3 ms·
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 al
by 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 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!