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

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SUMMARY

The discussion focuses on calculating the time required for 0.3 liters per square meter of hot water at 50°C to reach 0°C when placed on an ice rink at -5°C. The heat load is determined to be 165.9 kJ, leading to an evaporation moisture of 0.07 kg. Key factors include the heat transfer from water to ice, the latent heat of evaporation, and the impact of ambient temperature on evaporation rates. The complexity of the problem arises from the need to account for varying temperatures and heat loss through radiation.

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  • Familiarity with latent heat calculations
  • Knowledge of vapor pressure and partial pressure concepts
  • Basic thermodynamics, including specific heat capacities
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This discussion is beneficial for physicists, engineers, and students involved in thermodynamics, particularly those interested in heat transfer and phase change processes in water and ice systems.

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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

 
Last edited:
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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
 
Last edited:
Thank You very much
 

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