Cooking by drowning Imagine you drop a giant sack full of fish into the sea.
As the sack drowns deeper, the pressure will compress the air inside the sack.
This will make the temperature rise inside the sack.
Is it possible to cook something with this method?
Edit (tldr, don't propose machines):
I am specifically asking about the physics side of things.
e.g. will it be able to reach a high enough temp, sustain it, and cook the fish.
The mechanism is of little concern to me.
 A: Neat question.
You'd really want to run some numerics or solve some equations to be sure, but my guess is this.
The sack isn't isolated from the environment, in the sense that changes in temperature in the environment propagate through the sack. Furthermore, the sack is falling slowly enough that it can be modeled as always at thermal equilibrium with its environment, so the temperature inside the sack is roughly that outside it. It gets colder the further you go.
So my guess is you've found a ridiculous freezer rather than a ridiculous cooker.
A: I'm assuming that you have a sealed sack of fish. The source of pressure is the weight of water above the sack, which will increase with increasing depth.
I'm imaging a model here. Assuming an adiabatic process resulting in an increase in temperature as in $ T = (P+P_{atm})V/nR$ from the ideal gas equation (with $P_{atm} = 1$ bar or so), the product $(P+P_{atm})V$ will clearly depend on the external pressure / ocean depth. For the sack to remain stable, $P = P_{ext} = \rho gh +P_{atm}$. As $h$ increases, $P$ increases. Given the specific heat capacity of the air inside the sack, we can derive an expression for $V$ in terms of $P$ and $\gamma$. That is, $V = (k/(P+P_{atm}))^{1/\gamma}$.
$ T = (\rho gh+P_{atm})^{(\gamma -1)/\gamma}(k)^{1/\gamma}/nR$
A: I think you could consider the diesel effect if you design one portion of the chamber to collapse at sufficient pressure. This happens when submarine exceed crush depth as bulkheads collapse.
So if your cooking chamber has one volume with a movable wall as a piston that is exposed to the large amount of sea pressure due to rupturing a diaphragm or some other mechanism. Then you could calculate how hot the new volume of air  in a fairly precise way, and the cooking time would be determine by how well insulated the cooking chamber is with respect to the surrounding sea water.
A: Considering the large mass times heat capacity of the fish compared to that of the interstitial air, the temperature rise of the combination will be much much less than that obtained by adiabatically compressing air alone.  Here is a detailed analysis of that situation:
Consider that the flexible container is perfect insulation so that the compression takes place adiabatically.  Let $\phi_0$ represent the initial volume fraction of air in the container and let $V_0$ be the initial volume of air and fish.  Then the initial volume of air in the container is $\phi_0V_0$ and the initial volume of fish in the container is $(1-\phi_0)V_0$. In addition, the number of moles of air in the container is $n=\frac{\phi_0P_0V_0}{RT_0}$.
If we apply the first law of thermodynamics to the adiabatic reversible compression of this system, we obtain:  $$dU=\frac{\phi_0P_0V_0}{RT_0}C_{Va}dT+(1-\phi_0)V_0\rho_fC_{f}dT=-PdV_a=-\frac{\phi_0P_0V_0}{RT_0}\frac{RT}{V_a}dV_a$$where $V_a$ is the volume of air in the interstices, $C_{Va}$ is the molar heat capacity at constant volume of the air (=2.5R), $rho_f$ is the density of fish (assumed 1000 kg/m^3 for mostly water), and $C_{f}$ is the mass heat capacity of fish (assumed 4.184 kJ/kg.C for mostly water).  Multiplying the above equation by $\frac{T_0}{\phi_0P_0V_0T}$ yields:$$\left[\frac{C_{Va}}{R}+\frac{(1-\phi_0)T_0\rho_fC_{f}}{\phi_0P_0}\right]\frac{dT}{T}=-\frac{dV_a}{V_a}\tag 1$$From the ideal gas law, It follows that  that:  $$\frac{dV_a}{V_a}=\frac{dT}{T}-\frac{dP}{P}\tag 2$$ If we combine Eqns. 1 and 2, we obtain:  $$\frac{C_{Pa}}{R}\left[1+\frac{(1-\phi_0)RT_0\rho_fC_{f}}{C_{Pa}\phi_0P_0}\right]\frac{dT}{T}=\frac{dP}{P}\tag 3$$From this, it follows that the combined fish-air system temperature response to a pressure change would be the same as that for pure air compression if the heat capacity at constant pressure of the air had been higher by a factor of $\left[1+\frac{(1-\phi_0)RT_0\rho_fC_{f}}{C_{Pa}\phi_0P_0}\right]$.  For an initial void fraction of 0.2, an initial temperature of 293 K, and an initial pressure of 1 bar, this factor comes out to be equal to 15.  So, we have that $$\frac{T}{T_0}=\left(\frac{P}{P_0}\right)^{\frac{\gamma-1}{15\gamma}}=\left(\frac{P}{P_0}\right)^{0.019}$$which is much much less than compression of pure air.
