/sci/ - Science & Math

Let us assume for the moment that the water somehow manages to retain the intrinsic thermodynamic properties it has at STP. This is obviously unrealistic, but with the magnitudes of pressure we're talking about, it will be close enough.

If the water were compressed adiabatically:
[eqn]
\Delta Q = 0 \\
\Delta U = -\Delta W \\
\mathrm{d}\left( p V^\gamma \right) = 0
[/eqn]
Expanding out that total derivative:
[eqn]
V^\gamma \mathrm{d}p + pV^{\gamma - 1} \gamma \mathrm{d}V = 0
[/eqn]
Now by using the fundamental thermodynamic relation:
[eqn]
\mathrm{d}U = T \mathrm{d} S - p \mathrm{d} V
[/eqn]
Place it under isentropic conditions and rewrite it slightly:
[eqn]
\Delta W = p \mathrm{d} V
[/eqn]
Isentropic compressibility is defined as:
[eqn]
\beta_S = - \frac{1}{V} \left. \frac{\partial V}{\partial p} \right|_{\mathrm{d} S = 0}
[/eqn]
Therefore:
[eqn]
\frac{1}{V \beta_S} = -\frac{\partial p}{\partial V} \wedge p(V_0) = p_0 \\
p(V) = p_0 - \beta^{-1}_S \left( \ln V - \ln V_0 \right)
[/eqn]
Now with that done, we have to see what is the volume we need to compress the water to.

Assuming we start at STP, the amount of water in the OP has a mass of [math]998.207 \mathrm{kg}[/math]. The Schwarzschild radius is therefore:
[eqn]
r_s = \frac{2mG_C}{c^2} \approx 1.48247 \mathrm{ym}
[/eqn]
For reference, that yoctometer is [math]10^{-24}[/math] meters.

I assume you're not children, so how we go from that to a volume of [math]13.6472 \mathrm{ym}^3[/math] should be obvious.

The isentropic compressibility of water does not appear to be in the usual sources, but it can be calculated from the speed of sound [math]v[/math] and the density [math]\rho[/math]:
[eqn]
\beta_S = \frac{1}{\rho v^2} \approx 0.00456 \mathrm{MPa}^{-1}
[/eqn]

We conclude by integrating the work done to get from STP to this final volume:
[eqn]
W = \int_{V_0}^{V_1} p(V)\mathrm{d} V = \\
\left[ p_0 V - V \frac{\ln \frac{V}{V_0} - 1}{\beta} \right]^{V_1}_{V_0} \approx \\
4386.92\mathrm{MJ}
[/eqn]
...Maybe?