Heat is the total kinetic energy of all atoms of the system. When work is done on the system it means that a part of system kinetic energy is used to do the work, and this work makes the surrounding warmer. So "$\Delta U$" of the system is equal to "$Q$". And now, why we use the work of the system in: $\Delta U = Q + W$?

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    $\begingroup$ "Heat is the total kinetic energy of all atoms of the system." That's incorrect. Heat is the energy exchanged with the surroundings in the absence of work. When volume is constant and no other work is done, $\Delta U = Q_v$. More generally $\Delta U = Q + W$ $\endgroup$
    – Buck Thorn
    May 22, 2019 at 7:39
  • $\begingroup$ Heat is energy transferred because of a difference in temperature. You know from your own experience that if something is colder or warmer than your hand (say) you can feel the difference due to energy leaving your hand (if what you touch is colder) or gaining energy if warmer. If the temperature is the same there is no energy transferred, i.e. no heat. $\endgroup$
    – porphyrin
    May 22, 2019 at 8:12

1 Answer 1


The questions is a good one, and answering it winds up being central to a lot of key concepts in thermodynamics. A brief correction before starting, though I don't think it's central to the question, "When work is done on by the system it means that a part of system kinetic energy is used to do the work..."

The disconnect between your understanding and that used in thermodynamics is that in thermodynamics, the system and its surroundings are separate. So, when our system does work on the surroundings that energy won't come back to heat our system. If we do as you say and use our system to do work on the surroundings, turn that work into heat, and then apply that heat back into our system, the result is just as you say! The kinetic energy in our system isn't lost. Or put in another way when we're allowing heat to flow in and out of our system, all of that heat ultimately winds up as kinetic energy and $ΔU = Q.$

Let's look at a couple examples of putting $\pu{1 J}$ of heat into a system with different surroundings to see how this works out. Our system will be an insulated piston filled with an ideal gas (say helium), which can expand and push on one of 3 things in its surroundings: a mass, a vacuum, or a generator.

  • Mass: This follows $ΔU = Q - W$. Of the $\pu{1 J}$ of heat we apply, the portion that goes into warming up our system and increasing internal kinetic energy is $C_v$ (the molar heat capacity is $1.5R),$ whereas the portion of that instead goes into doing work on the surroundings is $R$ (the universal gas constant). So our system gains

    $$[\pu{1 J}]\cdot \frac{C_v}{C_v + R} = \pu{\frac{3}{5} J}$$

    of thermal energy, and does

    $$[\pu{1 J}]\cdot \frac{R}{C_v + R} = \pu{\frac{2}{5} J}$$

    of work on the surroundings by expanding and pushing up the mass. This $\pu{\frac{2}{5} J}$ of work energy won't return. These portions $C_v/(C_v + R)$ and $R/(C_v + R)$ show up a lot later on in thermodynamics as their inverses, $k$ and $k/(k - 1)$, respectively.

  • Vacuum: This follows $ΔU = Q$. There is no surroundings to do work on, so all of the expansion energy goes into slamming the piston against its backstop, and after everything settles this ultimately results in the full $\pu{1 J}$ of heat becoming thermal energy inside our piston.

  • Generator: This also follows $ΔU = Q$. It's the same situation as the mass, with $\pu{\frac{3}{5} J}$ of thermal energy and $\pu{\frac{2}{5} J}$ of work done on the surroundings, but that $\pu{\frac{2}{5} J}$ of work can be turned into electricity and then be used to heat our piston.

    Just like in the vacuum case, all $\pu{1 J}$ of heating ultimately winds up as thermal energy in the system, though it spent some time as work on the surroundings during the process. So, $ΔU = Q - W + W.$ I think this final case is the closest to matching your question.

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    $\begingroup$ Tech detail: please don't use \pu{…} everywhere in math mode. Its sole purpose is to handle physical units (hence the shortcut for the macros: \pu). $\endgroup$
    – andselisk
    May 23, 2019 at 6:50
  • $\begingroup$ Sorry, I don't quite understand what you're asking for. I can't seem to find a markup how-to page, but looking at your edits it looks like its best to handle equations by putting them within dollar signs to display them with TeX? Also, thanks for editing, it looks very much nicer! $\endgroup$ May 23, 2019 at 17:24
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    $\begingroup$ It's not excessive use of $$, it's using a macro for physical units on the entire expression. If you are familiar with LaTeX, the \SI{}{} command is similar to \pu of MathJax. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Also: MathJax basic tutorial and quick reference. $\endgroup$
    – andselisk
    May 23, 2019 at 17:33

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