# What would happen to a blob of mercury in space?

How long would it take for a blob of mercury (the size of a marble or so) to freeze in space? I'd emagine it would have to boil in some way first, and that would send pieces of it flying around, but the surface tension of mercury is much greater than that of water, or urine (as we'd seen evaporate and then freeze into fine crystals). I'd imagine the effects of mercury in space would be the similar to those of urine, except the crystals might be less fine perhaps? ..Or might the whole blob just freeze solid??

I cannot find any clear answer to this online, so I thought I'd ask because I'm no pro, and if somebody here can help me out, it would be greatly appreciated!

Mercury freezes at -38.83 °C. not very low, most of space is colder than this.

Mercury freezes at -38.83 °C. not very low, most of space is colder than this.
I understand that, but with there being nothing in space for the internal heat of the blob to escape into, do you think it would remain liquid for some time before it does? Or might the pressureless vacuum boil it so violently that its surface tension gives, and turns it into a mist before freezing?

I apologize for the obscurity of this specific question, but I'm writing a sci fi and I really don't want to make an obvious error with this here.

Terrakron
Bubbles are balanced by surface tension if R=2σ/P.
For water at 20 Celsius, P=2,339 kN/m2, σ=72,8 mN/m - so R is slightly above 60 μm.
For mercury at 20 Celsius, P=160 mN/m2, ρ=435 mN/m - so R is about 5,5 m.

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Terrakron, Malloy and mfb
Bystander
Homework Helper
Gold Member
Tom.G
Gold Member
Several Google searches found enough info for the following.

With a triple point of -39°C (the temperature at which it can be a solid, liquid, or gas) and the temperature of deep space about -270°C, it would rapidly cool to freezing by radiation. The vapor pressure of 10-6 to 10-7kPa is 25 000 times lower than the surface tension of 4⋅10-2kPa, so it won't boil, just freeze into a ball.

Cheers,
Tom

(p.s. Not really my field, so if I messed it up please jump in and give a couple lashings with a wet noodle.)

Terrakron and Malloy
SThe vapor pressure of 10-6 to 10-7kPa is 25 000 times lower than the surface tension of 4⋅10-2kPa, so it won't boil, just freeze into a ball.

Cheers,
Tom

(p.s. Not really my field, so if I messed it up please jump in and give a couple lashings with a wet noodle.)

Wet noodle beating. See a post above. Surface tension and vapour pressure have different dimensions.

Order of magnitude derivation for heat loss by evaporation:
Once you evaporate mercury, the vapour would escape to vacuum at a speed with magnitude of speed of sound, multiplied by a modest scale factor which I do not know.
Speed of sound in mercury vapour at 20 Celsius is about 125 m/s. So a square m of mercury surface exposed to vacuum should be giving off about 125 cubic m of mercury vapour per second.
The vapour pressure of mercury at 20 Celsius was found as 0,16 Pa, meaning 1,6 μbar.
Thus that 125 cubic m of mercury vapour at 1,6 μbar would be equivalent to 200 ml of notional mercury vapour at 1 bar and 20 Celsius. Since 1 mole of gas at 20 Celsius is about 24 l, it is 1/120 moles of mercury.
As the heat of evaporation of mercury is 59 200 J/mol, that 1 square m should be losing about 500 watts to evaporation of mercury.
Is that derivation correct, to the precision of the unknown scale factor?

Tom.G and Malloy
mfb
Mentor
As snorkack calculated it doesn't boil.
There will be a little bit of evaporation from the surface, speeding up the freezing a bit, but the main cooling mechanism (at room temperature) should be infrared radiation. It quickly freezes to a solid ball where sublimation gets even less common.
So a square m of mercury surface exposed to vacuum should be giving off about 125 cubic m of mercury vapour per second.
I would be surprised if that approach works. Local cooling as result of the evaporation should reduce the surface temperature.

