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Help with physics problem?
A lake 1km square in area is reduced in depth by 5 mm in the space of 30 minutes through the process of evaporation. What is the latent heat flux (in W/m^2) associated with this evaporation. You may assume that the latent heat of evaporation is 2260kJ/kg and the density of water is 1000kg/m^3.
2 Answers
- oubaasLv 77 years agoFavorite Answer
ΔVol = (10^4)^2*5*10^-2 = 5*10^6 dm^3 = 5*10^6 kg
E = 2.26*10^6*5*10^6 = 1.13*10^13 joule
P = E/t = 1.13*10^13/(1.8*10^3) = 6.28*10^9 watt
heat flux = 6.28*10^9/10^6 = 6.28*10^3 w/m^2
- ?Lv 77 years ago
area of lake = A = E3 m²
volume of lake water evaporated in 30 min = E3 m² x 5E-3 m = 5 m³
mass of water evaporated in 30 min = 5 x 1000 kg/m³ = 5000 kg = 5E3 kg
amount of latent heat energy reqd to evaporate 5E3 kg of lake water:
Latent heat reqd = Q = m(hv) = 5E3(2.26E6) = 11.3E9 J
Latent heat power reqd = P = Q/time = 11.3E9/30(60) = 11.3E9/1.8E3 = 6.28E6 watts
Latent heat flux = Φ = P/A = 6.28E6/E3 = 6.28E3 watts/m² ANS
