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:
-
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.
-
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.
-
The heat dissipated by the system equals the integral of the system's temperature times the change of the system's entropy.
-
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.
Best Answer
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.
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.
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
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).
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.