Malloy
As snorkack calculated it doesn't boil.
There will be a little bit of evaporation from the surface, speeding up the freezing a bit, but the main cooling mechanism (at room temperature) should be infrared radiation.
Solar constant is 1400 W/m2. Meaning that Earth is giving off average of 350 W/m2 of infrared. At room temperature, the cooling by evaporation should be comparable in size to cooling by radiation.
Of course, as temperature drops, the radiative cooling drops with fourth power of temperature, while the evaporative one drops exponentially.
I would be surprised if that approach works. Local cooling as result of the evaporation should reduce the surface temperature.
Mercury as a metal is a relatively good heat conductor, though a poor one for metal. I see the heat conductivity quoted as 8,3 W/(m*K), presumably near room temperature. In which case a mercury surface with 500 W/m2 heat flow should sustain a temperature gradient of 60 K/m - or 0,6 K/cm.

Malloy
Wow, thank you everyone for all your input! Seems to have gotten people thinking, and I'll try my best with that in return here –trying to comprehend all those terms which I'm only just now learning!

What I gathered; due to the fact that it's a highly viscous and dense metal, its sublimation period from liquid to solid would occur too quickly, and too much in the infrared for any evaporative action to overcome the surface tension in any case (unless it started off very hot?).

Now I'm taking a very inconfident guess here; as the surface's molar enthalpy is dropping the temperature through the triple point, there would be a (quick) moment of both gas and vapour emission, and from there it'd just continue irradiating heat (mostly) in the infrared?..

If I clearly don't know what the hell I am talking about here, lol, please forgive me and just for the sake of simplicity, how would you best describe what's happening to/around the mercury in plain visual terms? (I'm thinking of just adding vapour trails as some description in my script.)

What I gathered; due to the fact that it's a highly viscous and dense metal,
Neither viscosity nor density matter here.
its sublimation period from liquid to solid would occur too quickly, and too much in the infrared for any evaporative action to overcome the surface tension in any case (unless it started off very hot?).
Boiling, as in nucleation of bubbles inside of liquid, tends to take place even faster, or not at all.
Now I'm taking a very inconfident guess here; as the surface's molar enthalpy is dropping the temperature through the triple point, there would be a (quick) moment of both gas and vapour emission, and from there it'd just continue irradiating heat (mostly) in the infrared?..

If I clearly don't know what the hell I am talking about here, lol, please forgive me and just for the sake of simplicity, how would you best describe what's happening to/around the mercury in plain visual terms? (I'm thinking of just adding vapour trails as some description in my script.)
What would concentrate the vapour to trails, rather than spherically symmetrically all around the marble?
I found example diametre of a "marble" at 1 cm, though very different values exist, of course.
With that approximate assumption:
Volume of the marble - about 0,5 cm3 (π/6 - a better value is 0,523...)
Initial temperature close to 20 Celsius - I found more precise mercury vapour pressure data that give 160 mPa for 19 Celsius.
Heat loss components:
To infrared - 350 W/m2, from comparison to Earth - depends on the unknown infrared emissivity of mercury.
To vapour - 500 W/m2, derived above - so total of 850 W/m2. Supporting temperature gradient of 1 K/cm
850 W/m2 is 85 mW/cm2
And the area of the marble is π cm2.

Giving the total heat loss rate at 270 mW.
The heat capacity of mercury I see quoted as 0,14 J/K*g. So the 7 g marble has heat capacity of about 1 J/K and is cooling at a degree in under 4 seconds.
By the time the droplet (the temperature gradient of 1 K/cm over 0,5 cm radius of marble means the total temperature difference is under half a degree, plus scale factors due to geometry) has cooled to 0 Celsius, the infrared heat loss has dropped from 35 mW/cm2 to 26 mW/cm2, while the vapour pressure has dropped from 160 to 27 mPa - meaning the evaporative heat loss fell from 50 mW/cm2 to about 9. The total is now just 35 mW/cm2, or 110 mW - meaning the marble now takes 9 seconds to cool a degree, not 4, and took perhaps 2 minutes to cool through the 20 degrees in between.

