Finding how much heat released by evaporation of water

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The discussion centers on understanding the heat dynamics involved in the evaporation of water, particularly in relation to a specific figure of 120W. Participants clarify that evaporation itself does not release heat; instead, it requires energy absorption. The heat in question is derived from the human body, which provides the energy needed for water to evaporate. This exchange is crucial for understanding the cooling effect of evaporation. Overall, the conversation emphasizes the importance of energy absorption in the evaporation process.
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Homework Statement
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Relevant Equations
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For this,
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Dose anybody please know where they got the 120W from?

Many thanks!
 
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There must be more context to the question.
 
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haruspex said:
There must be more context to the question.
Thank you for your reply @haruspex!

I'm very sorry, but somehow, I missed the slide before. I think I understand where they got the 120W from now.

Thank you for your help!
 
What releases the heat? The water needs to absorb energy in order to evaporate. Evaporation does not release heat.
 
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nasu said:
What releases the heat? The water needs to absorb energy in order to evaporate. Evaporation does not release heat.
Thank you for your reply @nasu!

The question was about the heat released from the human body I recall. So I think the water absorbs the heat from the human.

Many thanks!
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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