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The earth receives \(1.8 \times 10^{14} \mathrm{kJ} / \mathrm{s}\) of solar energy. What mass of solar material is converted to energy over a 24 -h period to provide the daily amount of solar energy to the earth? What mass of coal would have to be burned to provide the same amount of energy? (Coal releases \(32 \mathrm{kJ}\) of energy per gram when burned.)

Short Answer

Expert verified
The mass of solar material converted to energy over a 24-hour period to provide the daily amount of solar energy to Earth is approximately \(1.727 \times 10^9 \mathrm{kg}\). To provide the same amount of energy, approximately \(4.860 \times 10^{14} \mathrm{kg}\) of coal would have to be burned.

Step by step solution

01

Calculate the total solar energy received by Earth over a 24-hour period.

To find the total energy received, we'll multiply the energy received per second by the number of seconds in a day. Total solar energy = Energy received per second × Number of seconds in a day Total solar energy = \(1.8 \times 10^{14} \mathrm{kJ/s} \times 24 \times 60 \times 60 \mathrm{s}\) Total solar energy = \(1.8 \times 10^{14} \mathrm{kJ/s} \times 86,400 \mathrm{s}\) Total solar energy = \(1.5552 \times 10^{19} \mathrm{kJ}\)
02

Find the mass of solar material converted to energy using Einstein's mass-energy equivalence formula.

Einstein's mass-energy equivalence formula is given by: \(E = mc^2\) Where: E is the energy in joules (we'll convert from kJ to J), m is the mass of the solar material in kilograms, c is the speed of light in m/s, which is approximately 3 × 10^8 m/s. First, let's convert the energy from kJ to J: \(1.5552 \times 10^{19} \mathrm{kJ} = 1.5552 \times 10^{22} \mathrm{J}\) Now, we'll rearrange the formula to find the mass, m: \(m = \frac{E}{c^2}\) Insert the values: \(m = \frac{1.5552 \times 10^{22} \mathrm{J}}{(3 \times 10^8 \mathrm{m/s})^2}\) \(m \approx 1.727 \times 10^9 \mathrm{kg}\) So, the mass of solar material converted to energy over a 24-hour period is approximately \(1.727 \times 10^9 \mathrm{kg}\).
03

Calculate the amount of coal required to produce the same amount of energy.

We are given that coal releases 32 kJ of energy per gram when burned, so we can find the mass of coal needed to produce the same amount of energy as follows: Mass of coal × Energy per gram of coal = Total solar energy in kJ Mass of coal = \(\frac{\mathrm{Total\,solar\,energy\,in\,kJ}}{\mathrm{Energy\,per\,gram\,of\,coal}}\) Mass of coal = \(\frac{1.5552 \times 10^{19} \mathrm{kJ}}{32 \mathrm{kJ/g}}\) Mass of coal = \(4.860 \times 10^{17} \mathrm{g}\) Since 1 kg = 1000 g, we can convert the mass of coal to kg: Mass of coal = \(4.860 \times 10^{14} \mathrm{kg}\) To provide the same amount of energy as the Sun, approximately \(4.860 \times 10^{14} \mathrm{kg}\) of coal would have to be burned.

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Key Concepts

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

Mass-Energy Equivalence
The concept of mass-energy equivalence is one of the most groundbreaking ideas in physics, introduced by Albert Einstein in his theory of relativity. It tells us that mass and energy are interchangeable. The famous formula for this is: \[E = mc^2\] Here, \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light, which is an incredibly high number at approximately \(3 \times 10^8\) meters per second. This equation shows that a small amount of mass can be converted into a tremendous amount of energy, because the speed of light squared is a very large number. In the context of solar energy, this concept helps us understand how massive amounts of energy can be generated from the conversion of small amounts of solar mass. When we receive solar energy on Earth, it is the result of nuclear reactions happening in the sun, where mass is converted into energy very efficiently, as explained by the mass-energy equivalence.
Coal Combustion Energy
Coal combustion has been a traditional method for generating energy. When coal is burned, it releases energy due to the chemical reactions between coal's carbon content and oxygen in the air. This process releases heat energy which can be used to produce electricity or provide heat. In the provided exercise, we understand that burning coal releases 32 kJ of energy per gram. The calculation of how much coal is needed to match the solar energy received by Earth involves a simple formula: \[ \text{Mass of coal} = \frac{\text{Total solar energy}}{\text{Energy released per gram of coal}} \]Thus, to produce the same energy as the solar input, a staggering amount of coal, specifically \(4.860 \times 10^{14}\) kg, would be required. This highlights the scale of solar energy compared to traditional coal-based energy, and underscores why transitioning to solar power could be more sustainable in the long run.
Solar Material Conversion
Solar material conversion refers to the process by which solar matter, primarily hydrogen in the sun, is transformed into energy. This conversion occurs through nuclear fusion, where hydrogen atoms in the sun's core join under immense pressure and temperature to form helium, releasing vast amounts of energy. This energy eventually makes its way to Earth, providing the light and warmth that sustain life. The mass-energy equivalence principle applies here as it shows that a tiny amount of solar mass – approximately \(1.727 \times 10^9\) kg over 24 hours – is converted directly into the abundant energy that reaches our planet.Understanding this process not only gives us insights into how sustainable solar energy is received naturally, but it also inspires the development of solar technology on Earth, which aims to capture and convert this energy into usable forms. Technologies like solar panels are examples of how we harness this natural phenomenon for renewable energy purposes.

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Most popular questions from this chapter

Many elements have been synthesized by bombarding relatively heavy atoms with high-energy particles in particle accelerators. Complete the following nuclear equations, which have been used to synthesize elements. a. \(\quad+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow$$\quad$$+6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+$$\quad$$\rightarrow \frac{260}{105} D b+4 \frac{1}{6} n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \frac{257}{153} \mathrm{Lr}+\)__________

Radioactive copper- 64 decays with a half-life of 12.8 days. a. What is the value of \(k\) in \(s^{-1} ?\) b. A sample contains \(28.0 \space\mathrm{mg}\space^{64} \mathrm{Cu}\). How many decay events will be produced in the first second? Assume the atomic mass of \(^{64} \mathrm{Cu}\) is \(64.0 \mathrm{u}\). c. A chemist obtains a fresh sample of \(^{64} \mathrm{Cu}\) and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below \(25 \%\) of the initial measured value. How long does she have to do the experiment?

Breeder reactors are used to convert the nonfissionable nuclide \(\frac{238}{92} \mathrm{U}\) to a fissionable product. Neutron capture of the \(\frac{238}{92} \mathrm{U}\) is followed by two successive beta decays. What is the final fissionable product?

Americium-241 is widely used in smoke detectors. The radiation released by this element ionizes particles that are then detected by a charged-particle collector. The half-life of \(^{241} \mathrm{Am}\) is 433 years, and it decays by emitting \(\alpha\) particles. How many \(\alpha\) particles are emitted each second by a 5.00 -g sample of \(^{241} \mathrm{Am} ?\)

In addition to the process described in the text, a second process called the carbon-nitrogen cycle occurs in the sun:a. What is the catalyst in this process? b. What nucleons are intermediates? c. How much energy is released per mole of hydrogen nuclei in the overall reaction? (The atomic masses of \(i \mathrm{H}\) and 4 He are 1.00782 u and 4.00260 u, respectively.)

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