The rain-soaked shoes of a person may explode if ground current from nearby lightning vaporizes the water. The sudden converse of water to water vapor causes a dramatic expansion that can rip apart shoes. Water has density 1000 kg/m^3 and requires 2256 kJ/kg to be vaporized. If horizontal current lasts 2.00 ms and enounters water with resistivity 150 ohms meter, length 12.0 cm, and vertical cross-sectional area 15x10^-5 m^2, what average current is required to vaporize the water?
To find the average current required to vaporize the water, we need to use Ohm's Law and the formula for power.
Let's break down the problem step by step:
Step 1: Calculate the resistance of the water.
The formula to calculate resistance is: R = (ρ * L) / A
where R is the resistance, ρ is the resistivity, L is the length, and A is the cross-sectional area.
Given:
Resistivity (ρ) = 150 ohm-m
Length (L) = 12.0 cm = 0.12 m
Cross-sectional Area (A) = 15 x 10^-5 m^2
R = (150 * 0.12) / (15 x 10^-5)
R = 1200 ohms
Step 2: Calculate the power dissipated in the water.
The power (P) dissipated can be calculated using the formula: P = I^2 * R, where I is the current and R is the resistance.
Given:
P = 2256 kJ/kg = 2256 x 10^3 J/kg (convert from kJ to J)
Since we want to find the average current, we are assuming a mass of 1 kg of water. Therefore, the power (P) required to vaporize 1 kg of water is 2256 x 10^3 J.
P = I^2 * R
2256 x 10^3 = I^2 * 1200
Step 3: Solve for the current (I).
Rearrange the equation to solve for I:
I^2 = (2256 x 10^3) / 1200
I^2 = 1880
Take the square root of both sides:
I = √1880
I ≈ 43.39 A
Therefore, the average current required to vaporize the water is approximately 43.39 Amps.