Until the steel of the bottle itself starts to vaporize. Still, this could at least provide a basis for some theoretical energy calculations.
The halving thickness for gamma radiation passing through steel is very close to 1 inch (2.5cm), according to Wikipedia), so a mere foot of steel should absorb nearly all the gamma released–only about 0.02% would get through. Alpha and beta, of course, are absorbed even more readily. So, for the sake of discussion, we can assume that all of the energy released by the bomb is being absorbed by the steel.
Some of that energy would do fun things like make the steel radioactive, but that doesn’t immediately contribute to the breakdown of the bottle integrity. Let’s assume that all the energy is directed to heating the steel. Different steels have different thermal properties, and not all of them are fully characterized. Let’s say it’s a carbon steel with the following characteristics:
Specific heat: 0.49 kJ/(kg K)
Full melting point: 1540C
Heat of fusion: 272 kJ/kg* (Extra energy required to liquefy the material after the melting point is reached.)
Boiling temperature: 2870C*
Heat of vaporization: 6090 kJ/kg* (Extra energy required to vaporize the material after the boiling point is reached.)
*Values are actually for pure iron, as I was unable to find values specifically for steel.
A little weather research suggests that an air temperature of ~19C would not be out of line for the site and time of the Trinity test, though I didn’t find a record of the actual temperature on the day of the test. The sheer thermal mass of the bottle and the temperature extremes involved make a few degrees of initial temp pretty irrelevant, so let’s set T[sub]0[/sub] to 20C.
So, to melt a kilogram of the steel, we’ll need to raise its temperature by 1520C (or K), then add another 272 kJ. Adding the heat of fusion to the basic formula, we have
Q = cmdT + m*Q[sub]f[/sub]
Plugging in specific heat ©, mass (m), temperature delta (dT), and heat of fusion (Q[sub]f[/sub]), we get:
Q = 0.49 kJ/(kg K) * 1 kg * 1520K + 1 kg * 272 kJ/kg = 1016.8 kJ. For reasons of precision and convenience, 1000 kJ to melt each kilogram of steel.
The Trinity test released ~84 TJ (terajoules) of energy. This offers us a simple, if intimidating calculation:
84 TJ = 84,000,000,000 kJ.
84,000,000,000 kJ/1000 kJ/kg = 84,000,000 kg of steel. 184,800,000 pounds, or 92,400 tons.
That’s the amount that could be completely melted by the energy release, absent any sort of compression wave or vaporization. So, you need an ablative layer of that much steel on the inside of your bottle, with more outside to contain your new vat of molten metal. Over 90% of the energy will be absorbed by the first 10 cm of steel, so let’s say that’s how thick you want to make your ablative layer. The density of carbon steel is ~7.8 g/cm[sup]3[/sup], or 7800 kg/m[sup]3[/sup], so you need ~11000 m[sup]3[/sup] of steel in your ablative layer.
The volume of a spherical shell is (4/3)π(R[sup]3[/sup]-r[sup]3[/sup]), where R and r are the outer and inner radii, respectively. We also know that (R-r) is 10 cm, or 0.1 m. From that, we see that:
11000 m[sup]3[/sup] = (4/3)π((r+0.1)[sup]3[/sup]-r[sup]3[/sup])
Solving for r gives us an internal radius of 93.5 meters, or ~309 feet.
Now, let’s see how much energy the first halving thickness of the ablative shell absorbs, to see if we have vaporization issues. It’ll absorb ~42 TJ throughout its total volume, which is a shell of volume:
(4/3)π(93.525[sup]3[/sup]-93.5[sup]3[/sup]) = 2747 m[sup]3[/sup]
That’s roughly 21,000,000 kg of steel, so each kilogram would be soaking ~2000kJ of energy. The first 1000 kJ/kg goes to melting it. The remaining 1000 kJ would first go to raise the temperature of the molten layer to the boiling point, an increase of 1330C. That would use
Q = 0.49 kJ/(kg K) * 1 kg * 1330K = 650 kJ
The remaining 350 kJ would go toward heat of vaporization without actually raising the temperature further. Our heat of vaporization is 6090 kJ/kg, so 350 kJ could potentially vaporize up to about 5% of our molten steel from the inner surface of the ablative layer–over 1,000,000 kg of steel vapor bursting from the inner surface. Looks like we have a little concussive force, after all.
So, you have an evacuated 10 cm-thick steel sphere over 600 feet across with a bomb in the center of it. That’s just the part that would be completely destroyed by the bomb; it would melt through and superhot steel vapor would burst out, which, let’s face it, would be embarrassing.
Internal structure can’t be relied on to support it, since it will all melt, so it has to be held up by superstructure, i.e., lots more steel–in fact, just making the shell thick enough to support itself over that kind of span might be enough to complete containment. I’ll leave that part as an exercise for the reader.
Feel free to check my math and/or my reasoning. I am not a physicist, and I am specifically not your physicist.