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`        . 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

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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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions