# What will happen if we boil water in the absence of gravity?

We know that, Convection is the mode of heat transfer due to the motion of individual molecules of a fluid (liquid or gas). The motion of the constituent particles arises due to difference in densities.

In the absence of gravity, water takes a spherical shape in order to minimize the energy due to surface tension, simply it wants to minimize its surface area for greater stability. Let us assume, we are able to manage to heat the centre of the water sphere by some kind of mechanism, how will the heat be transferred from the centre of the water sphere to its surface? Do convection currents, which are the major modes of heat transfer in the presence of gravity, take place in the absence of gravity? I think it will not take place. Is the heat transfer is due to conduction?

Further, it would be great if you could explain what will happen as time progress? Will the bubbles of dissolved gases formed inside stay where they are formed or move to the surface or they move towards the centre? How will the water boil (or it just explodes)?

For reference, this is how water looks in the absence of gravity (in the International Space Station):

Please note: Water is coloured to enhance visibility

• The drop should break apart. I see it like heating a spherical ball of a solid, somehow. Heating by conduction. Commented Oct 20, 2019 at 9:57
• It must make a big difference whether you are heating the water from one side or uniformly, as in a microwave oven. Commented Oct 20, 2019 at 11:14
• @BertBarrois, I am considering the case where heating is done by some mechanism which transfers heat to the centre of the sphere. The process is carried out slowly. Commented Oct 20, 2019 at 11:23

## 3 Answers

If you heat a sphere of water from the center in the absence of gravity, I see no reason for convective instability (in the usual sense of buoyancy of hot fluid). The heat transport should purely conductive, and the idealized heat conduction equation could be solved by the familiar Green function method. Hot liquid expands, so the sphere will inevitably grow. But there are many complications.

• A steam bubble will form around the heat source, and vaporization will use up 540 cal/g.

• A steam bubble might start to form on an impurity off-center, breaking the spherical symmetry.

• As the fluid and/or steam bubble at the center begins to expand, there will be a transient outward acceleration of fluid further out. This is the precondition for a Rayleigh-Taylor instability that could further break the spherical symmetry. (The classic example of R-T instability is water on top of oil in the presence of gravity. In the problem at hand, the outward acceleration mimics the effect of gravity.)

The heat flow will be radial. A bubble will form at the center of the sphere and increase in volume with time. Because of the much greater specific volume of the vapor compared to the liquid, the vapor will push the surrounding fluid outward radially, just as if a non-condensible gas were being released at the center of the sphere. So, within the surrounding fluid region, there will be radial convection of heat, with radial conduction superimposed on the radial convection. Is this good enough, or do you need to see the equations also?

• Thank you for your answer. I don't need equations. But, may I know how convection takes place, as it is the movement of particles due to the difference in density and in zero gravity there is no true notion of upwards or downwards? I am able to understand radial conduction but not the convection part of it. Commented Oct 21, 2019 at 2:46
• What you are referring to is called Natural Convection. This is due to density differences. But natural convection is not the only convection mechanism. There is also Forced Convection. Convection in general is defined as transport of heat due to fluid movement. In this case, the bubble is growing in size, and is forcing the liquid outside of it to move radially outward. The radial velocity is given by the equation $$v_r=v_br_b^2/r^2$$where $r_b$ is the radius of the bubble, $v_b$ is the radial velocity at the bubble surface, and r is a radial location outside the bubble. Commented Oct 21, 2019 at 12:10
• Thanks. So, as time progresses, the size of air bubble inside the water sphere increases. The heat transfer from the heating element in the centre to the inner surface of the water sphere takes place due to conduction through the water vapour region. This process continues till the pressure of water vapour inside the bubble is so large to contain within it. At this point it explodes, releasing steam. This is my conclusion based on your answer in the current state. Am I right? Commented Oct 21, 2019 at 12:35
• Pretty much so. Because of the radial velocity variation in the liquid, the temperature gradient in the liquid is constantly being enhanced, and this convective effect produces enhanced conduction through the liquid compared to the case of a rigid sphere. The pressure inside the bubble is actually going to be close to boiling pressure plus the surface tension throughout, but will decrease a little with time due to the decreased surface tension effect (as the radii of curvature increase). The bubble bursts through the liquid when it is thin enough for instabilities to play a role. Commented Oct 21, 2019 at 14:05

In absence of gravity water start to evaporate due to kinetic motion of water molecules.The making of water vapor absorbing latent heat from either near heat source or from water itself .So in this way water become solid ice which was locate far from heat source and other part become water vapor.