What is the differentiating factor between work and heat?'

Question

Although work and heat do the same thing (increase or decrease the internal energy of the system), There is still a fundamental difference between them. The definition of entropy necessitates the consideration of heat and not work. Second law of thermodynamics essentially states that heat cannot be converted to work completely by a cyclic engine. Is the word "Thermodynamic work" or "Hidden work" suitable for heat?

Edit (1) : If we admit the explanation that "heat" is the work which we do not see, or heat is the work done at molecular level, I feel we run into the problems of defining clearly what it means. For instance, someone with a super sense who can account for some of the work done at molecular level would calculate different amount of heat from someone who cannot account for it. It is possible that both do not agree on the allowed direction of the process due to this (which is essentially determined by second law, which relies on the definition of entropy that relies on the definition of heat). Does it all work out mathematically somehow?

Answer 1

This is a very good and fundamental question. It lies at the heart of thermodynamics and related disciplines. There is no general answer for all physical systems, unfortunately.

What we define as work and heat is highly dependent on the system in question. Take the classical example of gas confined by a movable piston, in contact with a heat reservoir. In this case, all energy transferred to/from the gas from the mechanical movement of the piston is work. All energy transferred to/from the heat reservoir is heat. We do not prove that the system only exchanges heat with the reservoir; we define it in this way.

The take-away point here is that we define heat and work at the outset, when we are developing our physical model. Where the concept of heat is useful is in keeping track of changes that result in changes in entropy, so we usually associate heat with entropy. Usually, the definition is of the form 'heat is energy exchanged with the heat reservoir', so we need to have some notion of a heat reservoir in our model, for which a temperature can be defined. Then:

$$ \partial S = \frac{\partial Q}{T} $$

Unfortunately many common definitions of heat that are given in various sources are circular in nature; for example "heat is the energy that is transferred due to temperature differences", or "heat is the energy that can't be used to do useful work."

Answer 2

Consider a gas inside a container. Every time a molecule hits the wall of the container there is work being done. If we were able to compute the force it did and the displacement it caused to the wall and then sum over all collisions we would have a macroscopic value for work and we would not need to talk about heat. The first law would be just $\Delta U+W=0$. Since we are not able to compute all those microscopic work we effectively call the missing quantity, $\Delta U+W\neq 0$, heat. Heat work done at molecular level.

Answer 3

There is a thermodynamic answer to this question. It is obtained by reasoning carefully, one step at a time, from the basic concepts.

First we need a way to define thermal isolation, i.e. the situation where there is no heat flow. We want to define it without first defining heat, so how can we? There is a way, by the following three steps:

(i). First define equilibrium as the state which an isolated system will tend to, over time, if the external constraints on it are not changing.

(ii). Next we need a definition of two things being in thermal equilibrium with each other. We consider two systems that have each reached equilibrium while apart. If, when they are brought into contact, no changes take place in either, then we say they are in equilibrium with each other.

(iii). Now we are ready to define thermal isolation. If we have two systems which are not in equilibrium with each other, then, if they continue being not in equilibrium with each other then they are thermally isolated from each other.

OK now we know how to define thermal isolation, so we can arrange to a have a system which is thermally isolated (we know what physical arrangements are needed) and we can experiment with it. In particular, we can discover the effects of doing work on such a system. We discover that a given state change requires a given amount of work, and we develop the equation $$ \Delta U = W $$ This expresses the fact that, in conditions of thermal isolation, the amount of work required to bring about a given state change is independent of the path. This is called the principle of adiathermal work. It allows us to deduce that there is a function of state, called $U$ (internal energy), which changes by the amount of work supplied if the change is brought about in conditions of thermal isolation. We can use such processes to map out, for any given system, what is the internal energy of each of its equilibrium states.

We're almost there! Finally, we allow processes on a system without thermal isolation. We define the heat passing into the system by the equation $$ Q = \Delta U - W. $$

This argument is presented more fully in the textbook on thermodynamics by myself, and I learned it mostly from the one by Adkins. The argument is not circular. It does, however, require us to know how to quantify work in an adiathermal process. This is done by bringing in the physics of whatever process is doing the work: a force pushing on a piston, a battery passing an electric current, a field magnetizing something, or whatever.

