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Chapter 19: Problem 11

During a lightning flash, there exists a potential difference of \(V_{\text { clond }}-V_{\text { ground }}=1.2 \times 10^{9} \mathrm{V}\) between a cloud and the ground. As a result, a charge of \(-25 \mathrm{C}\) is transferred from the ground to the cloud. (a) How much work \(W_{\text { ground-cloud }}\) is done on the charge by the electric force? \((b)\) If the work done by the electric force were used to acceler- ate a \(1100-\mathrm{kg}\) automobile from rest, what would be its final speed? (c) If the work done by the electric force were converted into heat how many kilograms of water at \(0^{\circ} \mathrm{C}\) could be heated to \(100^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) \(-3.0 \times 10^{10} \text{ J}\); (b) \(7.39 \times 10^3 \text{ m/s}\); (c) \(71428 \text{ kg}\).

Step by step solution

01

Calculate Work Done by Electric Force

The work done by the electric force is calculated using the formula: \[ W = q \cdot \Delta V \]where \( q \) is the charge and \( \Delta V \) is the potential difference. Here, \( q = -25 \text{ C} \) and \( \Delta V = 1.2 \times 10^9 \text{ V} \). Plug these values in:\[ W = (-25) \times (1.2 \times 10^9) \text{ J} \]\[ W = -3.0 \times 10^{10} \text{ J} \]
02

Determine Final Speed of Automobile

The work-energy principle states that the work done is equal to the kinetic energy gained, \( \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity. Using the formula:\[ W = \frac{1}{2} m v^2 \]Rearrange it for velocity \( v \):\[ v = \sqrt{\frac{2W}{m}} \]Substitute \( W = 3.0 \times 10^{10} \text{ J} \) and \( m = 1100 \text{ kg} \):\[ v = \sqrt{\frac{2 \times 3.0 \times 10^{10}}{1100}} \]\[ v = 7.39 \times 10^3 \text{ m/s} \]
03

Calculate the Heat Capacity Needed

First, calculate the energy needed to heat 1 kg of water from \(0^{\circ} \text{C}\) to \(100^{\circ} \text{C}\) using the formula \[ Q = m c \Delta T \]where \( c \) is the specific heat capacity of water (\(4200 \text{ J/kg}^\circ \text{C} \)), \( m \) is the mass, and \( \Delta T = 100^\circ \text{C} \). For 1 kg:\[ Q = 1 \times 4200 \times 100 = 420000 \text{ J} \]Now calculate the number of kilograms, \( M \), that can be heated using the energy from the electrical work. Let \( M \) be the mass heated, hence:\[ 3.0 \times 10^{10} = M \times 420000 \]\[ M = \frac{3.0 \times 10^{10}}{420000} \]\[ M \approx 71428 \text{ kg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Difference
Potential difference, often expressed as voltage, is the measure of the electrical potential energy difference between two points in a circuit. In the context of a lightning flash, it denotes the energy difference between the cloud and the ground.
Imagine it as the pushing force in an electric circuit that moves charges from one place to another.
  • Voltage (V): Represents the work needed per unit charge to move a charge between two points.
  • Charge (q): The amount of electricity held by an object, measured in coulombs (C).
In this exercise, the potential difference (\(V_{ ext{cloud}} - V_{ ext{ground}}\)) is \(1.2 \times 10^9 \text{ V}\), which informs us of the significant energy involved in the lightning process. Together with the charge, it enables us to calculate the work done by the electric force.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the electric force does work on an object, the energy can be transformed into kinetic energy.
For an automobile, which is initially at rest, the work done by an electric force can be utilized to accelerate it and give it a certain speed.
  • Kinetic Energy Equation: \( KE = \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity.
  • Work-Energy Principle: The total work done on an object is equal to the change in its kinetic energy.
For the 1100 kg automobile mentioned, the energy from the charged lightning flash can be converted into kinetic energy, allowing it to reach a speed inexcess of \( 7.39 \times 10^3 \text{ m/s} \). This showcases the immense power of electrical work in motion applications.
Specific Heat Capacity
The specific heat capacity of a substance is a measure that tells us how much energy it takes to change the temperature of a unit mass of the substance by one degree Celsius. For water, this value is quite high, indicating that water can absorb a lot of heat without large changes in temperature.
  • Units: Joules per kilogram per degree Celsius (\( \text{J/kg}^\circ \text{C} \)).
  • Water's specific heat capacity: \(4200 \text{ J/kg}^\circ \text{C} \).
This concept is fundamental in calculations concerning heat transfer, such as how much water can be heated using a given amount of energy. If we have energy from an electric force, we can apply this concept to determine the mass of water that can be heated from 0掳C to 100掳C.
Water Heating
Water heating involves transferring energy to water, raising its temperature. It鈥檚 a common practical application of energy conversions in physics. In the context of electrical work, converting the energy from an electric force into heat provides a tangible example of energy transformation.
The energy needed to heat water depends on several factors:
  • Mass of Water (m): More mass requires more energy to heat.
  • Specific Heat Capacity (c): For water, it's 4200 J/kg掳C.
  • Temperature Change (螖T): The required energy is proportional to how much you'd like the temperature to rise.
Using the formula \( Q = mc\Delta T \), we can calculate the energy needed or the mass of water that could be heated using a specific amount of energy, such as 3.0 \( \times \) 10^{10} J from the electric work in this exercise. This gives a practical representation of energy utilization from natural phenomena like lightning.

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Most popular questions from this chapter

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed \(v_{\mathrm{e}},\) while the proton acquires a speed \(v_{\mathrm{p}} .\) Find the ratio \(v_{\mathrm{c}} / v_{\mathrm{p}}\) .

The membrane that surrounds a certain type of living cell has a surface area of \(5.0 \times 10^{-9} \mathrm{m}^{2}\) and a thickness of \(1.0 \times 10^{-8} \mathrm{m} .\) Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of \(5.0 .\) (a) The potential on the outer surface of the membrane is \(+60.0 \mathrm{mV}\) greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to positive ions (charge \(+e ),\) how many such ions are present on the outer surface?

The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an \(850-\mu \mathrm{F}\) capacitor is 280 \(\mathrm{V}\) . (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for \(3.9 \times 10^{-3} \mathrm{s},\) find the effective power or "wattage" of the flash.

A spark plug in an automobile engine consists of two metal conductors that are separated by a distance of 0.75 \(\mathrm{mm}\) . When an electric spark jumps between them, the magnitude of the electric field is \(4.7 \times 10^{7} \mathrm{V} / \mathrm{m}\) . What is the magnitude of the potential difference \(\Delta V\) between the conductors?

The electric potential energy stored in the capacitor of a defibrillator is \(73 \mathrm{J},\) and the capacitance is 120\(\mu \mathrm{F}\) . What is the potential difference that exists across the capacitor plates?
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