Thermodynamics of a Fire Piston and the First Law I'm trying to understand how a Fire Piston works, in the context of the First Law of Thermodynamics.

The First Law states that
$$ \Delta U = q + w = 0 \text{ (for an isolated system)} $$
Where "isolated" means that there is no exchange of matter or energy with the surroundings.
We can construct an isolated composite system where system = surroundings + piston + tinder. Then:
$$ dU = dU_{surr} + dU_{piston} + dU_{tinder} $$
The fire piston is supposedly supposed to be understood as an adiabatic process so there is no transfer of heat energy between the surroundings and the piston. So both $q_{surr}$ and $q_{sys} = 0$. And since the tinder is essentially immoble and incompressible, we have $w_{tinder}=0$ So we have:
$$ dU_{sys} = w_{surr} + w_{piston} + q_{tinder} = 0 $$
The only way I can think of to make sense of how a fire piston works in this context is to say that $w_{surr} = 0$ (there is no work done by or on the surroundings), and the work for the piston is $w_{piston} = -P_{ext} \Delta V$, so:
$$ dU = -P_{ext} \Delta V + q_{tinder} = 0 $$
and since, for compression, $\Delta V < 0$:
$$ -P_{ext} \Delta V = -q_{tinder} > 0 \implies q_{tinder} < 0 $$
This makes sense, because the tinder combusts, releasing heat.
However, it seems odd to say that there is no work done by the surroundings, because aren't the surroundings (our hands slamming down on the piston) the thing doing the work on the piston? In which case, wouldn't $w_{surr}$ and $w_{piston}$ cancel out, leaving $q_{tinder} = 0$?
Or is the work done by the surroundings not only compression work on the piston, but also a different type of work (the work associated with slamming the piston down), and this is the work which is converted to heat in the tinder?
Is there something about this system that makes it not analyzable thermodynamically? i.e. is it a non-equilibrium process or something like that?
Where am I going wrong here (or am I)?
 A: You seem very confused about how to apply the first law of thermodynamics to this system.
First of all, the system is not isolated.  It is receiving work from its surroundings.  If the gas is the system, then its surroundings is the piston (and that is what is doing the work).  For an essentially massless, frictionless piston, the work done by your hand on the piston is equal to the work done by the piston on the gas. So, indirectly, your hand is doing the work.  You correctly noted that this work is $-P_{piston}\Delta V$.
So the correct application of the first law to this system reads:  $$\Delta U=nC_v\Delta T=-P_{piston}\Delta V$$where n is the number of moles of gas in the cylinder.  So the temperature rise of the gas (which is responsible for igniting the wood chip) is $$\Delta T=\frac{P_{piston}}{nC_v}(-\Delta V)$$Since $\Delta V$ is negative, the temperature rise of the gas is positive.
A: Your problem starts here:

We can construct an isolated composite system where system = surroundings + piston + tinder.

You've forgotten the air, which is very highly compressively heated by the piston to a point that exceeds the ignition temperature of the tinder. The air is not isolated from these other components.
In addition, it is not a very useful way to solve a thermodynamics problem to define a system that encompasses everything. I suggest defining a system at a much smaller and more homogeneous scale. Here, the air is well suited. You could idealize it as following $T=nR/PV$ and constant $PV^\gamma$ and estimate the minimum required compression of the piston.
