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Heat conduction is not considered to be a form of work, since there is no macroscopically measurable force, only microscopic forces occurring in atomic collisions.

This excerpt is from a 2007 Wikipedia selection, and I was curious as to why a distinction between microscopic forces and macroscopic forces result in something to not be considered a form of work.

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    $\begingroup$ I hope you get a nice answer cause it's an interesting question. My understanding is that it's a process that is "statistical" in nature (it is just "diffusion": diffusion of particles is usual diffusion, and diffusion of energy is heat conduction). It happens even without the application of any force as a consequence of the fact that typically it's just more probable to "spread" than to "concentrate" something. $\endgroup$
    – Quillo
    Dec 22, 2022 at 10:35
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    $\begingroup$ Possible duplicate of “Is there a formula of heat that expresses it in microscopic work done?”; if so, that question should really be directed here because @John’s answer is so good. $\endgroup$ Dec 22, 2022 at 12:08

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This is more or less a definition. The question is then why heat and work are defined as distinct channels of the increasing of internal energy. In my understanding the distinction is in the following sense.

Let $(S,V)$ be our thermodynamic variables. What does this mean microscopically? At the end of the day volume has to do with microscopic energy levels in a (quantum) system. Fixing volume you fix the microscopic energy spectrum $\{ E_i \}$ of your system. Entropy $S$ on the other hand is related to how these levels $\{ i \}$ are populated $S=\sum \frac{N_i}{N} \log\left( \frac{N_i}{N} \right)$ (we assume we have some fixed number of particles $N$ that can populate the levels in whichever way $\{ N_i\}$) .

The total energy of the system is $E_{tot}= \sum E_i N_i$. In general it can change either when $\{ E_i \}$ or $\{ N_i\}$ are changed (through volume or entropy change respectively): $$\delta E = X d V + Y d S$$ Clearly these two channels are qualitatively different. In the first case the energy changes due to "drift" of the levels. In the second, the particles hop around from level to level. The former is then a reversible(= macroscopically traceable) process while the latter is not.

The convention is that first term $X dV$ is referred to as work $\delta W$ and the second $Y dS$ as heat $\delta Q$ (obviously $X$ and $Y$ are pressure and temperature). I argue they correspond to very different microscopic processes and hence should not be mixed.

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  • $\begingroup$ How does fixing volume fix the microscopic energy spectrum of the system exactly? $\endgroup$ Aug 12, 2023 at 15:33
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To rephrase what I think has already been stated in the answer by @John:
In statistical physics we distinguish transfer of energy via macroscopic changes (which we call work) and by microscopic means (which we call heat). Thus, heat conduction is not work, simply because of how work and heat are defined in statistical physics.

Remarks
Following the discussion in comments, it is worth pointing the possibly misleading use of term conduction when speaking about electrical conduction and heat conduction:

  • Electrical conduction is work, performed by a macroscopic current, and resulting in increase of internal energy (misleadingly called Joule's heat.)
  • Heat conduction is propagation of energy on microscopic level, from regions with higher temperature to those with lower temperature. The equivalent to electrical conduction in a gas/liquid could be convection producing increase of internal energy, e.g., via viscous friction.
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  • $\begingroup$ "...and by microscopic means (which we call heat)." Are you suggesting that all energy transfer by microscopic means is by heat? $\endgroup$
    – Bob D
    Dec 22, 2022 at 18:00
  • $\begingroup$ @BobD Not sure what you are alluding to. Do you have a suggestion for improving this answer? $\endgroup$
    – Roger V.
    Dec 22, 2022 at 18:28
  • $\begingroup$ I'm not suggesting the answer needs improvement. I'm just asking if that is what you meant. $\endgroup$
    – Bob D
    Dec 22, 2022 at 19:28
  • $\begingroup$ What I mean is that energy is transferred either either via macroscopic changes or on the microscopic level. Usually the former is referred to as work, the latter - heat. This much is stated in my answer - your comment suggest that you have in mind something that does not fit this definition - if this is the case, please be more explicit about it. $\endgroup$
    – Roger V.
    Dec 22, 2022 at 19:35
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    $\begingroup$ Agreed. The same applies to so called “friction heating” from the rubbing of surfaces $\endgroup$
    – Bob D
    Dec 23, 2022 at 12:20
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From a purely classical view useful work is only performed when a "charge", say $K_n$, is dropped through a potential difference, say $\phi_1^n -\phi_2^n$, these partial useful work terms balance each other in a reversible process but produce excess entropy in an irreversible process $$\sum_n (\phi_1^n -\phi_2^n)K_n = \delta Q \ge 0 \tag{1}\label{1}$$ $\delta Q$ being the internally irreversibly generated heat. The index $n=0,1,2,..N$ refers to the possible interactions (thermal, mechanical, chemical, electrical, etc.), with $n=0$ denoting the thermal interaction, so that $\phi^0=T$ is temperature and $K_0=S$ is entropy.

