- #1

Jarfi

- 384

- 12

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?