A final comment. Some of the other answers take an interest in the fact that the distinction between work and heat is not clear-cut in some circumstances. That's correct. It is an example of the thermodynamic limit. The distinction between work and heat becomes clear-cut in the thermodynamic limit, which is the limit of systems with a large number of microstates.

Answer 4

Edit (1) : If we admit the explanation that "heat" is the work which we do not see, or heat is the work done at molecular level, I feel we run into the problems of defining clearly what it means. For instance, someone with a super sense who can account for some of the work done at molecular level would calculate different amount of heat from someone who cannot account for it. It is possible that both do not agree on the allowed direction of the process due to this (which is essentially determined by second law, which relies on the definition of entropy that relies on the definition of heat). Does it all work out mathematically somehow?

Heat is energy transfer on microscopic level, whereas work in thermodynamics is the energy change induced by the macroscopic variables. Thermodynamics/statistical physics aims at describing systems of many particles in terms of few macroscopic variables ("natural variables"), obtained via averaging over all the molecules constituting the system. Any energy change that can be presented in terms of these "natural" variables is called work. Obviously, some of the energy changes do not allow such a description, so the two kinds of energy transfer always remain distinct.

However, there is indeed some flexibility in choosing these thermodynamic variables, so the distinction between work and heat is not cut in stone: $$dU = TdS +\sum_iX_idx_i,$$ where the sum often includes terms like $-pdV$, $\mu dN$, but may also contain, e.g., $MdH$, if we account for the magnetization. The values of entropy will not be the same... because we are dealing with different ensembles.

In fact, the transition between different number of thermodynamic variables is extensively covered in statistical physics textbooks, but too obscured by these variables seeming too natural (pun intended) for an ideal gas, the ubiquitous example considered in most texts. Thus, in canonical ensemble we have only variables two variables: $$ dU = TdS - pdV,$$ whereas in a grand canonical ensemble we have $$ dU = TdS - pdV + \mu dN $$

Is the canonical entropy the same as the grand canonical entropy? Obviously they are not, since we consider different phase spaces: in the first case the number of particles fixed, whereas in the second case states with different particle numbers are possible, and we are interested in dependence on this particle number.

This dependence of different ensembles on the constraints used was explored by Jaynes, who showed how different thermodynamic ensembles follow from using different phase spaces and constraints together with the maximum entropy principle: see Information theory and statistical mechanics.

Example
Let us consider a system of $N$ non-interacting Ising spins, which can take values $\pm 1$. The number of states is $\Omega =2^N$, so that the entropy is $S=\log\Omega=N\log 2$. As a more general approach we could use the information-theoretical expression for entropy, where the probability of each state is $p_i=\Omega^{-1}$, so that $$ S = -\sum_i p_i\log p_i=\log \Omega.$$

Let us now introduce an additional variable $n$, that counts the number of spins $+1$. For every fixed $n$ there are $\Omega_N={N \choose n}$ possible configurations. Note also that $\sum_{n=0}^N{N \choose n}=2^N$. Now, if probability of state with $n$ spins up is $p_n$, and all the configurations for a fixed $n$ are equiprobable, we have: probabilities of states given by $$ P(i|N)=\frac{p_n}{\Omega_N}$$ and the entropy becomes $$ S=-\sum_{n=1}^N\sum_{i=1}^{\Omega_N} \frac{p_n}{\Omega_N} \log \frac{p_n}{\Omega_N}= -\sum_np_n\log p_n + \sum_n p_n\log{\Omega_N} \tag{1} $$ The second term is just the sum of entropies of the subsystems with different values of $n$, weighted with probabilities $p_n$. If we take probabilities $p_n$ to be equal to their weight in the total: $$ p_n=\frac{\Omega_N}{\Omega},$$ we will get back to the original value of the entropy. However, if we have an external parameter, controlling the occupation of macrostates with different $n$, the entropy will not be the same. So what makes entropy different is not the presence of an extra variable, but the fact that this variable may influence the probabilities of different macrostates, reducing the entropy. In other words, the reason for a different entropy is because the system is different - it is not fully described without taking into account the additional variable.

