Ice in space, sublimation time

In summary, it takes water ice H20 in space in our solar system to sublimate, say a basic ice cube, depending on the temperature and heat source. The article referenced provides a table of data for various sublimation processes. Keeping ice at 0 C in vacuum would take 5.6 MJ/m2*s.
  • #1

Albertgauss

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Looking for data about how long it takes ices to sublimate in space.
How long does it take water ice H20 in space in our solar system to sublimate, say a basic ice cube? It starts as a solid cube at the temperature of whatever space is above Earth and then completely turns to vapor. Just looking for ballpark situation here.

Does anyone know of a table or place where it is organized the data for the sublimation of the ices of other materials (CO2, N2, CH4, etc)?

I searched the thread for this topic, as I am certain it has come up before, but there was so much about "ice" I could not find what I was looking for.
 
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It depends on temperature and heat source.
I believe the Herst-Knudsen equation is what you are looking for.
Also, as sublimation occurs, is cools the ice that is left behind, so a steady source of warmth is helpful.

There is a table in this article.
 
  • #3
Confirm to see if I got the following correct for the Herst Knudsen equation referenced

The final units for the HK equation would be: particles/(secs*[meters][2])
Let alpha, sticking coefficient) be 0.5 (no other info to go on, could not find for water ice)
The gas pressure in space was looked up to be 1.322[10][-11] Pascal
Na is Avo's number 6.022[10][-23]
M is molar mass of ice, looked up to be 0.018 kg/mol
R is gas constant: 8.31 J/(mol*Kelvin)
T is temp above Earth, looked up to be 280 Kelvin

So then I got 2.46[10][11] particles/(secs*[meters][2])

Is this correct? What could I compare this number to in the realm of human experience? Is this fast on human timescales (secs, hours) etc or slow? I don't know that, if an astronaut drops an ice cube in space, will it sublimate quickly or slowly? For example, comets of water-ice are expected to last forever because their sublimation rate is so slow, but I don't know if the ice cube left outside the space shuttle is in the same catergory as a comet.
 
  • #4
.Scott said:
It depends on temperature and heat source.
I believe the Herst-Knudsen equation is what you are looking for.
Also, as sublimation occurs, is cools the ice that is left behind, so a steady source of warmth is helpful.

There is a table in this article.
I believe the article is much better than the equation. I independently came up with basically the same reasoning.
Consider the following reasoning:
In vacuum, a vapour will escape an evaporating surface at a speed at the order of the speed of sound, within a modest factor.
At 0 C, assume that it is 400 m/s for water vapour. Then at 0 C, 1 square metre of ice will give off 400 cubic m of water vapour per second.
But it is water vapour at the vapour pressure - 6,1 mbar. So if you compress it to STP - which won´t be stable, but I´m doing it for accounting reason to track the gas law - you would get 2,44 cubic m of steam.
Avogadro´s law states that 22,4 l is 1 mole at STP, so 2440 l at STP is about 110 mol. Which is 2000 g of water. One square m of ice at 0 c will lose 2,2 mm per second, 8 m per hour.
But ice´s latent heat of evaporation at 0 C is around 2830 J/g. Which means that keeping ice at 0 C in vacuum will take 5,6 MJ/m2*s
Sun at 1 AU will provide only under 1400 J/m2*s. You will need to approach 60 times closer, to within 2 400 000 km of Sun´s centre, that is, 1 700 000 km of Sun´s surface, to melt ice!

Alternatively, the ice would cool until its vapour pressure falls 4000 times or so compared to the vapour pressure at 0 C, and the speed of evaporation and loss of heat to evaporation decreases to below the heat input. My estimate is that at 1 AU, this would happen around -80 C or so.

Now, if the heat input is lower...
There will be two ways for heat to get out of the snowball. One is evaporation. This is close to proportional to vapour pressure, which falls close to exponentially with 1/T. The second is radiation, which falls with T4. So with sufficiently low T, the radiation becomes the main heat loss.
For high heat inputs, most of the heat is spent on evaporation and speed of the evaporation will be proportional to heat input. For low input, most of it is radiated away and speed of evaporation falls exponentially.

