Question

What is the total amount of heat required to completely vaporize a 100.0 g of

water at its normal boiling point?

Work too pls !!

Answer 1

The total amount of heat : 225.704 kJ

Further explanation

Heat of vaporization(ΔH vap) : heat needed to vaporize 1 mole of water , units : kJ/mol

For water at boiling point (100 °C) : ΔH vap = 40.66 kj/mol

mole of 100 g water :

`\tt (100)/(18.015)=5.551`

so total Heat :

`\tt 40.66* 5.551=225.704~kJ`