Viewed from this way, heat conduction is a special case when by definition $K_n=0$ for $n\ge1$ and thus $\phi_1^0-\phi_2^0=\delta T$ and $\delta Q=TdS$, or with the sign convention from higher to lower temperature $\delta T$ = -$dT$ and when combining this with $\delta Q = TdS$ we get energy conservation by $SdT=-TdS$ or $d(TS)=0$. In other words by excluding all the other work terms in $\eqref{1}$ and only keeping heat conduction total energy is conserved while entropy increases as it should in an irreversible process.

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In classical thermodynamics, the distinction between heat and work is always somewhat fuzzy. We usually know it when we see it, but we can't define it clearly.

From an atomistic and quantum perspective, however, the distinction is crystal clear:

  • A system (example: an ideal gas contained between two parallel plates) is made of many discrete microsystems (in the example, molecules).
  • Each microsystem has a number of discrete quantum states in which it can exist (in the example, the wavefunction of a molecule can have the component of its wavevector perpendicular to the plates such that the space between the plates contains one half wavelength, or two half wavelengths, or three half wavelengths, etc.)
  • Any particular quantum state of a microsystem has a characteristic stored energy (in the example, the characteristic stored energy of a quantum state is inversely proportional to the square of the wavelength).
  • One set of ways of transferring energy into or out of the system involves switching microsystems between different quantum states (in the example, a molecule might absorb a phonon from one of the plates, causing the molecule to switch up from the one-half-wavelength-between-the-plates state to the two-half-wavelengths-between-the-plates state, which has a shorter wavelength and therefore stores more energy). When energy is transferred that way, it's called "heat".
  • Another set of ways of transferring energy into or out of the system involves leaving the microsystems in the same quantum states, but changing the characteristic energies of the states (in the example, one might apply the necessary force to move the plates closer together, so that although molecules stay in states with the same number of half-wavelengths between the plates, the wavelengths get shorter and the stored energy increases). When energy is transferred that way, it's called "work".
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Some other answers veer close to saying this, but none outright.

The simple explanation is that in these old physical theories, the concepts are strongly (and today, quite obviously) anthropocentric.

"Work" is originally something that makes a machine move, in the sense that Victorian-era engineers would have known machinery.

Simple heat transfer is not "work", because it doesn't intrinsically make machinery move.

The fact that there may be microscopic alterations or movements in the substance is irrelevant - these were not generally visible in Victorian times as mechanical movements.

Devices like the bi-metallic strip were largely not in the consideration of theorists using thermodynamics to analyse steam engines and other large-scale industrial machinery.

The challenge for the Victorian engineer was typically to devise machinery that transforms a heat source efficiently into "work" - in other words, into movements of a kind which fulfil the human purposes for which the machinery was designed.

So what is "work" and what isn't, is essentially something that depends on the purpose of the process or machine being analysed.

The foundational thinkers in thermodynamics tended to define particular processes unconditionally as "work" - such as the generation of steam pressure, and pressure driving pistons - because they knew those processes were what drove the relevant machinery at the time.

Such processes were unlikely to arise accidentally, so the need to consider a pressure-buildup or the reciprocation of a piston as being non-work (i.e. an unwanted byproduct of a machine's operation) probably never arose.

Today, because thermodynamics tends to be taught to students abstractly (rather than in the context of an industrial application), and because many physicists would be loath to openly acknowledge the presence of such blatant human judgment in what many call a "hard science", the concept of "work" often ends up as a bit of a muddle.

But the essential difference between mere energy transfer, and "work", is that work is the energy transfer that is useful to the human purpose. A machine (made of metal and with parts designed to move), is "working" when it's fulfilling its purpose, not merely when it's thrashing about aimlessly, and certainly not when it is merely stationary and dissipating heat.

So to properly explain "work" in thermodynamics, you must be talking about a specific machine with a known purpose.

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  • $\begingroup$ reversible entropy transfer between two finite temperatures is always accompanied by macroscopic work $\endgroup$
    – hyportnex
    Dec 26, 2022 at 12:59
  • $\begingroup$ @hyportnex, I think you'll also find "entropy" and "reversible" require anthropocentric interpretation, and that (although not scientifically necessary or justified...) many insert cosmological assumptions about the supposed one-wayedness of physical processes. Again, this is largely a holdover of 19th century industrial realities which these theories analysed - you couldn't unburn coal, and considering the (re-)production of fossil fuels over eons did not make coal-burning "reversible" in any known human industrial process (even if it is ultimately reversible over cosmological timeframes). $\endgroup$
    – Steve
    Dec 26, 2022 at 13:45
  • $\begingroup$ A reversible Carnot cycle (isothermal-adiabatic-isothermal-adiabatic) is transport mechanism for entropy to be moved from a higher to a lower temperature whose net result is net work done $W$ equal to $A=(T_{high}-T_{low})S$ where $S$ is the amount of entropy absorbed at $T_{high}$ and also rejected at $T_{low}$ from the reservoirs. In general $A \ge W$ but are equal if and only if when the process is reversible. This is also an operational definition of reversibility. $\endgroup$
    – hyportnex
    Dec 26, 2022 at 13:59
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In thermodynamics, heat is defined as energy transfer, without transfer of mass, across the boundary of a system due solely to a temperature difference between the system and its surroundings. Work is defined as energy transfer, without transfer of mass, across the boundary of a system due to any intensive property difference other than temperature between the system and its surroundings.

