Problem 40
Question
If \(210 \mathrm\space{kg}\) of hydrogen could be entirely converted to energy, how many joules would be produced?
Step-by-Step Solution
Verified Answer
The energy produced would be \(1.89 \, \times \, 10^{19} \, \text{J}\).
1Step 1: Understanding the Problem
We need to calculate the energy produced from 210 kg of hydrogen when converted entirely into energy. This requires understanding and using Einstein's mass-energy equivalence formula.
2Step 2: Recall Einstein's Mass-Energy Equivalence
Einstein's mass-energy equivalence is given by the formula: \[ E = mc^2 \] where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light in a vacuum, approximately \(3 \, \times \, 10^8 \, \text{m/s}\).
3Step 3: Substitute the Values
Substitute the mass of hydrogen into the equation using \(m = 210 \, \text{kg}\), and substitute \(c = 3 \, \times \, 10^8 \, \text{m/s}\):\[ E = 210 \, \text{kg} \, \times \, (3 \, \times \, 10^8 \, \text{m/s})^2 \].
4Step 4: Calculate the Energy
First, calculate \(c^2 = (3 \, \times \, 10^8 \, \text{m/s})^2 = 9 \, \times \, 10^{16} \, \text{m}^2/\text{s}^2\). Then, use this to find the energy:\[ E = 210 \, \text{kg} \, \times \, 9 \, \times \, 10^{16} \, \text{m}^2/\text{s}^2 \]\[ E = 1.89 \, \times \, 10^{19} \, \text{J} \].
Key Concepts
Einstein's Theory of RelativityConversion of Mass to EnergyHydrogen Energy ConversionSpeed of Light
Einstein's Theory of Relativity
Einstein's Theory of Relativity revolutionized our understanding of the universe. It consists of the special and general theory of relativity. The special theory of relativity, introduced in 1905, includes the groundbreaking equation \(E = mc^2\). This equation implies that mass and energy are interchangeable.
The general theory, published in 1915, further explores gravity as a curvature of space-time, but it is the special theory that focuses on concepts such as how the laws of physics are the same for all non-accelerating observers. These theories help explain how energy and mass can be converted when objects move at speeds close to the speed of light.
The general theory, published in 1915, further explores gravity as a curvature of space-time, but it is the special theory that focuses on concepts such as how the laws of physics are the same for all non-accelerating observers. These theories help explain how energy and mass can be converted when objects move at speeds close to the speed of light.
Conversion of Mass to Energy
The conversion of mass to energy is a core concept in physics. Essentially, it describes how mass can be transformed into energy and vice versa. For this transformation, Einstein's mass-energy equivalence formula \(E = mc^2\) is often used.
The formula indicates that even a small amount of mass can produce a large amount of energy due to the speed of light squared \((c^2)\) being immensely large. This concept is foundational in understanding nuclear reactions, where mass is changed into energy, such as in the sun's core, enabling hydrogen atoms to fuse into helium, providing the sun's energy.
The formula indicates that even a small amount of mass can produce a large amount of energy due to the speed of light squared \((c^2)\) being immensely large. This concept is foundational in understanding nuclear reactions, where mass is changed into energy, such as in the sun's core, enabling hydrogen atoms to fuse into helium, providing the sun's energy.
Hydrogen Energy Conversion
Hydrogen is the most abundant element in the universe and a potential source of vast energy. Through a process known as nuclear fusion, hydrogen atoms can undergo conversion to form helium, releasing significant amounts of energy.
This process is similar to what occurs in the sun, where hydrogen nuclei combine under high pressure and temperatures. The energy released is mainly from the mass difference in this conversion, which is transformed into energy per the \(E = mc^2\) principle.
This process is similar to what occurs in the sun, where hydrogen nuclei combine under high pressure and temperatures. The energy released is mainly from the mass difference in this conversion, which is transformed into energy per the \(E = mc^2\) principle.
- Hydrogen fusion is a clean energy source, producing minimal waste products.
- The potential for hydrogen to be a widespread energy source on Earth is significant, though creating the necessary conditions for fusion reactions is technologically challenging.
Speed of Light
The speed of light, denoted as \(c\), is approximately \(3 \times 10^8 \text{ m/s}\). It is a constant in physics, crucial for calculations involving mass-energy equivalence.
The speed of light in a vacuum is one of the central postulates of Einstein's theory of relativity. It implies that nothing can travel faster than light, affecting how we perceive time and space.
In Einstein's mass-energy formula \(E = mc^2\), the speed of light's value illustrates why a small amount of mass can be converted into a large amount of energy. Because \(c\) is so large, the energy \(E\) derived from even a tiny mass \(m\) becomes substantial. This concept shapes modern physics and our understanding of the universe.
The speed of light in a vacuum is one of the central postulates of Einstein's theory of relativity. It implies that nothing can travel faster than light, affecting how we perceive time and space.
In Einstein's mass-energy formula \(E = mc^2\), the speed of light's value illustrates why a small amount of mass can be converted into a large amount of energy. Because \(c\) is so large, the energy \(E\) derived from even a tiny mass \(m\) becomes substantial. This concept shapes modern physics and our understanding of the universe.
Other exercises in this chapter
Problem 35
What other phenomena are like the Sun's energy in that they are lifegiving and also potentially harmful and destructive?
View solution Problem 38
How long, in years, would it take a passenger jet to arrive at the Sun's photosphere, assuming it travels at \(800 \mathrm{km} / \mathrm{h}\) ?
View solution Problem 44
The temperature of a sunspot is 0.66 as high as the surrounding photosphere. What is the ratio of its brightness to that of an equal-sized area around it?
View solution Problem 33
What is helioseismology, and what useful information does it yield?
View solution