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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. 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? Is it the limitation of Newtonian mechanics that it never accounted for something like heat which could also change the energy of the system? Is the word "Thermodynamical work" or "Hidden work" suitable for heat?

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  • $\begingroup$ In general heat is kind of considered a loss, especially for friction, so it's a bad thing where work is generally a good thing. So generally the 2 terms work and heat don't go together. $\endgroup$ – PhysicsDave Sep 11 '18 at 3:57
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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."

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  • $\begingroup$ dQ = TdS seems like a perfectly good definition once the macrostates are defined. $\endgroup$ – Ryan Thorngren Sep 11 '18 at 7:21
  • $\begingroup$ The equation you provided only applies to a reversible path, unless you are referring to the entropy exchanged across the interface with the surroundings. $\endgroup$ – Chet Miller Sep 11 '18 at 12:36
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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.

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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

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