Answer:
[tex]Q_{net} = 15469.5\,J[/tex]
Explanation:
Let assume that liquid water has an initial temperature of 25°C and heating and evaporation process occurs under a pressure of 1 atmosphere. Then, the heat absorbed is equal to the sum of sensible and latent components:
[tex]Q_{net} = Q_{s} + Q_{l}[/tex]
[tex]Q_{net} = (6\,g)\cdot \left[\left(4.186\,\frac{J}{g\cdot ^{\circ}C}\right)\cdot (100\,^{\circ}C-25\,^{\circ}C)+2264.3\,\frac{J}{g} \right][/tex]
[tex]Q_{net} = 15469.5\,J[/tex]
