If we are moving a block on a rough surface extremely slowly (quasi-statically) with the help of an external horizontal force, then is it the case that no heat will be produced, but the work done by friction force will still be equal to the work done by the external force for any given displacement?

The reason this question comes to my mind is, in thermodynamics, if we say that a process is reversible when it is slow, then there should not be any heat production despite having frictional/viscous forces in the system when the process is slow.

So, this implies that the existence of a reversible process rests on the idea that friction should not produce any heat.

So, it is correct to assume that friction can do work without producing heat when the process is extremely slow? Because if in all the examples we see around, we see that work by friction is always accompanied by heat production.

Kindly help.

  • 6
    $\begingroup$ Whether friction exist in the quasistatic limit is a hot debate that professors will fight over! Im of the type to say that friction is going to still generate entropy and heat in the quasistatic limit because 1) there is no time dependence on the frictional force function 2) the actual mechanism by which friction works is a fascinating thermodynamic rearrangement of, say, the impurities on the junction surface of the two bodies, and this rearrangement will happen regardless of the slowness at which you do things. It is rather manifestly an entropy generating process. $\endgroup$ Commented Aug 18, 2023 at 6:48
  • 1
    $\begingroup$ Why should the friction interaction be something that you can slow down? If we have a block sliding across another block in the xy plane, we can slow down the xy translation of the block, but it seems to me we can't slow down the electric field or the electron exchange interaction as molecules and atoms bang into each other on the z axis under the influence of the normal force. $\endgroup$
    – g s
    Commented Aug 18, 2023 at 7:21
  • 1
    $\begingroup$ Friction at the Nanoscale - - Relative motion between two solids in contact $\Rightarrow$ loss of mechanical energy. $\endgroup$
    – Farcher
    Commented Aug 18, 2023 at 8:20
  • 1
    $\begingroup$ In viscous friction, the friction force is proportional to the velocity difference, so doing it slowly generates less viscous dissipation. With dry friction, the friction force does not depend on the velocity difference (to a first approximation). So dry friction is always accompanied by heat release at the interface. $\endgroup$ Commented Aug 18, 2023 at 11:02

5 Answers 5


The parts added as edit to the original answer are in italics.

There is a misunderstanding about the conditions of a reversible process. Quasistatic is a necessary condition to ensure the well-definiteness of the thermodynamic state of the system. However, it is not sufficient. Even a quasistatic process becomes irreversible in the presence of dissipation (aka production of entropy).

A quasistatic process is a process slow enough (with respect to the relevant relaxation times of the system) to guarantee that, at each time, the system is as closest as possible to a thermodynamic equilibrium state. Missing such a condition, it would not be possible to characterize the process through well-definite thermodynamic state functions. The few-variable description of thermodynamics should be abandoned in favor of locally evolving dynamics described by many degrees of freedom. Think, for example, of the case of a fluid, where a non-quasi-static process would require the full set of local hydrodynamical variables to describe phenomena like sound waves, shock waves, and so on.

In addition to the need for a local field description, non-quasi-static phenomena usually involve internal mechanisms generating entropy in the system. Therefore, a quasistatic process is necessary for reversibility.

The opposite is not true: not every irreversible process corresponds to non-quasi-static processes. Actually, we may have quasi-static irreversible processes as soon as some additional mechanisms of production of entropy are present. One such case is the presence of dissipation.

Looking at this issue from a microscopic way, dissipation implies a transfer of energy into microscopic motion. Whether such a process is slow or fast doesn't matter. It will affect only the power dissipated (energy per time unit).

Whether such energy transferred to microscopic degrees of freedom should be called heat or otherwise is, in a way, a matter of conventions without effects on the phenomena.

Such a statement may look odd, but it is present in the literature (see, e.g., Mungan, C. E. (2007). Thermodynamics of a block sliding across a frictional surface. The Physics Teacher, 45(5), 288-291 where the author concludes by writing "while Q and W have important roles in the introductory teaching of the reversible thermodynamics of simple systems such as ideal gases, students should eventually be brought to realize that it is not always convenient nor necessary to categorize all channels of energy transfer as either “heat” or “work.” )

As an illustration, I can notice that if we define heat every variation of internal energy of the system not amenable to macroscopic work or matter transfer (i.e., the definition used by Caratheodory, Planck, Born, and others), the answer to the original question still depends on what we want to consider as the thermodynamic system. If the surface where friction is present is not considered part of the thermodynamic system, friction at the wall should be accounted as heat. If it is part of the system, we must evaluate friction forces' work.

However, what is physically meaningful is the variation of internal energy of the system due to the interaction with its environment.

