel that studying conductive heat transfer is not a useful pursuit, neglect it at your own risk.

3. Dec 24, 2015

### trekkiee

More than one person has objected to "unwanted heat loss", which is apparently the wrong thing to say in this context. I'm referring to the idea that heat transfer through a finite temperature difference can contribute to the irreversibility of processes (Moran, John (2008). "Fundamentals of Engineering Thermodynamics", p. 220. John Wiley & Sons). I should have been more clear in the OP. I'm talking about a real process (not idealized or quasistatic, etc.) of heat conduction across some finite temperature difference (not infinitesimal), and therefore not reversible. What I mean by unwanted heat losses is heat loss that causes an irreversible increases in entropy, either in the system or the surroundings, and that these heat losses that cause irreversible entropy increases imply that heat, even if it's being treated as some sort of idealized fluid, is not a conserved quantity.

A long-needed review of thermodynamics started me thinking about this. Of the many ways that work and heat are different, one that struck me was that work in a conservative field could be path independent, whereas it seems that heat is never path independent in a real process. What seemed to me to be the only process which might come close to path independent heat was Fourier's conduction law: a vector quantity = -gradient (a scalar), which is similar in form to the definition of a conservative field: field vector=-gradient(potential). So I wondered if the vector quantity in Fourier's law, conductive heat flux, can be considered to be a conservative field with potential= the scalar quantity in Fourier's law =temperature. It seems to me that all heat transport texts, when they discuss conductive heat transport, treat heat in this fashion - as an idealized, conserved fluid that flows from higher temperature surfaces to lower temp. surfaces.

But if real heat transport across a finite temp. diff. is irrev. then heat must be lost to irrev. increases in entropy (syst. or surr.), implying heat is never a conserved quantity, even when we try to pretend it's an idealized fluid.

I realize that there are irrev. processes, which therefore cause irrev. increases in entropy, but involve no heat in/out of the system- e.g., free expansion vs a vacuum. But free expansion isn't a heat transfer process (conduction, convection, radiation). It seems to me that any real heat transfer process across a finite temp. diff. must be irrev., and therefore incur irrev. entropy losses (in the syst. or surr.), and that these irrev. entropy differences must be cause by heat in or out, which I was calling unwanted heat losses.

I'm not stating these ideas as fact. In fact, I do not completely understand them, and I'm trying to understand them. That is why I posted.

Chestermiller: Thanks for your reply. I glanced through an online copy of Carslaw and Jaeger, Conduction of Heat in Solids, but there was no mention of rev. or irrev. processes, or of entropy, etc., just conductive heat transport. I looked at your post, understanding-entropy-2nd-law-thermodynamics, but It didn't seem to address the idea I'm trying to understand in this thread.

4. Dec 24, 2015

### Staff: Mentor

As I said in my previous post, heat losses and irreversible increases in entropy are mutually exclusive effects. Even when heat is a conserved quantity, entropy can increase. If heat is transferred between a hot body and a cold body, there is no heat loss, but the entropy of the hot body decreases and the entropy of the cold body increases. If the transfer is reversible, then the increase equals the decrease, and there is no net change in entropy. If the transfer is irreversible, heat energy is still conserved (i.e., there are no heat losses), but the sum of the entropy increase in the hot body and the entropy decrease in the cold body do not cancel, and there is a net increase in entropy. But the entropy generation takes place within each of the bodies by dissipation of temperature gradients, and the heat lost by the hot body is gained by the cold body. You need to read my Insights article more carefully.

No. No. No. Irreversibility and conservation of energy are two entirely separate concepts.
Of course it is. The rate of heat transfer is zero.
If you are talking about the combined system and surroundings, then there are never any entropy losses. In irreversible processes, entropy always increases for system plus surroundings. So where are these alleged heat losses associated with irreversible processes?

For a given system, entropy can increase in an irreversible process by two different mechanisms: (1) Heat transfer at the boundary of the system with the surroundings and (2) entropy generation within the system as a result of dissipation of temperature gradients and viscous heating (this is the only entropy effect that occurs in your free expansion example). Mechanism (1) is present in both reversible and irreversilbe processes. Mechanism (2) is present only in irreversible processes.
The reason I gave you two references (the best ones I know of on these effects) is that one of them (Carslaw and Jaeger) addresses all your doubts regarding conservation of thermal energy, while the other one (my Insights post) focuses on your gaps in understanding of entropy and irreversible processes. I was hoping that together, you would be able to obtain a correct picture of both concepts and how they relate. So far, your misunderstanding remains.

Let's drop the discussion about conservation of thermal energy (heat) temporarily, and focus on reversible and irreversible processes, and entropy. Please read my Insights article again, and get back to me with any questions you may have. After that is cleared up, we can get back to conservation of thermal energy.

Chet

5. Dec 25, 2015

### trekkiee

Thanks! I understand it much better now.