# Why work $W$ and heat $Q$ are different concepts?

I understand heat as the flow of energy (through radiation, convection or conduction) from one body to another. When I think about conduction (for example) I visualize particles that jiggle a lot bouncing against particles that jiggle less and transferring heat to them progressively.

Thus, with this collisions and change in momentum $\Delta p$ over a period of time $\Delta t$, we have a force $F$ and a displacement $d$ of particles. As the formula of work is defined as $W = F \times d$ why can't we consider heat the same as work at the atomic level?

If this kind of work is not what we define as heat, why don't we take this kind of energy into consideration when it comes to use the formula of internal energy $U = W + Q$ ?

The distinction between heat and work only comes about in statistical physics. The idea is that while work is a transfer of energy through the macroscopic degrees of freedom which are described by macroscopic (i.e. thermodynamic) quantities (like pressure and volume), heat is a transfer of energy through the remaining degrees of freedom which are ignored in the macroscopic description.

In the above sense, the distinction between work and heat is in some way artificial. It is induced by our choice of which degrees of freedom we want to consider in our description of a physical system and which ones we choose to be ignorant about. This also means that temperature and entropy are basically dependent on the convention one chooses to describe the system. This perfectly matches the idea from information theory where the amount of information contained in a message is characterized by the so called Shannon entropy which is exactly the same thing as the physical entropy, except for the factor of $k_B$: a message like "The OP's name is 21Brunoh." contains exactly zero information to you because you already knew that. That same message may however be considered quite informative by someone unaware of this thread. Thus, information is an observer-dependent concept. Entropy characterizes the lack of information.

In thermodynamics however, there is a natural ("canonical") choice of the quantities of a physical system one can know about (like pressure, number of particles) and those which are inaccessible (like the positions and momenta of all the individual particles). The latter lead to a lack of information, characterized by the (physical) entropy. Changes in the observable/macroscopic quantities are called different forms of work, e.g. changes in pressure and volume are referred to as mechanical work. Changes in temperature and entropy are referred to as heat.

As an additional remark, please note that heat and work are meaningful only as changes of the internal energy. I'd always avoid writing $U=W+Q$ and instead use $\Delta U=W+Q$: changes in the internal energy are due either to work or heat. There is no such thing as the heat content or work content of some object, but one can talk about the energy content. You wouldn't talk about the amount of cash vs. the amount of credit card money you have in your bank account. Both paying in cash and by credit card are ways of changing the balance of your account, though.

• About notations : I avoid $\Delta W$ and $\Delta Q$ as transferred work and heat may not be differential forms
– Pen
Mar 10, 2017 at 7:40
• @DSuchet You're absolutely right of course. I changed that. In terms of differentials, I've seen $\mathrm d U=\delta Q + \delta W$ where $\delta$ indicates an inexact differential. Put simply, they are 'infinitesimal' amounts of heat and work but they cannot be integrated to give a path-independent state function. Mar 10, 2017 at 8:56

You are right. Microscopically, work and heat are just about the same. Both involve molecular collisions transfer energy from one object to the other.

Work involves a kind of "coherent" transfer in a manner of speaking, in which the collisions are predominantly, and to an extreme degree, in one direction. Also, typically the force is applied to one location on the object. And importantly, the boundary of the system is displaced. (E.g. translation or deformation)

On the other hand, transfer of energy by heat is "incoherent", many directions, and typically in all directions. And importantly, the boundary of the system is not displaced.

Finally, everyday phenomena fall into one or the other category, and they differ in their macroscopic behavior. Loosely speaking, when heat is transfered the temperature rises. When work is done, the boundary of the system changes. Of course adding heat generally also results in the boundary moving [say, expanding], and work generally results in a temperature change [Joule's experiment]. I'm trying to motivate the macroscopic separation between work and heat without a lengthy discussion.

To the engineers who first worked all of this out, the very existence of atoms was unknown. To them, the separation between heat and work was very clear. They had little reason to view them as manifestations of the same microscopic process. They thought that heat was a physical fluid. In any event, their remarkable achievements have stood the test of time.

It may be useful to have a look at the problem from a more quantitative perspective. Let $E$ denote the total internal energy of the system. For a thermodynamic system, it is defined as an ensemble average: $E = \sum_nE_np_n$, where $E_n$ and $p_n = e^{-E_n/k_BT}/Z$ are the energy and the realisation probability of the $n$th micro-state. Here $Z = \sum_{n}e^{-E_n/k_BT}$ is the partition function. I have used a sum, which may actually represent an integral (depending on the spectrum of the system). Now comes the crunch: apart from the obvious dependence on temperature $T$ via $p_n$, $E$ can also depend on other experimental knobs such as volume, external magnetic field and so on, because $E_n$ can depend on them. Namely, we may put $E_n(x_1,x_2,...)$, where $x_i$ is the $i$th external knob. As a result, $E$ must also depend on these arguments in addition to $T$, $E(T,x_1,...)$. The change in $E$ then reads, $\delta E = \delta Q + \delta W$, where $\delta Q = \delta T \partial E/\partial T$ and $\delta W = \sum_i\delta x_i \partial E/\partial x_i$. The former we call 'heat' while the latter we call 'work'. So, in this respect, heat refers to the energy change due to variation in $p_n$ whereas work to variation in $E_n$.

• The trouble is, changing some macroscopic variables changes $E$ but does not involve work. For example, changing the pressure at constant volume would involve a term $\delta P \partial E/\partial P$ in your sum for $\delta W$, but this would generally be considered heat rather than work.
– pwf
Dec 14, 2019 at 5:47

Classical thermodynamics deals with objects on the macroscopic level. So by $Q$ we mean the energy transferred as a result of temperature gradient $\Delta T$ only, and $W$ has of course the same definition we use in physics.

A recent paper by the physicist Carlo Rovelli (Where was past low-entropy?), puts the type of answer given by @Jonas quite succinctly:

Heat versus work. Thermodynamics has developed as the science describing the exchanges of heat and work between a system and its environment. Both are exchanges of energy: what determines the difference between heat and work? Intuitively, it is simple: work is a form of mechanical energy. But so is heat, at the microscopic level. When we give a macroscopic account of a process, heat is not anymore considered mechanical energy only because the relevant degrees of freedom are not directly accessible. The distinction between heat and work is therefore subtle: heat refers to the energy in the microscopic variable, while work refers to the energy in the macroscopic variables. The distinction between heat and work depends only on what we call macroscopic.

Fascinatingly, in the same paper, Rovelli uses the perspectival nature of macroscopic variables to show that the low entropy past of our universe is inherent in the set of macroscopic variables we interact with, not inherent in the evolution of the microstates of the universe.