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.
