When does a specific equipment at a certain temperature start to freeze in interstellar space?

This is my first post here. I'm working on a project which involves a number of problems and some parts of that are related to empty space and space exploration. I try to be focused and reflect one of my problems here.

We know that the interstellar space is mostly empty. The average temperature of empty space between celestial bodies is calculated at 2.7 Kelvins. So, the two mechanisms of conduction and convection have noting to do with heat transfer in empty space and only radiation matters there. I've been thinking about a spaceship with a specific inner temperature floating in empty space. To idealize the problem I simplify my question as: If I release a hot, small ball (at 373 Kelvins) from my spaceship, when does its surface start to freeze (reaching at 273 Kelvin)? And how long does it take for the center of the ball to reach 273 Kelvins? Knowing these is important for interstellar travels.

Thank you very much in advance!

• All formulas you will need for this are given in my answer to this question: space.stackexchange.com/questions/41353/… Mar 28, 2021 at 19:54
• @Polygnome, Thank you very much. It was helpful (+1 there and also here). Mar 28, 2021 at 20:06

The magic words you're probably reaching for are "radiative cooling".

If your ball is in a vacuum far from any star and doesn't generate any heat of its own, this isn't too difficult a problem to solve. There's even a handy calculator for it (and no doubt many others elsewhere) but I'll summarise the key bits here.

The cooling rate in terms of power per unit area is defined by the Stefan-Boltzmann law, and given a nice spherical black body radiator you can work out the radiated power as $$P = 4\pi r^2 \sigma(T_{ball}^4 - T_{ambient}^4)$$ where $$\sigma$$ is the Stefan-Boltzmann constant. Given that this experiment is being run in deep space and is stopping at merely the freezing point of water, $$T_{ambient} \ll T_{ball}$$, so you can quietly disregard the ambient term and not go too far wrong.

The hyperphysics calculator suggests using a nice simple model for the energy content of your hot ball as $$E = N \frac{3}{2}k_BT$$ where $$N$$ is the number of particles, $$k_B$$ is Boltzmann's constant and $$T$$ is the temperature of the object.

Throwing these together and doing a little calculus that I won't try to repeat here gets you a nice simple cooling rate equation:

$$t_{cooling} = \frac{Nk_B}{8\sigma \pi r^2}\left[ \frac{1}{T_{final}^3} - \frac{1}{T_{initial}^3} \right]$$

You can get $$N$$ from $$\frac{mN_A}{M}$$ where $$m$$ is the mass of your sphere, $$N_A$$ is Avogadro's constant and $$M$$ is the molar mass of whatever your ball is made of.

For a worked example, lets imagine your sphere is made of some ideal material with the approximately the properties of ruby... density ~4000kg/m3, molar mass ~100g/mol and a magical emissivity of 1. If it were a sphere 1m in radius, it would have mass ~16755 kg and so $$N$$ is basically 1029.

Throwing all that into the cooling equation gets you a cooling time of about 8 hours.

Obviously using real materials is more problematic. Handling emissivity is easy, but the simple thermal energy model breaks down, especially with awkward stuff like water. As a starter for 10 though, it should do.

• Thanks for your nice answer (+1) and also thanks for introducing that handy calculator. Applying these and also many other variables into my problem is really a challenge! Your answer was helpful for me. Mar 28, 2021 at 20:26