To see how this is related to microcanonical ensemble, we note that probability of a microstate is $$ p(i,N)=Z^{-1}e^{-\beta E_i(N)+\beta\mu N}=p(i|N)p_N$$ So that entropy is $$ S=-\sum_{N=0}^{\infty}\sum_{i=0}^{\Omega_N}p(i|N)p_N\log \left[p(i|N)p_N\right] $$ If we assume that all the states for the same $N$ are equiprobable, i.e., all of them have the same energy $E(N)$, then we can write $$ p(i|N) = \frac{1}{\Omega_N}, p_N=\frac{\Omega_N}{Z}e^{-\beta E(N)+\beta\mu N}$$, and we arrive at the expression very similar to (1). If the energy of all configurations is the same and the chemical potential is zero, then all the configurations are equiprobable, and the entropy becomes the log of the total number of states.

Related
Notion of Entropy as a functional - is it neccessary?
Microcanonical ensemble through Maximum Entropy method
How Subjective is Entropy Really?

Answer 5

It turns out that terms representing the kinds of energy in a system generally are products of two factors. One factor is an extensive variable (depends on how much of the system you consider), and the other is intensive (does not). For example, $PV$: $P$ is intensive, $V$ extensive, or $\mu N$: $\mu$ (the chemical potential of a species) is intensive, $N$ (the number of molecules of the species) is extensive, or $VQ$: $V$ (not volume, but electrical potential) is intensive, $Q$ (charge) is extensive. These pairs are called "thermodynamically conjugate."

An infinitesimal change in energy of the system can occur either with a change of the extensive variable or the intensive one: e.g. $-PdV$ or $-(dP)V$. For any extensive variable other than entropy, when the extensive variable changes, work is being done (e.g. $-PdV, \mu dN, V dQ$). This is analogous to (and sometimes equal to) a force acting over some distance. Any additional change of internal energy (e.g. $dU + PdV - \mu dN - VdQ -$ (other changes associated with extensive variables)) is associated with heating, and equals $TdS$.

Answer 6

Although work and heat do the same thing (increase or decrease the internal energy of the system), There is still a fundamental difference between them. The definition of entropy necessitates the consideration of heat and not work. Second law of thermodynamics essentially states that heat cannot be converted to work completely by a cyclic engine. Is the word "Thermodynamic work" or "Hidden work" suitable for heat?

I describe a view here that goes back 200 years now to Carnot's "Reflections on the Motive Power of Fire," published in 1824, which after Colding and Mayer's discovery of energy conservation has largely been suppressed, misunderstood or ignored. Ove the intervening years many historians and physicists have claimed that Carnot was the originator of the concept of entropy, although his use of the words caloric (or calorique) and chaleur itself was sometimes ambiguous or contradictory.

Putting aside historical priorities, we can define thermodynamic work as the movement of thermodynamically relevant stuff by thermodynamically relevant force. Here the various stuff and forces are the subject of mechanics, electricity, magnetism, chemistry, etc. and the goal is to investigate how the movements of the various form of stuff and forces are related to each other.

In the simplest model of a thermodynamic interaction, we imagine that the system that can participate in several forms of interactions (gravitational, mechanical, chemical, electrical, etc.) are surrounded by pairs of work reservoirs of the same kind, a pair of mechanical, a pair of electrical, etc. Each reservoir is imagined to be so large and the stuff in the system being moved to and from to be so small as compared to the reservoir's size that the reservoir's characteristic potential does not change by the interaction with the system. We might call the reservoirs of infinite extent.

If in the $k^{th}$ interaction, an amount of stuff $\Delta X_k$ is moved from a reservoir storing an infinite amount of $X_k$ at potential $Y^1_k$ to another reservoir, also storing an infinite amount of $X_k$ now at potential $Y^0_k$ then, by definition, the amount of work $(Y^1_k$ -$Y^0_k)\Delta X_k$ was lost between them. This in complete analogy the way gravitational work is done on a mass of "human" size in the gravitational field of the Earth because the field is not affected by the work. Similar work can be done on electric charges in an electrostatic potential field or chemical compound in a chemical potential field, etc.

Now Carnot's calorique or what we nowadays call entropy is similar in the sense of being the stuff of work in that if an amount of entropy $\Delta S$ is moved from a "heat reservoir", better called an "entropy reservoir" at temperature $T^1$ to another at temperature $T^0$ then thermal work in the amount of $(T^1-T^0)\Delta S$ is lost.

