Question

Use the exact values you enter in previous answer(s) to make later calculation(s).

(a) How much energy is required to evaporate all the water in a swimming pool of area 114 m² and depth 2.3 m on a typical summer day?

(b) The intensity of sunlight is about 1,000 W/m². If all this energy is absorbed by the water in the pool, how long will it take the water to evaporate?

Answer 1

(a) volume of water = area of pool X depth of pool = 114m^2 X 2.3 m= 262.3 m^3

Density of water = 1000Kg/m^3

Mass of water in pool = 262.3 m^3 X 1000 Kg/m^3 = 262.3 X 10^3 Kg

Heat of vaporiztion of water = 2257 X 10^3 J/Kg

Total Heat required to vaporize water = 2257 X 10^3 J/Kg X 262.3 X 10^3 Kg = 5.92 X 10^11J

(b)

Heat of sunlight = 1000 Watts/m^2

Heat of sunligt per second per m^2 = 1000 J/m^2/sec

Total heat of vaporiztion needed /m^2 of water = 5.92 X 10^11 /114 m^2 = 0.0519 X 10^11 = 5.19 X 10^9 J/m^2

Total time required for evaporation = 5.19 X 10^9 J/m^2 / 1000 J/m^2/sec = 5.19 X 10^6 seconds