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Chrome File Edit View History Bookmarks People Window Help D D 4) 80% (a, Sun 5:33 AM Q EE O O O http C https:// black board.olemiss.edu 1/courses/Chem_106 Web 1 DAVIS 2014-2015 SUM2/cp 3.pdf E App untitled folder App cloud If Facebook Twitter W wikipedia Y Yahoo! News Popular Imported From Safari Chemistry 106 Challenge Problem 3 Consider a one acre field with a 100 ft high layer of air exactly over the field. Water vapor condenses and forms a 0.1 mm layer of dew on the surface of the field. If the air temperature was 60 OF when the dew forms, what is the final temperature of this layer of air (in F)? Assumptions and Values: Let the surface of the field be perfectly flat. Let the air absorb all heat from the formation of the dew. Let the heat of vaporization of water be AH 44 mo. Let the density of the air be p 1.2 mL Let the specific heat of the air be C 1.01 g. OC Kind of amazing isn't it?

Explanation / Answer

1 acre = 43560 sqft = 4046.86 sqm

Volume of the air field = area of the field*depth of layer = 43560*100 = 4356000 cuft

Now, 1 cuft = 0.03 cum

THus,. volume of air = 4356000*0.03 = 130680 cum

density of air = 1.2 g/ml = 1200000 g/cum

Thus, mass of air = volume*density of air = 130680*1200000 = 1.568*1011 g

Now, mass of water vapor = Volume*density = area*depth*density = 4046.86*0.0001* = 0.405*1000000 = 4.05*105 g

Now, molar mass of water = 18 g/mole

Thus, moles of water = mass of water/molar mass = 22482.56 moles

Now, heat lost/absorbed = mass of the substance*specific heat capacity of the substance*change in temperature

Now, Let the tempertaure of air after dew formation = T

Thus, heat absorbed by air = (1.568*1011)*1.01*(T-60)...........(1)

Heat lost by vapour for conveting to dew = moles of water*delta Hvapourization =

22482.56*44000 = 9.89*108 J................(2)

Now, From the law of conservation of energy , heat lost = heat absorbed

Thus, (1) = (2)

or, (1.568*1011)*1.01*(T-60) = 9.89*108

or, T = 60.006 0 C = 140.0108 0F

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