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Chapter 19: Problem 61

For how many years could all the energy needs of the world be supplied by the fission of \(\mathrm{U}-235\) ? Use the following assumptions: The world has about \(1.0 \times 10^{7}\) metric tons of uranium ore, which are about 0.75\% U-235. The energy consumption of the world is about \(4.0 \times 10^{15} \mathrm{~kJ} / \mathrm{y}\) and does not change with time. The fission of U-235 releases about \(8.0 \times 10^{7} \mathrm{~kJ} / \mathrm{g}\) of \(\mathrm{U}-235\).

Short Answer

Expert verified
Answer: The fission of U-235 could supply all the energy needs of the world for about 1,500 years.

Step by step solution

01

Calculate the total amount of U-235 in the world

We are given that the world has about \(1.0 \times 10^{7}\) metric tons of uranium ore, and 0.75\% of it is U-235. To find the amount of U-235, we can multiply the total amount of uranium ore by the proportion of U-235: U-235 = (Total uranium ore) 脳 (Proportion of U-235) U-235 = (\(1.0 \times 10^{7}\) metric tons) 脳 (0.0075) U-235 = \(7.5 \times 10^{4}\) metric tons
02

Calculate the total energy that can be generated from the fission of U-235

We are given that the fission of U-235 releases about \(8.0 \times 10^{7} \mathrm{~kJ} / \mathrm{g}\) of U-235. First, we need to convert the amount of U-235 from metric tons to grams: \((7.5 \times 10^{4} \mathrm{~metric~tons}) \times (10^{6} \mathrm{~g/metric~ton}) = 7.5 \times 10^{10} \mathrm{~g}\) Now, we can calculate the total energy generated from the fission of all available U-235: Total energy = (Amount of U-235) 脳 (Energy released per gram of U-235) Total energy = (\(7.5 \times 10^{10} \mathrm{~g}\)) 脳 (\(8.0 \times 10^{7} \mathrm{~kJ/g}\)) Total energy = \(6.0 \times 10^{18} \mathrm{~kJ}\)
03

Calculate the number of years the world's energy needs can be met by U-235 fission

We are given that the world's energy consumption is about \(4.0 \times 10^{15} \mathrm{~kJ/y}\). To find out for how many years the energy needs can be met, divide the total energy generated from the fission of U-235 by the world's energy consumption per year: Years = (Total energy) / (Energy consumption per year) Years = (\(6.0 \times 10^{18} \mathrm{~kJ}\)) / (\(4.0 \times 10^{15} \mathrm{~kJ/y}\)) Years = 1.5 \(\times 10^{3}\) years The fission of U-235 could supply all the energy needs of the world for about 1,500 years.

