What quanity of heat is required to vaporize 38.5 of liquid water at 100 degrees C?

Question

i need the answer in kJ with significant numbers

Answer 1

40.657 kJmol^-1 is the enthalpy of vaporisation of water at 100 degrees, according to the CRC handbook.
You haven't given units for water, but I assume you either mean millilitres or grams. Fortunately, these are the same for water.
Molar mass of H2O = 18 gmol^-1. Therefore, 38.5g/18gmol^-1 = 2.13 moles. So, the energy needed = 2.13 mol x 40.657 kJmol^-1, = 86.6 kJ.

Answer 2

you choose the molar mass of water to end this. that's approximately 18 g/mole, so which you have approximately 21/18 = 7/6 of a mole. Multiply that via 40.7 kJ/mole and you ought to get around 40 seven kJ. this is *no longer* the three-important-digit accuracy you like for the difficulty, however.

Answer 3

Latent Heat of vaporisation of water = 2260kJ/kg
or 2260J/g

Your question : - 38.5 what units