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 oF)?

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

Data.

Area = 1 acre = 4046.86 m2

h = 100 ft = 30.48 m

V_air = 4046.86 m2 x 30.48 m = 1.2335 x 10⁵ m3

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h_dew = 0.1 mm = 0.0001 m

T_dew = 60 °F = 15.56 °C

V_dew = 4046.86 m2 x 0.0001 m = 0.4047 m3

--

∆H_vap = 44 kJ/mol

ρ_air = 1.2 kg/m3

C_air = 1.01 J/g.°C

T_f = ?? °C

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Heat absorbed.

q_air (KJ) = m_air x C_air x ∆T

where;

m = air mass, g

C = specific heat of air, J/g.°C

∆T = (Tf - To), change in temperature °C

--

m_air = ρ_{air x} V_air = (1.2 kg/m3) x (1.2335 x 10⁵ m3)

m_air = 1.48 x 10⁸ g

m_vap = ρ_{vap x} V_vap = (1000 kg/m3) x (0.4047 m3)

m_vap = 4.05 x 10⁵ g

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Heat absorbed from the formation of the dew.

q_air = ∆H_vap x moles_(dew)

q_air = 44 kJ/mol x [(4.05 x 10⁵ g)/(18 g/mol)]

q_air = 9.89 x 10⁵ KJ

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Let's find the missing temperature:

9.89 x 10⁸ J = 1.48 x 10⁸ g x 1.01 J/g.°C x (T2 - 15.56 °C)

6.61 x 10^-3 °C = T2 - 15.56 °C

T2 = 15.56661 ° C = 60.019898 °F

It all looks that this process acts as an isothermally one!