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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) × (Proportion of U-235) U-235 = (\(1.0 \times 10^{7}\) metric tons) × (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) × (Energy released per gram of U-235) Total energy = (\(7.5 \times 10^{10} \mathrm{~g}\)) × (\(8.0 \times 10^{7} \mathrm{~kJ/g}\)) 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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Most popular questions from this chapter

When a positron and an electron collide, they annihilate each other and produce two gamma photons, which carry the same amount of energy. What is the wavelength (in nanometers) of these photons?

Consider the fusion of boron- 10 with an alpha particle. The products of the fusion are carbon-13 and a proton. (a) Write a nuclear reaction for this process. (b) How much energy is released when \(1.00 \mathrm{~g}\) of \(\mathrm{B}-10\) is fused with an \(\alpha\) -particle?

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