Given,
The latent heat of vaporization of water, L=2.26×10⁶ J/kg
The specific heat capacity of the water, c=4190 J/kg/K
The density of the water, ρ=1000 kg/m³
The volume of the water, V=100 m³
The initial temperature of the water, T₁=18 °C
The final temperature of the water, T₂=100 °C
The mass of the given volume of water is,
[tex]\begin{gathered} m=\rho V \\ =1000\times100 \\ =100\times10^3\text{ kg} \end{gathered}[/tex]
The heat needed to raise the temperature of the water to 100 °C is given by,
[tex]Q_1=mc(T_2-T_1)[/tex]
The heat needed to vaporize the water at 100 °C is given by,
[tex]Q_2=mL[/tex]
Thus the total heat energy needed is given by,
[tex]\begin{gathered} Q=Q_1+Q_2 \\ =m\lbrack c(T_2-T_1)+L\rbrack \end{gathered}[/tex]
On substituting the known values,
[tex]\begin{gathered} Q=100\times10^3\lbrack4190(100-18)+2.26\times10^6\rbrack \\ =2.6\times10^{11}\text{ J} \end{gathered}[/tex]
Thus the heat energy needed to evaporate the given amount of water is 2.6×10¹¹ J
