Answer
The average current that is required to vaporize the water is $~~13.0~A$
Work Step by Step
We can find the mass of the water:
$m = \rho~V = \rho~A~L = (1000~kg/m^3)(15\times 10^{-5}~m^2)(0.120~m) = 0.018~kg$
We can find the required energy:
$E = (2256~kJ/kg)(0.018~kg) = 40.6~kJ$
We can find the required power:
$P = \frac{E}{t} = \frac{40.6~kJ}{2.00~ms} = 20.3~MW$
We can find the resistance:
$R = \frac{\rho~L}{A} = \frac{(150~\Omega\cdot m)(0.120~m)}{15\times 10^{-5}~m^2}$
$R = 1.20\times 10^5~\Omega$
We can find the required current:
$P = i^2~R$
$i^2 = \frac{P}{R}$
$i = \sqrt{\frac{P}{R}}$
$i = \sqrt{\frac{20.3\times 10^6~W}{1.20\times 10^5~\Omega}}$
$i = 13.0~A$
The average current that is required to vaporize the water is $~~13.0~A$
