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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

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