ANSWER
q = 336,943.2J
EXPLANATION
Given that;
The mass of water is 132 g
The temperature of water is 30 degrees Celcius
Follow the steps below to find the amount of energy required
Step 1; Calculate the heat required to warm water from 30oC to 100oC
[tex]\text{ q = mc}\Delta T[/tex]Substitute the given data into the formula above
[tex]\begin{gathered} \text{ c = 4.18 J/g}\degree C \\ \text{ q}_1\text{ = 132 }\times\text{ 4.18}\times(100\text{ - 30\rparen} \\ \text{ q}_1\text{ = 132 }\times\text{ 4.18 }\times\text{ 70} \\ \text{ q}_1\text{ = 38,623.2J} \end{gathered}[/tex]Step 2; Calculate the amount of heat required vapourize the water steam
[tex]\text{ q = m}\Delta H_{vap}[/tex][tex]\begin{gathered} \text{ }\Delta H_{vap\text{ }}=\text{ 2260J.g}^{-1} \\ \text{ q}_2\text{ = 132}\times2260 \\ \text{ q}_2\text{ }=\text{ 298320 J} \end{gathered}[/tex]The total energy is
q = q1 + q2
q = 38,623.2 + 298320
q = 336,943.2J