4 ms·

Very cool. One quibble with the title. This is what happens in zero G, I'd love to see the behaviour in space (outside the capsule) too. Does anyone know off

Very cool. One quibble with the title. This is what happens in zero G, I'd love to see the behaviour in space (outside the capsule) too. Does anyone know off the top of their head the state/behaviour of water in the low temperature and low pressure of space? My prediction for the Zero G experiment was that that it would spray in all directions. [Partial spoiler] I was wrong because I forgot about a critical behaviour of water.

icegreentea 14y ago
Apparently, depends on how much sun you get. If you're in relative darkness, the water would disperse and then freeze into droplets. If you have some sun, the ice would vapourize. http://www.space.com/7274-mystery-explained-glow-night-sky-astronaut-urine.html http://www.space.com/7274-mystery-explained-glow-night-sky-a...
claudius 14y ago
The phase diagram[0] for water suggests that it will be frozen at 0 Pa and anything less than 200 K, so it will likely form small ice crystals on the cloth. Then trying to wring out the frozen cloth could prove interesting, though. [0] http://www.lsbu.ac.uk/water/phase.html http://www.lsbu.ac.uk/water/phase.html
- 4ad 14y ago
  The water won't have less than 200K when you start the experiment though (else you won't have a wet cloth but a frozen cloth), so all the water will boil off almost instantly.
  - spinonethird 14y ago
    It would start boiling but this very fact would very quickly cool it off to freezing point: water has a very high heat of vaporization.
    - 4ad 14y ago
      ~72% of the initial water would freeze as ice. Phase #1, water cooling and boiling: mc dT = λv dm dT = λv/(mc) dm | ∫ Tƒ - T₀ = λv/c ∫ (dm/m) = λv/c (ln mƒ' - ln m₀) = = λv/c ln (mƒ'/m₀) ln (mƒ'/m₀) = c/λv (Tƒ - T₀) mƒ'/ m₀ = exp[c/λv (Tƒ - T₀)] mƒ'/m₀ = exp[ 4192J/(kg K) * 1/(2257 * 10^3J/kg) * (-100K)] = = exp[ -4192/2257 * 10^-3 * 10^2 ] = = exp[ -4192/2257 * 10^-1 ] ≈ ≈ 83%. Phase #2, water freezing, remaining water still boiling: mƒ = mƒ' - mv |Qced| = Qabs mƒ λc = mv λv mƒ = λv/λc mv mƒ = λv/λc (mƒ' - mƒ) = λv/λc mƒ' - λv/λc mƒ mƒ (1 + λv/λc) = λv/λc mƒ' mƒ = λc/(λc + λv) * λv/λc mƒ' = = λv/(λc + λv) mƒ' mƒ/m₀ = 2257kJ/kg * 1/(2257kJ/kg + 335kJ/kg) * 0.83 ≈ ≈ 2257/2592 * 0.83 ≈ ≈ 72%
      - kvprashant 14y ago
        Well that escalated quickly!
timroman 14y ago
It would probably boil off before it would freeze.

icegreentea 14y ago

Apparently, depends on how much sun you get. If you're in relative darkness, the water would disperse and then freeze into droplets. If you have some sun, the ice would vapourize. http://www.space.com/7274-mystery-explained-glow-night-sky-astronaut-urine.html http://www.space.com/7274-mystery-explained-glow-night-sky-a...

claudius 14y ago

The phase diagram[0] for water suggests that it will be frozen at 0 Pa and anything less than 200 K, so it will likely form small ice crystals on the cloth. Then trying to wring out the frozen cloth could prove interesting, though. [0] http://www.lsbu.ac.uk/water/phase.html http://www.lsbu.ac.uk/water/phase.html

4ad 14y ago
The water won't have less than 200K when you start the experiment though (else you won't have a wet cloth but a frozen cloth), so all the water will boil off almost instantly.
- spinonethird 14y ago
  It would start boiling but this very fact would very quickly cool it off to freezing point: water has a very high heat of vaporization.
  - 4ad 14y ago
    ~72% of the initial water would freeze as ice. Phase #1, water cooling and boiling: mc dT = λv dm dT = λv/(mc) dm | ∫ Tƒ - T₀ = λv/c ∫ (dm/m) = λv/c (ln mƒ' - ln m₀) = = λv/c ln (mƒ'/m₀) ln (mƒ'/m₀) = c/λv (Tƒ - T₀) mƒ'/ m₀ = exp[c/λv (Tƒ - T₀)] mƒ'/m₀ = exp[ 4192J/(kg K) * 1/(2257 * 10^3J/kg) * (-100K)] = = exp[ -4192/2257 * 10^-3 * 10^2 ] = = exp[ -4192/2257 * 10^-1 ] ≈ ≈ 83%. Phase #2, water freezing, remaining water still boiling: mƒ = mƒ' - mv |Qced| = Qabs mƒ λc = mv λv mƒ = λv/λc mv mƒ = λv/λc (mƒ' - mƒ) = λv/λc mƒ' - λv/λc mƒ mƒ (1 + λv/λc) = λv/λc mƒ' mƒ = λc/(λc + λv) * λv/λc mƒ' = = λv/(λc + λv) mƒ' mƒ/m₀ = 2257kJ/kg * 1/(2257kJ/kg + 335kJ/kg) * 0.83 ≈ ≈ 2257/2592 * 0.83 ≈ ≈ 72%
    - kvprashant 14y ago
      Well that escalated quickly!

timroman 14y ago

It would probably boil off before it would freeze.