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

Question

Consider a one acre field with a 100 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 degree F when the dew forms, what is the final temperature of this layer of air (in degree 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 Let the density of the air be P air Let the specific heat of the air be C air = 1.01 g J/g degree C Kind of amazing isn't it?

Explanation / Answer

1 acre = 4046.86 sqm

Now, thickness of air = 100 mm = 0.1 m

Thus, volume of air = area*thickness = 4046.86*0.1 = 404.686 cum

Now, 60 0F = 15.556 0C

density of air = 1.2 g/ml = 1200 kg/cum ; specific heat capacity of air = 1.01 J/g/0C = 1.01 KJ/g/0C

Let the final temperature of the layer of air be 'T' 0C

Now, heat absorbed by air = mass of air*specific heat of the air*rise in temperature

Mass of air = volume of air*its density = 404.686*1200 = 485623.2 kg

Thus, heat absorbed by air = 485623.2*1.01*(T - 15.556) = 490479.432*(T - 15.556) KJ.................(1)

Heat lost by water vapor = moles of water converted to dew dew*heat of vaporisation of water

volume of dew formed = area*thickness of dew = 4046.86*0.0001 = 0.40469 cum

density of water = 1000 kg/cum

Thus, mass of dew = volume*density = 404.686 kg = 404686 g

molar mass of water = 18 g/mole

Thus, moles of water = mass/molar mass = 404686/18 = 2248.556

Thus, heat lost by water vapor = 22482.556*44 = 989232.445 kJ...........(2)

Now, heat absorbed = heat lost

therefore (1) = (2)

490479.432*(T - 15.556) = 989232.445

or, T = 17.573 0C = 63.6314 0F

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