Problem 48 - Fundamentals of Physics

Question

The rain-soaked shoes of a person may explode if ground current from nearby lightning vaporizes the water. The sudden conversion of water to water vapor causes a dramatic expansion that can rip apart shoes. Water has density \(1000 \mathrm{~kg} / \mathrm{m}^{3}\) and requires \(2256 \mathrm{~kJ} / \mathrm{kg}\) to be vaporized. If horizontal current lasts \(2.00 \mathrm{~ms}\) and encounters water with resistivity \(150 \Omega \cdot \mathrm{m},\) length \(12.0 \mathrm{~cm},\) and vertical cross-sectional area \(15 \times 10^{-5} \mathrm{~m}^{2},\) what average current is required to vaporize the water?

Step-by-Step Solution

Answer

The average current required is approximately 130 A.

Step 1: Calculate the Volume of Water

First, calculate the volume of water present in the shoes. We use the formula for volume, which is the length multiplied by the area of cross-section:\[V = L imes A = 0.12 \, \text{m} \times 15 \times 10^{-5} \, \text{m}^2 = 1.8 \times 10^{-5} \, \text{m}^3\]

Step 2: Calculate the Mass of Water

Next, calculate the mass of the water. The mass is given by the density of water multiplied by its volume:\[m = \rho \times V = 1000 \, \text{kg/m}^3 \times 1.8 \times 10^{-5} \, \text{m}^3 = 0.018 \, \text{kg}\]

Step 3: Calculate Energy Required to Vaporize Water

The energy required to vaporize the water is calculated using the mass of water and its vaporization energy:\[E = m \times \text{Latent Heat} = 0.018 \, \text{kg} \times 2256 \, \text{kJ/kg} = 40.608 \, \text{kJ} = 40,608 \, \text{J}\]

Step 4: Calculate Resistance of the Water

Using the resistivity formula, calculate the resistance of the water. Resistance is given by:\[R = \rho \times \left(\frac{L}{A}\right) = 150 \, \Omega \cdot \text{m} \times \left(\frac{0.12 \, \text{m}}{15 \times 10^{-5} \, \text{m}^2} \right) = 1200 \, \Omega\]

Step 5: Relate Energy to Current using Joule's Law

From Joule's law, we have that the energy expended is given by:\[E = I^2 \times R \times t\]Substituting the known values and solving for the current \(I\):\[40,608 \, \text{J} = I^2 \times 1200 \, \Omega \times 2 \times 10^{-3} \, \text{s}\]\[40,608 = I^2 \times 2.4\]

Step 6: Solve for Average Current

Finally, solve for the current \(I\):\[I^2 = \frac{40,608}{2.4} \approx 16,920\]\[I \approx \sqrt{16,920} \approx 130 \, \text{A}\]

Key Concepts

Electric Current

Electric current is a fundamental concept in physics, especially in the study of electromagnetism. It refers to the flow of electric charge, generally through a conductor like a wire. This flow of electrons is what powers our world, from lighting our homes to charging our phones.

In this exercise, we're dealing with a specific scenario where current flows through water soaked in shoes. When lightning strikes, it can cause massive amounts of current to pass through a wet surface. It’s important to understand how this current behaves and is measured. The unit of current is the ampere (A), and in the exercise, we calculate an average current of about 130 A. That’s a lot of electricity and is why such events can be so dramatic.

Understanding how current flows and its magnitude helps in solving real-world physics problems involving electrical safety and energy transfer.

Volume Calculation

Calculating the volume is a critical step in many physics problems. Volume refers to the amount of space that an object or substance occupies. Here, the internal volume of the water in the shoes is essential for determining how much energy is required to vaporize it.

In our exercise, we calculate the volume of water using the formula:

• Volume ( V ) = Length ( L ) × Cross-sectional Area ( A )

Here, the length is 12.0 cm (converted to meters for SI unit consistency), and the area is given. The volume then comes out to be 1.8 × 10^-5 m³.

This step allows us to find out how much water we're working with, which is crucial for further calculations like determining mass and energy needs.

Energy Conversion

Energy conversion is all about changing one form of energy into another. In physics, this is a key principle that describes how energy changes form without being created or destroyed.

In this problem, we want to convert the ground current’s electrical energy into enough heat to vaporize water in the shoes. This requires a clear understanding of the energy needed to make this phase change happen.

The exercise reveals the energy needed as 40,608 J, based on the mass of water and the latent heat of vaporization. This latent heat is the amount of energy per kilogram needed to turn liquid into vapor, showcasing a direct conversion of electrical energy into thermal energy.

Resistivity

Resistivity is a material property that indicates how strongly a material opposes the flow of electric current. It is measured in ohm meters (Ω·m). High resistivity means that a material does not allow electricity to flow through it easily, whereas low resistivity does.

In this scenario, water with a resistivity of 150 Ω·m is acting as the medium through which the current flows. Resistivity helps in calculating the resistance, which is vital in understanding how much voltage is needed for a certain current to pass through a given medium.

Using the resistivity formula:

• Resistance ( R ) = Resistivity ( ρ ) × ( L / A )

we calculate the water's resistance to be 1200 Ω. This important step aids in determining the current needed using the principles of Joule's Law.

Joule's Law

Joule's Law is significant in understanding how electrical energy is converted into heat. It states that the energy ( E ) emitted by a current through a resistance ( R ) for a specified time ( t ) is equal to the current squared multiplied by the resistance and time.

In mathematical terms, Joule's Law is expressed as:

• E = I^2 × R × t

In our example, we utilize this formula to find the necessary current to vaporize water. By reorganizing this equation, we solve for current in terms of known quantities such as resistance, time, and energy.

This calculation yields an approximate average current of 130 A, highlighting that Joule's Law provides a straightforward method for relating energy, current, resistance, and time in practical applications.