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I am a bit confused about an aspect the "work" part in the first law of thermodynamics, which says that the change in the internal energy of a system is the work done on the system + the heat transferred to the system. Here's my question:

If I do work on, say, a stone, causing it to gain a large total kinetic energy, then according to the first law of thermodynamics (TD), the internal energy has increased. But internal energy simply means the energy contained within the stone, not external energies, such as gravitational potential energy, or, more importantly in this case, the overall kinetic energy of the system. How is the first law of TD consistent with this definition of internal energy? The same problem arises, if, say, I also raise a stone by doing work on it, thus increasing the gravitational potential energy.

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    $\begingroup$ Together with ValterMoretti's answer below, I recommend that you check some good textbook that explains the first law in the broader perspective of mechanics. Check for example the first chapter in Astarita's Thermodynamics: he explains in detail the relationship between the action of forces, work, kinetic energy, internal energy, heat, with examples. $\endgroup$ – pglpm Dec 25 '20 at 13:04
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The general formulation of the first principle for a closed system says that $$L+Q= \Delta K + \Delta U + \Delta u$$ Where $L$ is the total non-conservative work done on the system. $Q$ is the heat entering the system. $K$ is the macroscopic kinetic energy, $U$ the macroscopic potential energy and $u$ the internal thermodynamic energy. Usually $U$ and $K$ are disregarded, the former because $-\Delta U$ can be viewed as further work on the system due to macroscopic conservative forces, the latter in particular because one usually deals with initial and final states where all macroscopic parts of the system are at rest. Exploiting this equation the processes you consider can be coherently discussed. In particular, if your action on the stone does not imply deformations of its form with production of internal dissipative stresses and there is no flux or production of heat (a completely mechanical kick), there is no variation of $u$ but only of $K$.

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  • $\begingroup$ Really clear and understandable answer. Thank you :) $\endgroup$ – Felis Super Dec 25 '20 at 18:35
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This confusion arises from the definition, first law states that for a closed system, change in internal energy is due to "thermodynamic" work done on/by the system (depends on how you choose the signs) and heat transfer.

Thermodynamic work is defined in terms of quantities that describe the states of materials, which appear as the usual thermodynamic state variables, such as volume, pressure, temperature, chemical composition, and electric polarization. (https://en.wikipedia.org/wiki/Work_(thermodynamics)#:~:text=In%20thermodynamics%2C%20work%20performed%20by,lift%20a%20weight%2C%20for%20example.)

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  • $\begingroup$ Thank you for the answer, I was thinking the answer had something to do with what you said , but I wasn't sure what the precise definition of "work" was in this case. However, in my thermal physics book, it said somewhere that a simple method for converting work into heat is by lifting a weight, thus applying work on it, and then dropping it, causing the ground to heat up slightly as the weight slams into the floor, converting some of the potential energy into heat. In this case, then surely the author isn't referring to thermodynamic work? $\endgroup$ – Felis Super Dec 25 '20 at 11:45

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