4 ms·
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. The
by 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 agoThe 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 agoIt 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 agoWell that escalated quickly!