What would happen if a solid ball of helium suddenly appeared on the bottom of a deep sea? [closed]

Imagine that a solid ball of helium (at zero Kelvin) appeared at the bottom of a deep-sea on Earth? Say the ball has a radius of 100 meters and that the sea is 10 kilometers deep. The temperature of the sea is 10 degrees Celsius. Will the sea freeze (will a shell of ice form?), after it has delivered heat to the ball, thereby making the expansion of the helium (into a superfluid state after which is evaporates) more difficult when it transforms into gas? Will this freezing stop the expansion? Will superfluidity play a role (though I don't see exactly how)?

Say that the helium is free to do whatever it "wants". I'm not asking what would happen if it stayed in its form.

• I’m voting to close this question because this sufferts from the same problems as your earlier question ( now closed ) and is another "what if..." scenario. May 27 at 22:35
• Frankly this seems like a troll question in response to the closing of your previous one. May 27 at 22:35
• At 1 atmosphere He does not solidify. May 27 at 22:51
• There are literally no end to the artificial variations of this question you could write, and the purpose of the site is not to answer random what-if scenarios posed by the entire internet. We're trying to develop a knowledge base of physics concepts explained and dispell misconceptions people have in learning physics. Your questions related to this one do not seem (IMO) to match that goal and are a poor fit here. There's no apparent rational for this question other than "I had a question closed so I'll beat the system" - not one that I can see. May 28 at 4:53
• Understanding how a physical scenario unfolds, including what kinds of effects (thermal, mechanical, fluid dynamics) are relevant and what they predict, is a part of physics and often quite illuminating. That the root cause in the scenario is somewhat arbitrary does not mean answering what happens is arbitrary or pointless. May 28 at 8:47

You don't say why the helium is solid, but I guess you are also assuming it to be at some cyrogenic temperature where the solid form is stable.

Helium, even in liquid or solid form is still not very dense. It is much less dense than water and would rise in the column. There might be some ice formation initially, but the helium will have very little heat capacity. Any heat it pulls in to form the ice is going to cause vaporization. Any ice that forms is unlikely to stick to the sphere. Ice also isn't very strong. The amount that could form would have no ability to resist the pressure as the helium boils off.

My speculation is that the sphere would pretty quickly crack, increasing the surface area for heat transfer. As the helium would constantly be pulled into water that hasn't been cooled, I think it would rapidly warm, break up, and vaporize. It would complete the rise to the surface as gas bubbles.

• I am assuming that the solid can freely interact with the water. So it can make phase transitions. May 27 at 22:50

Solid helium is unusually compressible for a solid, so the underwater pressure would compress it into a smaller sphere. That heats it up, presumably turning it back into a gas.

The fact that it is very cold does not mean it freezes a lot of water. Like in the question about the 0 K iron cube in the living room, the thermal capacity of water is much bigger than the thermal capacity of the solid helium and there is far more of it. The pressure is definitely high enough to make even a thick spherical ice shell buckle and implode immediately.

So the sphere implodes and heats up, and the implosion becomes turbulent, mixing the water and the helium which increases the heat transfer. At this point things become very complicated as I would expect some explosive bounces, but in the end it turns into supercritical liquid (since at the 108 MPa pressure in the Marianas Trench at a few Kelvin this is the equilibrium state; see the phase diagram). Note that this is a normal (critical)liquid rather than superfluid.

The bubbles of helium start ascending. They quickly reach terminal velocity, which is $$v_t = \sqrt{\frac{4g(\rho_L-\rho_b)d^2}{3C_d \rho_L}}$$ If we use $$\rho_b\approx 125$$ kg/m$$^3$$, $$\rho_L=1000$$ kg/m$$^3$$, $$d=1$$ cm, $$C_d\approx 1$$ we get $$v_t \approx 0.03$$ m/s. At that rate it takes about 102 hours to get to the surface. However, at the start the helium will be compressed and later expand, there will be big turbulent blobs and whatnot producing currents, likely making a plume dragging water with it, so the actual speed is likely a bit higher.