- #1
Jarfi
- 384
- 12
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?
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?