The purpose of the thermodynamic system sandwiched between these pairs of reservoirs is to facilitate possible work interactions among these reservoirs by allowing that the system participate in all the possible interactions but not necessarily simultaneously. The usual Carnot cycle is an engine that can connect to a pair entropy reservoirs and a pair of other reservoirs, usually mechanical but can also be magnetic as used in paramagnetic cooling.

A single equation/inequality based on pure observation, physics and chemistry, summarizes almost everything with $k=1,2,..,n$ denoting the various nonthermal interactions:

$$\sum_{k=1}^n (Y^1_k-Y^0_k)\Delta X_k + (T^1-T^0)\Delta S =\mathcal D \ge 0 \tag{1},$$ where $\mathcal D$ denoted the total lost, i.e., dissipated work, that is never negative.

A reversible process is defined by the equality $\mathcal D=0$, and irreversibility obtains when it is not reversible that is $\mathcal D >0$. For a reversible transport of stuff $\Delta X_k$ and $\Delta S$ we see the the various lost work terms are in balance, just as a see-saw in a gravitational field, or any mechanical system with friction and other types non-mechanical "losses". You can also see that if one ignores certain non-thermal processes, interactions, that are actually happening, then it may appear as dissipation of the work of the other work processes that are being taken into account. This is the origin of one major branch of irreversible thermodynamics, where these dissipating processes are called internal or concealed, and I think this may be what you are referring to as "hidden work" in your question.

This concept of "hidden work" can also be combined with another major assumption, you may call it axiom, according to which the dissipation, $\mathcal D$, the total lost work is actually a constitutive quantity characteristic of the system itself through which the transport of the various work stuff is taking place.

So far, we have only discussed work conservation if $\mathcal D=0$ or work non-conservation if $\mathcal D >0$, this is the first half of the 2nd law. When this is rounded up with the 1st law we will also give meaning to $\mathcal D$. Specifically we can rewrite Eq (1) as energy conservation. Define $\Delta U^i = \sum_{k=1}^n Y^i_k\Delta X_k + T^i\Delta S;i=1,2$, then:

$$\Delta U^1=\Delta U^0 + \mathcal D\tag{2}$$

If we assume that the system has returned to its original state at the end of the interactions, then $\mathcal D$ ends in the reservoirs of index $i=0$, and all experience shows that it ends in the entropy reservoir at temperature $T^0$, and never in anything else. This is the other "half" of the 2nd law. Specifically, that if the system performs a cycle then the entropy reservoir at $T^0$ will absorb $$\Delta S^0= \Delta S + \frac{\mathcal D}{T^0} \tag{3}.$$

This Eq (3) is sometimes phrased as that the system generates entropy in the amount of $$\sigma = \frac{\mathcal D}{T^0} \tag{4}.$$

In this view of thermal interactions, "heat" is only the dissipation $\mathcal D$ and is "evolved heat"; "heat" or "evolved heat" is not work and it is not energy, per se. Instead, thermal work is the transport of entropy from one temperature to another, and entropy is a para-conserved quantity, meaning that the entropy participating in any process cannot decrease, it can increase in an irreversible process. Electric charge is conserved, gravitational mass is conserved, inertial mass is conserved, volume is conserved, surface area is not conserved, etc., but entropy is para-conserved, and each of these quantities has an associated work process.

Thermal work, that is entropy transport between two different temperatures can convert to other types of work but cannot convert if the transport is isothermal. This does not mean that during isothermal entropy transport the system cannot work just that the source of such work is not thermal. It is not "heat" that cannot be converted to work completely but that entropy is a para-conserved quantity and thermal work is the transport of that para-conserved entropy between temperatures.

Of course, evolved heat, ie., dissipated work, can be used for thermal work but at another, lower temperature; entropy is just entropy, if its temperature dropped you move it to a lower temperature and then you obtain thermal work.

Lastly, for thermodynamic systems that cannot be reasonably described as being sandwiched between pairs of work reservoirs; the answer is still basically Eq (1). We break up the system into small pieces so that we have $Y^1_k$ and $Y^0_k$ infinitesimally close and the flow $\mathbf J_k dt \approx \Delta X_k$ being so small that its transport does not disturb the potential field $\nabla Y_k \cdot \delta \mathbf r \approx Y^1_k-Y^0_k$, and then $\mathbf J_k\cdot \nabla Y_k$ is the working (ie., work rate) per unit time and per unit volume lost in the $k^{th}$ interaction ($X_0=S, Y_0 = T$).

