The amount of heat absorbed by 1g of water while vapourising?

Heat absorbed while vapouraisation = m×Cp×ΔT + m×Lv .............(1)

where m is mass of water = 1 g = 10-3 kg

Cp is specific heat of water = 4200 J/(kg.K)

ΔT  is temperature difference = (100-20) °C = 353 K ( initial temperature is assumed as 20 °C)

Lv is latent heat of vapouraisation = 2260 kJ / kg

 

Substitute all the data in Eqn.(1) to get the answer 3743 J