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 oF when the dew forms, what is the final temperature of this layer of air (in o 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 ∆Hvap = 44 kJ mol • Let the density of the air be ρair = 1.2 kg m3 • Let the specific heat of the air be Cair = 1.01 J g · oC

Answer 1

The heat relesed by the dew layer = The heat absorbed by the air

Since condensing water vapor is the opposite process of vaporization,

The heat relesed by the dew layer = ∆Hvap x moles of dew layer

                                                  =    44 kJ mol^-1 x (4046.86 m² x 1x10^-9 m x 1000 kgm^-3/ 18x10^-3 kg mol^-1)

                                                  =    9.88 kJ

The heat absorbed by the air        =     mass of air x specific heat of the air x temperature difference

                                                 =     30.38 m x 4046.86 m² x 1.2 kgm^-3 1.01 kJ kg^-1 K^-1x temperature difference

                                                 =     149007 kJ x temperature difference

Therefore,

9.88 kJ = 149007 kJ x temperature difference

temperature difference = 6.63 x 10^-5 K = 6.63 x 10^-5 oF

t₂ - t₁ = 6.63 x 10^-5 oF

t₂ - 60 oF = 6.63 x 10^-5 oF

t₂ = 60.0000663 oF

• Note that temperature difference is same in any units either Celcius,kelvin or faranheit
• here cres and feets have been converted to meters