Answer 7

If I understand your question correctly, you're asking about what fundamentally distinguishes work from heat. The key difference between work and heat lies in how they affect the system’s microstates.

Work doesn’t directly lead to a significant increase in microstates, unless irreversibilities are involved (e.g., rapid compression causing friction or turbulence).

Heat transfer inherently increases microstates by introducing disordered energy distribution, always tied to entropy.

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Work

When work is done on a system, how the microstates change depends on how the work is applied. For instance, compressing a gas decreases spatial microstates as volume decreases, while momentum microstates increase as internal energy rises. The total number of microstates is given by:

$$ \Omega_{\text{total}} = \Omega_{\text{spatial}} \times \Omega_{\text{momentum}} $$

Spatial microstates scale with the volume:

$$ \Omega_{\text{spatial}} \propto V^N $$ where $ N $ is the number of particles. A smaller volume means fewer spatial microstates.

Momentum microstates scale with internal energy:

$$ \Omega_{\text{momentum}} \propto U^{3N/2} $$

As internal energy increases due to work, the momentum microstates increase. The overall change depends on whether the process is slow (quasistatic) or rapid (introducing irreversibilities).

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Heat

In contrast, adding heat always increases microstates predictably, raising entropy through the relation:

$$ S = k_B \ln \Omega $$

For a reversible process, the entropy change is:

$$ dS = \frac{\delta Q_{\text{rev}}}{T} $$

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Example: Compressing a Gas

• Work (Compression): Compressing a gas reduces spatial microstates but increases momentum ones as internal energy rises. The total number of microstates changes based on the speed of the compression.
• Heat Addition: Adding heat consistently increases microstates, raising entropy predictably.

Answer 8

What is the differentiating factor between work and heat?'

Both heat and work are means of energy transfer. The main differentiating factor is heat is energy transfer driven solely by temperature difference whereas work is energy transfer driven by force acting through a distance.

For example, The way in which entropy is defined is a very good way to differentiate between work and heat. But, why is there such a distinction between the two things?

Entropy is only transferred when there is a transfer of heat. However entropy can be generated by irreversible work or by an irreversible transfer of heat. So entropy is not unique to heat transfer.

Is it the limitation of Newtonian mechanics that it never accounted for something like heat which could also change the energy of the system?

Newtonian mechanics accounts for macroscopic changes in the energy of a system relative to an external (to the system) frame of reference, such as changes to the kinetic energy and gravitational potential energy of an object as a whole. Heating such an object would not change its velocity (kinetic energy) or its elevation (gravitational potential energy).

Thermodynamics, in contrast to Newtonian mechanics, accounts for changes in the internal energy of an object, i.e., changes in the kinetic and potential energy of the object at the microscopic (molecular) level, which may occur as a result of work or heat transfer.

Is the word "Thermodynamical work" or "Hidden work" suitable for heat?

I wouldn't say that "thermodynamical work" or "hidden work" are suitable terms for heat. But it is true that when one heats an object or does thermodynamics work on an object one can not "see" the molecules of the object speed up or slow down (change kinetic energy) or see the molecules move closer or farther apart (change potential energy) as a result of the energy transfer. On the other hand, when one does Newtonian work on an object we can see its velocity change (change in kinetic energy), or see its elevation change (change in gravitational potential energy). So, in that sense, the effects of heat and thermodynamics work might be considered "hidden".

Hope this helps.

Answer 9

Heat is a form of energy. And if we go into the basic definition of energy, It is the capacity to do work. They are interrelated. But mathematically they are one and the same. Their SI units is the same ie. Joules . The only thing that changes is the physical interpretation .

Think of a heat engine in a car. It is ready to do work because it contains a certain amount of stored energy . When work is done, energy is transferred between systems or from one form of energy to another.

Thus Heat energy is what the engine contains ( or possesses) while work is what has happened with that heat energy , is it lost to the surrounding, or is it converted to kinetic energy? If yes, how much is what is Work . You will be knowing the formulas that work is the difference in potential energy or sometimes difference in kinetic energy.

So finally, Heat is what you already have and Work is what you do with it