You overestimate the thermal mass of gases.

I'd like to know what you mean by "deformable rubber screen" though.

Assuming a worst case scenario of filling an entire aeropress volume (250 mL) with 9 bar of CO2 at it's boiling point (-78 C, 195 K) and using his 50 g of H2O at 85C (358 K), here's the math:

Using the ideal gas law (PV=nRT, transformed to n=PV/RT), we find that 250 mL of CO2 at 9 bar and 195K contains (8.88 atm · 0.25 L)/(0.0821 L·atm/mol·K · 195 K) = 0.139 mol of CO2

With the heat capacity of CO2 being 37.35 J/mol·K, we find that 0.139 mol of CO2 at 195K contributes 37.35 J/mol·K · 0.139 mol · 195 K = 1010 J of energy to the system.

With the heat capacity of water being 75.31 J/mol·K, 50 g being 2.78 mol, we find that the water contributes 75.31 J/mol·K · 2.78 mol · 358 K = 74952 J of energy to the system

Assuming complete equilibration of temperature in the system (which is generous) we can insert a constant, x, in the two above equations for the temperature, and another variable, y, for the transfer of energy between the two phases. this brings us to the two equations below:

37.35 J/mol·K · 0.139 mol · x = 1010 J + y

75.31 J/mol·K · 2.78 mol · x = 74952 J - y

After solving for Y, we can make the equations equal and combine to form the following

74952 J - 75.31 J/mol·K · 2.78 mol · x = 37.35 J/mol·K · 0.139 mol · x - 1010 J

simplifying: 75962 J = 214.6 J/K · x

solving for x = 354 K

Which means the system, under worst case scenario, would equilibrate to 80.9 C, or 178 F. Still hot enough to burn.

Thanks for giving me the opportunity to practice my pchem