Every day a certain amount of water evaporates from Earth’s oceans, lakes, and land surface and forms water vapor and clouds in the atmosphere. Every day a certain amount of rain falls back to Earth. Make the reasonable assumption that, on average, the energy absorbed by the evaporation and lifting of the water is equal to the energy released by its condensation and falling back to earth. The evaporation of one mole of water requires approximately 40.6×103 J of heat. An equal amount of heat is released when a mole of water condenses.

QUESTION: The annual volume of rainfall on Earth is approximately 4.9×105 km3 and the average cloud altitude is 8.8 km above Earth’s surface. How much energy, in joules, is required every day to evaporate and lift the water?

Since one mole of water requires 40.6 × 10³ J of heat to evaporate, and the volume of annual rainfall is 4.9 × 10⁵ km³ = 4.9 × 10⁸ m³, we need to find the amount of heat required to evaporate this volume of rainfall.

Amount of heat required to evaporate this volume of rainfall.

So, E = nH where

n = number of moles of water in rainfall = m/M where
m = mass of water = ρV where
ρ = density of water = 1000 kg/m³ and
V = 4.9 × 10⁸ m³ and
M = molar mass of water = 18 g/mol = 1.8 × 10⁻² kg/mol and
H = energy required to evaporate one mole of water = 40.6 × 10³ J/mol

So, E = nH

E = mH/M

E = ρVH/M

Substituting the values of the variables into the equation, we have

E = ρVH/M

E = 1000 kg/m³ × 4.9 × 10⁸ m³ × 40.6 × 10³ J/mol/ 1.8 × 10⁻² kg/mol

E = 198.94 × 10¹⁴ J-kg/mol/ 1.8 × 10⁻² kg/mol

E = 110.52 × 10¹⁶ J

E = 1.1052 × 10¹⁸ J

The amount of energy required to raise this volume to an altitude of 8.8 km

Now, the amount of energy required to raise this volume to an altitude of 8.8 km is E' = mgh where

m = mass of water = ρV where
ρ = density of water = 1000 kg/m³ and
V = 4.9 × 10⁸ m³,
g = acceleration due to gravity = 9.8 m/s² and
h = average cloud altiude = 8.8 km = 8.8 × 10³ m

So, E' = mgh

E' = ρVgh

Substituting the values of the variables into the equation, we have

E' = ρVgh

E' = 1000 kg/m³ × 4.9 × 10⁸ m³ × 9.8 m/s² × 8.8 × 10³ m

E' = 422.576 × 10¹⁴ J

E' = 4.22576 × 10¹⁶ J

The total amount of energy required

So, the total energy required to evaporate and raise the volume of water is E" = E + E'

E" = 110.52 × 10¹⁶ J + 4.22576 × 10¹⁶ J

E" = 114.74576 × 10¹⁶ J

E" = 1.1474576 × 10¹⁸ J

E" ≅ 1.15 × 10¹⁸ J

The total amount of energy required everyday to evaporate and lift the water is 1.15 × 10¹⁸ J

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