How much energy is required to vaporize 10 grams of water at its boiling point in joules. explain.

Energy required to vaporize 10 grams of water at its boiling point is 22,600 J or 22.6 k J

Explanation:

In order to find the energy required for the vaporization of water at its boiling point, we must understand about heat of vaporization.

                        Heat of Vaporization of Water Hv = 2260 J /g

Heat of vaporization is the amount of heat required to convert unit mass of liquid into vapor without change in temperature.

To convert 1 g of water at boiling point  to 1 g of steam at 100 ºC, 2260 J of heat must be absorbed by the water.

The formula for finding energy is given by

                                                  q= m Hv

Here m=10 g

Energy required q= 10*2260=22600 J (or) 22.6 k J.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8