Amount of water evaporated by laser, questionable results.

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A 30kW laser can evaporate approximately 13-14 grams of water per second when aimed at a boiling pot, based on calculations using the heat of vaporization. This rate is considered quite fast, as it would take about 76 seconds to boil away one liter of water. However, the assumption that all laser light is absorbed by the water is questionable, as absorption depends on the laser's wavelength. Factors such as power loss through air, reflection, and beam dispersion could further reduce the actual evaporation rate. Overall, while the calculations are correct, practical considerations may lead to lower evaporation results.
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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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