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?