Evaporation Moisture: 0.3L/m2 Hot Water at 50°C

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To determine how long it takes for 0.3 liters/square meter of hot water at 50°C to reach 0°C on an ice rink at -5°C, the heat load is calculated to be 165.9 kJ. This results in an estimated moisture release of 0.07 kg when considering the evaporation heat of water. The problem involves complex calculations, including heat conductivity, latent heat, and the varying temperature of the water as it cools. Additionally, heat loss due to radiation must be factored in, requiring knowledge of emissivity. The rate of evaporation is also complicated, necessitating integration of vapor pressure and ambient conditions over time.
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Homework Statement


- if I put 0.3 litres/squaremeter of hot water (50 C) on ice rink (-5 C), how long will take to the water to be 0 grade C ? How much evaporation moisture will be in the air (5 C) and how long will be the time for evaporation ?

Homework Equations


Heat load= 1000*0.3(liters/sm)*(4.2*(50-0)+333+2*(0-(-5))
Heat load=165.9 kJ
- evaporation heat of water =2270(kJ/kg)
so the moisture will be: 165.9/2270=0.07 kg


The Attempt at a Solution

 
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I would start by assuming a thin layer of water above the ice, so that it's of uniform temperature but still with finite heat capacity. Then you have to use the conductivity of water and ice, latent heat of the latter, and specific heats of both, to determine the rate at which heat leaves the water layer. Not easy, since the ice is melting to some extent. (Eventually, all the water should wind up as ice even though the ambient temperature is +5C.)

Also, you should proably take into account the loss of heat of the water to radiatiion : eσ(T^4 - Ta^4) where Ta is +5 deg C and e for water has to be looked up (emissivity). T will vary from +50 to 0C.

As for rate of evaporation, the formula I found is
(mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) ). Has to be integrated since T is not constant. The rersult should give you both the amount of evaporate and the time to evaporate all the water.

This is a very complicated problem which may explain why you're not getting any responses from anybody else, and why mine is so flimsy. :-)

http://van.physics.illinois.edu/qa/listing.php?id=1440
 
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Thank You very much
 
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