In July in Los Angeles, the incident sunlight at ground levelhas a power density

Question

In July in Los Angeles, the incident sunlight at ground levelhas a power density of 1.2 kW/m^2 at noon. A swimming pool of area50 m^2 is directly exposed to the sun. What is the maximum rate ofthe loss of water assuming that all the radiant energy isabsorbed?
1 105g/s 2 37g/s 3 25g/s 4 0.20g/s In July in Los Angeles, the incident sunlight at ground levelhas a power density of 1.2 kW/m^2 at noon. A swimming pool of area50 m^2 is directly exposed to the sun. What is the maximum rate ofthe loss of water assuming that all the radiant energy isabsorbed?
1 105g/s 2 37g/s 3 25g/s 4 0.20g/s 1 105g/s 2 37g/s 3 25g/s 4 0.20g/s

Explanation / Answer

1.2 kW/m^2. So on 50 m^2 area, we get 1.2*50 kW heat = 60 kW = 60,000 J/s To vaporize water, 540 cal/g heat is reqd. 540cal = 540*4.2 J pergram So 60,000 J can vaporize 60000/(540*4.2) g water = 26.45g/s. I have assumed that water is already at 100 deg C . In realsituations, water will be at a lower temperature so even less watercan be evaporated. So I'm getting 26.45 g/s....doesn't match with the given optionsthough

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