Consider a one acre field with a 100 ft high layer of air exactly over the field

Question

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 degrees F 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 : DeltaHvap = 44 kJ/mol Let the density of the air be: air = 1.2 g/mL Let the specic heat of the air be Cair= 1.01 J/gxC

Explanation / Answer

As the complete field is perfectly flat and has auniform layer of air we can consider any small piece of the field say 1sq.m for simplification.

dew formed over 1 sq.m

1*1.0e-4 cu.m

mass of the dew = 1.e-4 *1000 = 0.1 kg

when the dew formation started the temperature of the air is 60 degC

water vapor has condensed to air

molar mass of water = 2+16 = 18 gm

number of mols in 1 sq.m of dew = 100/18

heat given out by condensation of water vapor = 44 kJ*(100/18)

                                                                            = 244 kJ

This heat givne out by the water vapor and raises the temperature of the aire column above the 1sq.m of field

100ft = 100*0.305 m = 30.5 m

volume of air = 30.5*1 = 30.5 cu.m

density of air = 1.2g/mL , it s given wrong

it must be 1.2 g/L not mL

    = 1.2 kg/cu.m

mass of air = 30.5*1.2 = 36.58 kg

specific heat of air = 1.01 J/g-C

                             = 1.01 kJ /kg-C

if T is the raise in temperature of the air then

1.01kJ*36.58*T = 244kJ

T = 244/(36.58*1.01) = 6.60 deg C

       = 11.89 F

Final temperature of the air = 60+11.89 = 71.89 F

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