Would you like more trying the details?
 
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  • #5
Albertgauss said:
but I don't know if the ice cube left outside the space shuttle is in the same catergory as a comet.
Probably not. The radiation from the Sun, when the cube is near Earth, would be something like 1kW per m squared. That keeps us at around 300K equilibrium temperature. Comets spend most of their time around the Kuiper Belt, which is 50 AU from the Sun. The radiation would be (1/50)2 = 1/2500 of what earth gets and comets spend most of their time at or beyond that distance so their situation would be a lot different. I guess that the rate of heating would be 1/2500 so the time taken for the ice cube to sublimate could be something like 2500 times as long. The short amount of time comets spend near the Sun would reduce that time but not a lot, I suspect.

A dirty comet could have a higher surface temperature than a pristine piece of ice.

@snorkack introduces the idea that the radiation loss is relevant. If the temperature is low then that may dominate because of the probability of a molecule escaping and taking potential energy from the surface.
 
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  • #6
Yes, working through what you said. I think it will be the answer I am looking for.
 
  • #7
But all the above doesn't discuss how water is accreted into comets in the first place. That must be relevant. It suggests to me that the difference between them is that an ice cube beside the Shuttle will be shrinking but a comet in the Kuiper Belt will probably be growing??? (They have to be, or have been, growing somewhere.)

Something 'we' tend to ignore is the vast range of timescales that operate out in space. Huge stars taking only a few million years to form and red dwarves taking tens of billions of years . I suspect that activity in the Kuiper Belt will be very slow indeed; objects in 50 AU orbits. Would the 'bits' eventually form planets, left to themselves?
 
  • #8
sophiecentaur said:
Probably not. The radiation from the Sun, when the cube is near Earth, would be something like 1kW per m squared. That keeps us at around 300K equilibrium temperature. Comets spend most of their time around the Kuiper Belt, which is 50 AU from the Sun. The radiation would be (1/50)2 = 1/2500 of what earth gets and comets spend most of their time at or beyond that distance so their situation would be a lot different. I guess that the rate of heating would be 1/2500 so the time taken for the ice cube to sublimate could be something like 2500 times as long.
The short amount of time comets spend near the Sun would reduce that time but not a lot, I suspect.

A dirty comet could have a higher surface temperature than a pristine piece of ice.

@snorkack introduces the idea that the radiation loss is relevant. If the temperature is low then that may dominate because of the probability of a molecule escaping and taking potential energy from the surface.
Yes, and then the rate of sublimation drops far below the rate of heating.
So, 1400 W per m squared evaporates about 0,5 g ice per square m per s. In a day, it makes 43 kg - about 5 cm layer. In a month, it makes 150 cm. A month is the order of magnitude that a comet spends near perihelion. If the comet approaches to Venus´ orbit at 0,7 AU, it will be losing 10 cm per day but also will orbit faster. If the comet retreats to Martian orbit, it will be evaporating slower but orbiting slower.
But what happens at lower temperatures?
https://www.researchgate.net/public...ressure_above_Ice_at_Temperatures_below_170_K
offers
log P = -3059/T + 14.88 (P in Pa and T in K)
So...
At 170 K (-103 C), log P comes as -3.12
Which means 130 million times less than atmospheric pressure. 400 cubic m/s at that pressure would make around 3 ccm of "STP steam", which is 2,4 mg. And evaporating 2,4 mg ice per second takes just 6,7 W/m2
At 170 K, a black body will radiate 5.67*(1.7)4=47,3 W/m2. So the total heat loss 54 W/m2 will be supplied by Sun at about 5 AU... an insulated black ice surface turned to Sun at Jupiter´s orbit would warm to about -103 C, spend 12% of the sunlight on evaporation and lose 7 mm ice per month.
Cool the surface to 160 K, and P will decrease 13 times, so evaporation loss falls to 0,5 W/m2. Whereas the black body radiation decreases to just 37,2 W/m2.
 
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