Energy transfer can occur due to either heat or work, and both are treated as essentially the same as far as an energy balance is concerned (first law of thermodynamics), but we must distinguish between the two since according to the second law of thermodynamics we cannot completely transform heat into work (but we can completely transform work into heat). A simple viewpoint is heat is associated with randomness and work is associated with non-randomness. See discussions on entropy in a thermodynamics textbook.

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Work is, by definition, the quantity that appears in the Work-Energy Theorem: $\Delta W=\int\vec{F}\cdot d\vec{x}$. For a fluid system, for which the force is described by a pressure (force per unit area on a boundary surface), this can be re-expressed as $\Delta W=-\int p\,dV$ (for the work done on the fluid). This is all defined at the level of classical mechanics; nothing has yet entered from thermodynamics. Moreover, the work is expressed entirely in terms of the macroscopic variables—the pressure and volume of the fluid as a whole—not the microscopic position and momentum variables describing the locations of the constituent molecules making up the fluid.

The First Law of Thermodynamics states that there is another mode of energy transfer for macroscopic systems, different from work. That is heat, and the energy transfer into a system is $\Delta U=\Delta W+\Delta Q$.* Heat transfer does not involve a change in the macroscopic position of the fluid, so it is not considered work on the macroscopic fluid.

Think of what happens when molecules in a container are exposed to a source of heat energy. In particular, say that the surface of the container is heated up to a temperature that is hotter than the fluid itself. How does that thermal energy contained in the walls transfer itself to the moving fluid molecules? A typical sequence of events might run as follows:

  • An incoming molecule strikes the wall and is physically adsorbed onto the surface. (That is, it is briefly stuck in place.)
  • After a short time adhering to the hot surface, the particle is dislodged by thermal vibrations in the surface and leaves again, returning to the fluid bulk.
  • On average, after this happens, the molecule has more kinetic energy than it started with.

At the microscopic level, work is done on the molecule, by the thermal vibration of the surface. And indeed, in a completely microscopic description of energy, there is no such thing as heat; kinetic energy is only transferred to particles via a force doing work. However, at the macroscopic level, we do not see the surface of the container moving, and so the energy transfer is not counted as work, but instead as heat.

*The use of the $\Delta$ notation here is standard, as I previously noted in another answer:

The use of a $\Delta$ in an expression like $\Delta Q$ or $\Delta W$ does not indicate that $\Delta W$ is the difference between two values of $W$, which would be impossible, since W is not a state function. Instead, $\Delta W$ is just defined as an integral of $\not dW$. Another similar notation that is frequently used in engineering is $\dot{Q}$, meaning heat transfer per unit time, even though it is not the time derivative of anything.

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Yet another way to reach the same conclusion as others have done in various ways. In (classical mechanics) work is something that will result in change of the kinetic energy of its target. (A implies B.) Thus, logically, if such change of kinetic energy is not observed, there was no work either. (Not B implies not A.)

For a single atom hitting another one, there is clearly a force-mediated change in energies. But when looking at a whole object being heated, only its constituent particles have their average kinetic energy changed; the system itself has not changed its state of motion at all (it's center of mass will have conserved the previous motion) and the kinetic energy of the whole system is the same as before. ... thus there could not have been work.

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Why isn't heat conduction considered a form of work?

In the sense that, on average, the molecules of the higher temperature object do work via collisions on the molecules of the lower temperature object, thus transferring molecular KE from the higher to the lower temperature object, heat conduction involves net work at the molecular level.

However, this average transfer of energy by means of molecular work is due, in the first place, to the fact that the average KE of the molecules of one object is greater than the average KE of the other, i.e., is due to the temperature difference between the objects. In other words, it is only a consequence of the temperature difference that net molecular work is done by one object on the other by conduction.

If the two objects were the same temperature, there would still be molecular work done by above average KE molecules of one object on below average KE molecules of the other, and vice versa, but there would no be net molecular work done by one object on the other.

Hope this helps.

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There are already a lot of great in depth answers, so I'll provide a very simple and concise one.

Heat is a form of work. In thermodynamics, work is divided into "proper work" and "heat". But they are both work.

The distinction was originally from an engineering point of view (can or can't power a machine). Later, it was related to statistics. But on a more fundamental level there is no difference.

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