1st law of thermodynamics for free-falling object I have a debate with my friend. Suppose that an apple with weight of 10N fell from 1m high. We want to apply the 1st law of thermodynamics for this situation.
My friend thinks that $W$ should be considered as work done by the apple to the Earth, and thus $W=-10J$. Meanwhile the kinetic energy of the apple increased by 10J, so $\Delta E = 10J$. Finally, $Q = 0J$ so $\Delta E = Q - W$.
Meanwhile, I think that $W$ should be considered as work done by the apple to the surrounding air. So if the free-fall action is happening in vacuum, $W=0$. Since energy is conserved inside a conservative field, $\Delta E = \Delta K + \Delta U = 0$. Finally, $Q = 0J$ so $\Delta E = Q - W$.
Whose interpretation is right? Thank you.
 A: If the system is just the apple, then in vacuum gravity does work on the system, thus increasing it's energy, so the first explanation is mostly right. The second is not; there is definitely work being done on the apple.
Be careful with work though. Forces, not objects, do work. The apple doesn't do work on the Earth or the air. If your system is the apple, then just look at the forces acting on the apple. Gravity acts on the apple as it falls. Therefore, gravity does work. This work increases the kinetic energy of the apple. If you want to include air resistance, then that is another force that would do work on the apple, and its effect on the kinetic energy can be handled accordingly.

Now, I didn't bring up the first law of thermodynamics above because I don't think it is needed explicitly, especially in the form discussed below and in other answers. But since they are being used in other answers, and because it appears like we are saying different things, I figured I should explain them in more detail so there is no confusion. We are all essentially saying the same thing, just in different ways.
To address the differences between my answer and other answers here, like BobD's and Chet's, let's be careful with how we define our terms.
At the base level, the first law of thermodynamics is just energy bookkeeping. At the very least, we know that if a system's energy is changing, it must be due to some sort of energy transfer through work or heat:
$$\Delta E=Q+W$$ (I like thinking in terms of the system, so I am going to use the $+W$ convention here). This equation is fine, but it can be convenient to subjectively break these terms up.
First, the energy of the system. First, as BobD alludes to, we can think of energy in terms of "internal energy" $\Delta U$ (energy dealing with "microscopic" interactions between parts of the system, as well as kinetic energies of parts of the system relative to the center of mass). This is in contrast with what BobD and Chet note as $\Delta(KE)$, which would be the kinetic energy associated with the motion of the center of mass of the system ("macroscopic" movement). Taking these into account, we end up with
$$\Delta U+\Delta(KE)=Q+W$$
Second, the work term $W$. As we have our equation now, this term encapsulates work done by all external forces acting on the system. However, it is often convenient to break up this work into conservative and non conservative forces: $W=W_\text{c}+W_\text{nc}$. Furthermore, we know that the work done by a conservative force on a system is equal to the negative change in potential energy of that system: $W_\text{c}=-\Delta(PE)$. Moving this potential energy term to the left side of our equation we end up with
$$\Delta U+\Delta(KE)+\Delta(PE)=Q+W_\text{nc}$$
Which is what BobD and Chet have in their answers. This is why I say that there is work being done, whereas Chet says $W=0$. We are saying the same thing, it's just that Chet has taken into account the work done by gravity in the $\Delta(PE)$ term, whereas I am including as $W_\text{c}$.
A: 
My friend thinks that $W$ should be considered as work done by the
apple to the Earth, and thus $W=-10J$.

The apple doesn't do the work. If it did, it would be at the expense of its kinetic energy. When the apple falls gravity does work on the apple. Since the direction of the force of gravity is the same as the displacement of the apple, based on the definition of work the work done by gravity is positive.

Meanwhile the kinetic energy of the apple increased by 10J, so $\Delta
 E = 10J$. Finally, $Q = 0J$ so $\Delta E = Q - W$.

In the version of the first law you are using ($\Delta E=Q-W$) $\Delta E$ is the change in the internal energy of the apple. The internal energy consists of the kinetic and potential energy at the microscopic level (random movement of the molecules and atoms for kinetic energy, intermolecular forces for the potential energy). The kinetic energy of the center of mass of the apple falling is not part of the apples internal energy. (Though some may argue otherwise). It is the kinetic energy of its center of mass with respect to an external frame of reference (the earth), as opposed to microscopic kinetic energy with respect to the center of mass reference frame.
The longer form of the first law is
$$Q-W=\Delta E + \Delta KE+\Delta PE$$
The change in the apples kinetic energy while falling is the $\Delta KE$ term.

Meanwhile, I think that $W$ should be considered as work done by the
apple to the surrounding air. So if the free-fall action is happening
in vacuum, $W=0$. Since energy is conserved inside a conservative
field, $\Delta E = \Delta K + \Delta U = 0$. Finally, $Q = 0J$ so $\Delta E = Q - W$.

The equation $\Delta E = \Delta K + \Delta U = 0$ does not apply when there are  dissipative forces involved, e.g., air friction. Friction reduces the increase in kinetic energy.
And, BTW, it's debatable that $Q=0$ since air friction may involve some heat transfer from the air in contact with the falling apple into the apple, increasing its internal energy.
Hope this helps.
A: It is best to consider this problem first without air drag.  The expanded version of the first law for a closed system experiencing significant changes in kinetic and potential energy is $$\Delta U + \Delta (KE)+\Delta (PE)=Q-W$$In this case, the decrease in potential energy is balanced by an increase in potential energy.  So $$\Delta (KE)+\Delta (PE)=0$$In addition, no deformational work is done by the apple on its surroundings and no heat transfer occurs.  So, Q = W = 0.  So there is no change in internal energy:  $\Delta U=0$.
The case with air drag is just a small perturbation on this.