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Key Concepts

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

Uranium Ore
Uranium ore is a naturally occurring rock that contains uranium minerals, which are the source of nuclear energy. Uranium is the heaviest naturally occurring element used as fuel in nuclear reactors. In its raw state, uranium ore consists of minerals like uraninite, which contain uranium along other elements.
To extract uranium for use in reactors, the ore undergoes a refining process to enrich the uranium content, specifically increasing the concentration of uranium-235.
  • Uranium ore is mined from the earth and then crushed to prepare it for extraction.
  • It contains a very small percentage of uranium-235, usually about 0.7% to 0.75%, which is the isotope used in nuclear fission.
Understanding uranium ore is essential because it serves as the starting material for generating nuclear energy.
Energy Consumption
Energy consumption refers to the amount of energy used by human activities, and it varies across different regions and industries. The global demand for energy is increasing due to population growth and industrial development.
The exercise mentions that the world's energy consumption is approximately \(4.0 \times 10^{15} \text{ kJ/year} \), a figure representing the total annual energy needs of the planet. This encompasses all forms of energy used in transportation, manufacturing, heating, and electricity production.
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  • To supply all energy needs, efficient and sustainable energy sources are critical.
  • Nuclear energy, as shown, can greatly contribute by providing a high energy yield per unit of fuel.
    • Monitoring and optimizing energy consumption is key to addressing both economic and environmental challenges.
    Nuclear Energy
    Nuclear energy is a powerful source of energy generated through the process of nuclear fission. This involves splitting the nucleus of an atom, such as uranium-235, which releases a tremendous amount of energy.
    The given exercise highlights nuclear energy's potential to meet global energy demands due to its high output. One gram of uranium-235 can release around \(8.0 \times 10^{7} \text{ kJ}\), illustrating its efficiency compared to fossil fuels.
    • Nuclear power plants use controlled nuclear reactions to produce electricity.
    • They provide a continuous and reliable supply of energy with low greenhouse gas emissions.
    Despite its advantages, it鈥檚 important to address nuclear waste management and potential risks associated with nuclear energy to ensure its safe utilization.
    Nuclear Reactions
    Nuclear reactions encompass processes that alter the composition of an atom's nucleus, and they are at the heart of nuclear energy production. One of the most significant types of reactions for energy production is nuclear fission.
    In the fission of uranium-235, a neutron collides with the uranium nucleus, causing it to split into smaller atoms while releasing energy, additional neutrons, and radiation. This chain reaction is controlled in nuclear reactors to produce energy safely.
    • Fission reactions produce a large amount of energy from a small amount of fuel.
    • The validity of nuclear energy as a long-term solution encompasses both efficient fuel use and managing the byproducts.
    Understanding nuclear reactions is crucial for developing and improving technologies that harness nuclear power effectively and safely.

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    Most popular questions from this chapter

    Consider the reaction $$2{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}$$ (a) Calculate \(\Delta E\) in kilojoules per gram of deuterium fused. (b) How much energy is potentially available from the fusion of all the deuterium in seawater? The percentage of deuterium in water is about \(0.0017 \%\). The total mass of water in the oceans is \(1.3 \times 10^{24} \mathrm{~g}\). (c) What fraction of the deuterium in the oceans would have to be consumed to supply the annual energy requirements of the world \(\left(2.3 \times 10^{17} \mathrm{~kJ}\right) ?\)

    It is possible to estimate the activation energy for fusion by calculating the energy required to bring two deuterons close enough to one another to form an alpha particle. This energy can be obtained by using Coulomb's law in the form \(E=8.99 \times 10^{9} q_{1} q_{2} / r\), where \(q_{1}\) and \(q_{2}\) are the charges of the deuterons \(\left(1.60 \times 10^{-19} \mathrm{C}\right), r\) is the radius of the He nucleus, about \(2 \times 10^{-15} \mathrm{~m}\), and \(E\) is the energy in joules. (a) Estimate \(E\) in joules per alpha particle. (b) Using the equation \(E=m v^{2} / 2\), estimate the velocity (meters per second) each deuteron must have if a collision between the two of them is to supply the activation energy for fusion \((m\) is the mass of the deuteron in kilograms).

    A 35-mL sample of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(35 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) NaI labeled with I-131. The following reaction occurs. $$\mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{AgI}(s)$$ The filtrate is found to have an activity of \(2.50 \times 10^{3}\) counts per minute per milliliter. The \(0.050 \mathrm{M}\) NaI solution had an activity of \(1.25 \times 10^{10}\) counts per minute per milliliter. Calculate \(K_{\text {sp }}\) for AgI.

    Write balanced nuclear equations for the bombardment of (a) U-238 with a nucleus to produce Fm-249 and five neutrons. (b) Al-26 with an alpha particle to produce P-30. (c) Cu-63 with a nucleus producing \(\mathrm{Zn}-63\) and a neutron. (d) Al-27 with deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) to produce an alpha particle and another nucleus.

    Cobalt-60 is used extensively in medicine as a source of \(\gamma\) -rays. Its half-life is \(5.27\) years. (a) How long will it take a \(\mathrm{Co}-60\) source to decrease to \(18 \%\) of its original activity? (b) What percent of its activity remains after 29 months?
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