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Grade: 12th pass
. It is claimed that large trees can evaporate as much as 900 kg of water per day. (a) Assuming the average rise of water to be 9.0 m from the ground, how much energy must be supplied to do this? (b) What is the average power if the evaporation is assumed to occur during 12 h of the day?
3 months ago

Answers : (1)

Atharva Vaidya
14 Points
Though a different question I think this will be th answer.
900kg per day means 450kg per 12hrs.
So, to lift 450kg by 9m, the work done is 450*9*10=40500J {asssuming g=10}
Also to convert it into vapour form assuming the tempereture of atmosphere be 25C, the water is not heated so only energy required will be latent heat which is m*L=450*2260000=1017000J {Latent heat of vapourization is 2260000J/kg}
Adding this up we get 1017000+40500=1057500J=1057.5kJ.
For power divide energy by time in seconds.
one month ago
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