Answer:
2465 J/g
Explanation:
The amount of energy required to boil a sample of water already at boiling point is given by
[tex]Q=m\lambda_v[/tex]
where
m is the mass of the water sample
[tex]\lambda_v[/tex] is the specific latent heat of vaporization of water
In this problem, we know
[tex]Q=813,600 J[/tex]
[tex]m = 330 g[/tex]
Solving the equation for [tex]\lambda_v[/tex], we find
[tex]\lambda_v = \frac{Q}{m}=\frac{813600}{330}=2465 J/g[/tex]