Amount of water evaporated by laser, questionable results.

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Discussion Overview

The discussion revolves around the amount of water that can be evaporated by a 30kW laser directed at a boiling pot. Participants explore the calculations related to the heat of vaporization of water and the assumptions involved in the evaporation process, including the absorption of laser light by water.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates that a 30kW laser could evaporate approximately 13.2 grams of water per second based on the heat of vaporization of water.
  • Another participant agrees with the calculation but emphasizes that 13.2 g/s is a significant rate of evaporation, noting it would take about 76 seconds to boil away one liter of water.
  • Some participants highlight the importance of understanding the units involved in the calculations, suggesting that a better grasp of energy and power units could clarify the results.
  • There is mention of the latent heat of water being around 2.2 kJ/g, which aligns with the earlier calculations, indicating that the evaporation rate is consistent across different contributions.
  • Concerns are raised regarding the assumption that all laser light is absorbed by the water, with participants noting that the absorption depends on the wavelength of the laser and that it is unlikely to be 100% effective, especially for visible light.
  • The original poster acknowledges additional factors that could affect the evaporation rate, such as power loss through air, reflection of EM waves, and the beam's alignment with the pot.

Areas of Agreement / Disagreement

Participants generally agree on the calculations regarding the evaporation rate, but there is disagreement about the assumptions made regarding laser light absorption and its implications for the actual evaporation process. The discussion remains unresolved regarding the extent of these assumptions' impact.

Contextual Notes

Participants note several assumptions that could affect the calculations, including the efficiency of laser light absorption by water, potential power loss over distance, and the alignment of the laser beam. These factors introduce uncertainty into the evaporation rate derived from the initial calculations.

Jarfi
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Assume a laser with 30kW output energy is pointed at a boiling pot, how much water would it evaporate per second?

30kW=30kJ/s.

I assume the water is at boiling point and get:

Water heat of vaporization (40,65 kJ·mol−1).

Thus we get:

(30kJ/s)/(40,65kJ/mol)=0,74mol/s evaporated.

This gives us(using the molecular mass of water)

0,74mol/s*18g/mol=13,2 g/s.

Results are: 13 grams of water are evaporated per second if a 30kW laser is heating the pot.

I find this very little and have assumed more water would have been evaporated, are any of these calculations wrong?
 
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That looks correct. The problem is you thinking that 13,2 g/s is not much. It is an incredibly fast evaporation: it would take 76 s to completely boil away one liter of water. Your pasta won't have time to cook!
 
Your calculations are correct.
The problem is you still don't have a good sense about units you're using.
Try to "understand" what is a joule, a second, a watt!
 
I always keep in mind the order of magnitude to evaporate water: 2.5 GJ/ton = 2.5 kJ/g.
So, you are right.
 
The latent heat of water is about 2.2 kJ/g.
Your laser provides 30 kJ in one second. So it may evaporate about 13-14 g /s.
Your calculation seems OK.

But you assume that all the laser light is absorbed by water, which is questionable.
The fraction absorbed depends on the wavelength of the laser. But ti is not very likely to be 100%.
If it's visible light, won't be too much absorption.

edit. So many people answering at the same time. :smile:
 
nasu said:
The latent heat of water is about 2.2 kJ/g.
Your laser provides 30 kJ in one second. So it may evaporate about 13-14 g /s.
Your calculation seems OK.

But you assume that all the laser light is absorbed by water, which is questionable.
The fraction absorbed depends on the wavelength of the laser. But ti is not very likely to be 100%.
If it's visible light, won't be too much absorption.

edit. So many people answering at the same time. :smile:

Thanks for the help guys,

It is light in the infrared to ultraviolet spektrum.

I am glad my calculations are ok and fully aware that I was making assumptions,

not including: Loss of power through distance traveled through air, amount of EM-waves reflected and amount of the beam that would miss the pot(depending on the distance from the laser.)

Which would make even less water boil !
 

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