  • 2
    $\begingroup$ There are many professors out there asserting that quasistatic is sufficient for reversibility because they do not think that dissipation / generation of entropy happens when processes are quasistatic. $\endgroup$ Commented Aug 18, 2023 at 8:01
  • 1
    $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90: Thanks for your response. The core question is, will friction produce heat if the process is slow/quasistatic? If Yes, there is probably no way in which a process can be reversible. If No, then only there is a possibility of a reversible process. The core of the question is about friction. $\endgroup$ Commented Aug 18, 2023 at 8:20
  • 3
    $\begingroup$ @DevanshMittal I thought it was clear, but I'll make more clear my answer. In short, yes, friction/dissipation always produces heat. $\endgroup$ Commented Aug 18, 2023 at 9:58
  • 1
    $\begingroup$ @DevanshMittal Thanks, I would like to know what is the reason for the downvotes. $\endgroup$ Commented Aug 18, 2023 at 14:14
  • 1
    $\begingroup$ GiorgioP, I agree with you and I also would like to know the reasons for downvotes, any downvotes for that matter. Maybe it should be kept anonymous, if it is so wished, but from a downvote by itself nobody learns anything, be it right or wrong and this is a perfectly correct answer. If it were up to me, for any downvote, I would demand at least a sentence long reasoning, the same way as closing a question must be justified to some extent. $\endgroup$
    – hyportnex
    Commented Aug 18, 2023 at 15:12

When thermodynamic work is done, that is an amount of physical quantity is transported from one potential to another, and that work is not converted to another work, such as in friction, then the same is dissipated, ie., evolved, into heat irrespective of the speed at which the process occurs. This happens when a stone is dropped from a higher potential to a lower one and stops at the bottom, or the same slides down ever so slowly on a slope. If you move a stone on a flat horizontal surface with friction then the relevant potential is not the gravitational potential but whatever force induces that motion. If the stone drops at a high speed but its energy is still converted to, say, electric work, then it is not dissipative; think of a water turbine.

Without dissipation, heat evolution, all work converts to work, and then the total work by itself is being conserved irrespective of its speed, and the process is reversible. The reason for the concept of a "quasi-static process" is that, for example, in gasses if the relaxation rate of the macroscopic inhomogeneities approach the sound velocity then it becomes more and more dissipative. But there are also inherently dissipative processes irrespective of their speed for which one example is plastic deformation.


is it the case that no heat will be produced

"Heat production" refers to increase of internal energy of some body associated with dissipation of energy, e.g. via chemical reaction, or irreversible work. It does not refer to transfer of energy by heat. Since block temperature and its internal energy increases due to dissipation of energy, "heat production" does occur. But in physics education we don't like to use the word "heat" in this way (heat production).

"Heat" in thermodynamics is primarily a mode of energy transfer, where energy moves via chaotic motion of microscopic particles, which are hard to detect, and can't be expressed as macroscopic work.

Heat in this thermodynamics sense will occur too, but in the opposite direction than you would expect - heat goes from the block to the rough surface (table). This is because the block gets hotter than the table and heat can only move from hotter to colder body (or to a body of same temperature, during isothermic heat exchange).

the work done by friction force will still be equal to the work done by the external force for any given displacement?

Yes in magnitude, but with opposite sign.

in thermodynamics, if we say that a process is reversible when it is slow, then there should not be any heat production despite having frictional/viscous forces in the system when the process is slow.

No, this is all wrong. "Slow process" is not necessarily a reversible process. We talk about making processes slow in thermodynamics because the slower the macroscopic work is done, the less entropy is produced in the system per unit displacement (e.g. piston compressing a gas, or a paddle stirring a liquid) and the better the assumption that the process is reversible. This is because dynamic friction in fluids is proportional to difference of velocities, so if that difference goes to zero, dynamic friction goes to zero.

But this does not apply to dynamic friction of dry surfaces; friction force there is proportional to normal force, almost independent of velocity difference and its entropy production per unit displacement does not go to zero when velocity difference goes to zero.

Reversible process implies there is no dynamic friction at all. Dynamic friction is always irreversible, it dissipates energy into microscopic degrees of freedom, into chaotic motion, and creates additional entropy.

So, this implies that the existence of a reversible process rests on the idea that friction should not produce any heat.

No, but it means there is no dynamic friction.

  • $\begingroup$ Thank you. You perfectly got the core of my question and responded accordingly. I have accepted your response. $\endgroup$ Commented Aug 19, 2023 at 12:24
  • $\begingroup$ I have noticed that you understand my questions in the spirit I ask them and respond accordingly. I have a few small conceptual questions which are not getting answered anywhere. May I request some of your time at some forum/place where I can chat with you and get resolutions to my doubts? I would be highly grateful. Thanks in advance. $\endgroup$ Commented Aug 19, 2023 at 12:28

We say that a process is reversible when it is slow ...

This is incorrect. Being slow (or "quasi-static") is a necessary condition for a reversible process but it is not sufficient. We can tell that the example you give (a block pushed by a small force and moving extremely slowly on a rough surface) is not reversible because the motion of the block does not reverse if we reverse time - we would also have to reverse the direction of the force. Contrast this with the motion of a block pushed by a small force on a smooth surface. In this case, if we reverse time (and hence also reverse the velocity of the block) then the motion does reverse - the force - still acting in the same direction - now decelerates the block.


There are some cases where the apparent friction is not a real friction - e.g. the dynamic brakes in the electric vehicles (where the motor is controlled in a way to mimic brake friction while it actually produces electrical energy and only a little heat as a side effect).

Whatever you do, the conservation of energy applies. If you fail to account for some energy or work, it most probably became heat while you didn't look carefully.

And in regard to reversibility - friction is not reversible. That's the whole point of it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.