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I often get confused when I try to grasp what the terms in the first law of thermodynamics mean and what the sign conventions mean, because they usually refer to undefined concepts such as work done by or on the system. I think the same confusion arises regarding absorption and dissipation of heat.

Say a system consists of a meteor. The surroundings consist of a huge planet. My assumptions are:

  1. The work done by the system equals the line integral of the gravitational forcefield that the meteor generates times the displacement of the huge planet.

  2. The work done on the system equals the line integral of the gravitational forcefield that the huge planet generates times the displacement of the meteor.

  3. The heat dissipated by the system equals the integral of the system's temperature times the change of the system's entropy.

  4. The heat absorbed by the system equals the integral of the surroundings' temperature times the change of the surroundings' entropy.

What coefficients should be in the first law according to the above assumptions:

$$d E_{system} = a_0*\text{heat dissapated by system} + a_1 * \text{heat absorbed by system} + a_2*\text{work done on system} +a_3*\text{work done by system}$$

Edit: by change i mean $E_{end} - E_{start}$.

Edit: i guess the second law might introduce $a \leq$ (or is it $\geq$?) instead of an equal sign since I talk about entropy.

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  • $\begingroup$ Your system is a somewhat bad choice. For example it is open, its far from equilibrium, but most importantly: Heat isnt that much of a issue there. You would simply say "total energy is conserved". So whatever change in internal energy ("work done") there is in the meteor is the same as the change in internal energy of the planet up to signs $\endgroup$
    – Bort
    Sep 11, 2015 at 9:24
  • $\begingroup$ It's just a toy system so I can remember the meanings and sign conventions when I do "real" stuff. $\endgroup$
    – Emil
    Sep 11, 2015 at 9:31
  • $\begingroup$ undefined concepts such as work done Is work an undefined concept? Is it more undefined than kinetic energy or heat or alike? $\endgroup$
    – Steeven
    Sep 11, 2015 at 12:42
  • $\begingroup$ Work done ON A SYSTEM or BY A SYSTEM, I have never, ever seen a definition of. Work done by a force I have seen defined, on the other hand. $\endgroup$
    – Emil
    Sep 11, 2015 at 12:49
  • $\begingroup$ My guess is: work done by a system, is the thermodynamical force generated by the system multiplied by its conjugate variable on the surroundings. As in my example here: a gravitational forcefield is generated by the system on the surroundings, and the conjugate variable displacement on the surroundings is multiplied by it. $\endgroup$
    – Emil
    Sep 11, 2015 at 12:55

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In the situation sketched in the question, only point 1. and 2. are relevant, as there is not any significant heat transfer between the bodies.

What coefficients should be in the first law according to the above assumptions:

There is no single/unique answer to this - you can choose the coefficients as you like.

The 1st law of thermodynamics says that

$$d U=Q-W$$

which is the same as you describe ($U$ is internal energy, $Q$ is heat, and $W$ is work). But, I could just as well have written it as

$$d U=Q+W$$

or with signs added in other ways. It depends on what I mean by $Q$ and $W$ in the specific case. To choose for example the version $d U=Q+W$, I have to state at the same time that $Q$ is heat absorbed and $W$ is work done on the body (by an external force). If I used $d U=Q-W$, then $W$ would have been work done by the system.

The method to remember it: Consider energy positive when it is entering (or absorbed by or added to) a body or system, and consider energy negative when it is leaving. Choosing this sign convention makes energy simple to add. If I lift a book up to a shelf, then I do work on the book (+W) by adding potential energy to the system. If I push a box across the floor, then I do work on it (+W) by adding kinetic energy to it. If on the other hand, the heavy box comes sliding towards me and hits me, so I am pushed away, then the box did work on me ($-W$). The work the box did on me is energy lost from itself; the amount of kinetic energy remaining in the box will necessarily be smaller after the collision.

Edit: I guess the second law might introduce a≤ (or is it ≥?) instead of an equal sign, since I talk about entropy.

The usual equation for entropy of a system is $S\geq\int\frac{dQ}{T}$. It states that entropy can never decrease in a process, if the whole isolated system is considered.

From comments

Work done ON A SYSTEM or BY A SYSTEM, I have never, ever seen a definition of. Work done by a force I have seen defined, on the other hand

You are correct, that only forces do work. The statement, "work done by a system" simply means that the system applies a force, which is doing work (on something else). In your meteor example, the meteor as the system is doing work on the planet by pulling in it toward itself through its gravitational force. (Note: The work done by the meteor on the planet is very, very small, since the displacement of the planet will be very, very small).

My guess is: work done by a system, is the thermodynamical force generated by the system multiplied by its conjugate variable on the surroundings. As in my example here: a gravitational forcefield is generated by the system on the surroundings, and the conjugate variable displacement on the surroundings is multiplied by it.

Yes, this is the mathematical definition, in short written as:

$$W=\vec F \cdot \vec s$$

or in general for non-constant forces:

$$W=\int \vec F \cdot d\vec s$$

where $\vec s$ is the displacement vector.

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  • $\begingroup$ Thanks, it solved one part of my understanding issues. So dU_system = F(generated by surroundings) * dX(system) at least. Not sure where F (generated by system)*dX (system) and F (generated by system)*dX (surroundings) and F (generated by surroundings)*dX (surroundings) should go though. $\endgroup$
    – Emil
    Sep 11, 2015 at 14:01
  • $\begingroup$ A system cannot do work on itself. It cannot add energy to itself. In $W=\vec F \cdot \vec s$, $\vec F$ is the force that does work, $W$ is the work done, and $\vec s$ is the displacement of the body that work is done on. So, $F_{\text{generated by system}}*dX_{system}$ and $F_{\text{generated by surroundings}}*dX_{surroundings}$ are impossible, while $F_{\text{generated by system}}*dX_{surroundings}$ is the same as the first one you mention, just the other way around - now the force done by the system on the surroundings is considered. $\endgroup$
    – Steeven
    Sep 11, 2015 at 14:08
  • $\begingroup$ So if I understand correct Force(systemA)dX(systemB) + Force(systemB)dX(systemA) = 0, that feels like it should be a law in itself, it is slightly reminiscent of newtons law but since it is work it is probably something else. Is the TdS term that usually pops up T(surroundings)*dEntropy(system) by the way? $\endgroup$
    – Emil
    Sep 11, 2015 at 14:42
  • $\begingroup$ No no, the two values for work are not the same! For example, the meteor exerts the same force on the planet, as the planet exerts on the meteor (because of Newton's 3rd law). But the meteor is displaced much more than the planet is. So the work $W=F \cdot s$ done by the gravitational force on the meteor is much bigger than the work done on the planet. $\endgroup$
    – Steeven
    Sep 11, 2015 at 14:53
  • $\begingroup$ But, I thought you said dU_A = Q_A+W_BA = Q_A + F(B)dX(A) or dU_A = Q_A - W_AB = Q_A - F(A)dX(B). Maybe the Q_A should be Q_BA=T(B)*dS(A) in the first case and -Q_AB in the second case I guess $\endgroup$
    – Emil
    Sep 11, 2015 at 15:07

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