By the time the droplet reaches -39 degrees, the infrared heat loss is down to 14 mW/cm2 - but evaporative one is relatively small, so within rounding error, 14 is the total as well. Over the surface, it makes up 44 mW, so the marble cools 1 degree in 22 seconds, Taking perhaps 15 minutes total to cool from 20 degrees to freezing, just 2 minutes of them above 0 where evaporation is a large fraction of heat loss.
The latent heat of freezing I see quoted as 11,3 J/g. That means the total latent heat is about 80 J, and at 44 mW, it takes about 30 minutes.
All in all, three-quarters of an hour. No boiling. Heat loss largely by quiet evaporation from the smooth mirror surface in the first minute or two when the drop is still above zero, thereafter mostly radiation. Quarter an hour to cool to freezing, then half an hour to freeze through.

Can anyone go about improving my guesstimated values?

Malloy and mfb
Chestermiller
Mentor
It seems to me all the solid mercury would eventually evaporate if it were receiving any non-zero amount of radiation.

Bystander
It seems to me all the solid mercury would eventually evaporate if it were receiving any non-zero amount of radiation.
If it gets 1400 W/m2 radiation - solar constant at 1 AU - it should lose 350 W/m2 in total. Which, as noted above, it does at 0 Celsius - where it radiates 260 W/m2 and evaporates 90. Thus 1/4 of the solar radiation goes to evaporate it.
I found the triple point pressure of mercury in the meantime, though. 0,165 mPa.
Thus at -39 degrees, mercury would be losing 140 W/m2by radiation and just 0,5 W/m2 by evaporation.
And that total of 140 W/m2 to lose, 560 W/m2 to get, would be reached at 1,6 au - perihelion of Mars.
Now, remember the derivation of 1/120 moles of Hg vapour from square metre per second?
That 1/120 moles is about 1,7 g.
A 1 mm layer on 1 square m is 1 l, mass 13,6 kg. So about 8000 seconds to evaporate 1 mm Hg at 19 Celsius.
At 0 Celsius, it is about 44 000 seconds - the marble exposed to sunshine on Earth orbit, assuming it absorbs it rather than reflect it back, would lose 2 mm in a day, and since the radius was assumed 5 mm, it would evaporate in 2,5 days.
At -39 Celsius, it is about 8 million seconds - 1 mm in 3 months, or the marble with 5 mm radius evaporates in a bit more than Earth year.

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Malloy and Tom.G
Staff Emeritus
2021 Award
It seems to me all the solid mercury would eventually evaporate if it were receiving any non-zero amount of radiation.

Nothing magic about mercury. That argument is equally valid for rock. Yet there are rocks in space. Ergo "eventually" is a really, really long time.

Malloy
Nothing magic about mercury. That argument is equally valid for rock. Yet there are rocks in space. Ergo "eventually" is a really, really long time.

The argument even applies to He. While zero point motion does allow it to stay liquid, and the solubility of He-3 in He-4 stays nonzero, the vapour pressure of even He-3 in vacuum does drop to zero.

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Malloy
Okay, I'm very glad I asked, and I'm sorry if I caused any headaches with this! I'm thinking about scrapping the idea for my story, sadly, but you might've saved me from some potential embarrassment! I thank everyone for their effort here, it's MUCH appreciated!!

I wonder how far into the IR and UV the surface of mercury continues to be a mirror - it certainly absorbs very few photons in the visible wavelengths. The rate of sublimation would be slower than expected, to the extent that the flux of solar photons is reflected rather than absorbed. Note also that a mirror surface is a poor emitter of radiation as well, so that evaporative cooling (the ejection of "hot" surface atoms) becomes more important relative to radiative cooling. The high surface tension of mercury suggests that very few atoms will be hot enough to escape from a very cold surface.

A phase diagram for mercury is surprisingly difficult to find on the Internet - anybody have any luck? All I could locate was one with high-pressure data.

Tom.G
Gold Member
A phase diagram for mercury is surprisingly difficult to find on the Internet - anybody have any luck? All I could locate was one with high-pressure data.

EDIT: Oops! Please disregard diagram. As @mfb pointed out in the next post, this is a phase diagram for water. I should have paid more attention. The Google search below is still valid though.

end Edit

Above, and others, found with: https://www.google.com/search?&q=mercury+pressure+temperature+phase+diagram

Cheers,
Tom

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mfb
Mentor
That is a phase diagram for water.

Tom.G
Tom.G