Consider a one acre field with a 100 ft high layer of air ex

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

Solution

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