Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

Answer to Problem 10P

The mass of uranium in reserve is $3.1\times {10}^{10}\text{g}$.

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70\%$.

The amount of uranium reserved is $4.4\times {10}^6$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m_{{235}_U}=\Delta pm$

• $m_{{235}_U}$ is the difference in time between the cars to reach the destination
• $\Delta p$ is the percentage of mass reserved m is the speed of the slower car
• m is the mass of uranium in reserve

Substitute $0.70\%$ for $\Delta p$ and $4.4\times {10}^6$ for m to find $m_{{235}_U}$.

$\begin{array}{c}{m}_{{235}_U}=\left(0.70\%\left(\frac{1}{100\%}\right)\right)\left(4.4\times {10}^6\text{metric}\text{ton}\right)\left(\frac{{10}^3\text{kg}}{1\text{ton}}\right)\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =3.1\times {10}^{10}\text{kg}\end{array}$

Thus, the mass of uranium in reserve is $3.1\times {10}^{10}\text{g}$.

Conclusion:

The mass of uranium in reserve is $3.1\times {10}^{10}\text{g}$.

(b)

To determine

The number of moles and atoms in reserve.

Answer to Problem 10P

The number of moles in reserve is $1.3\times {10}^8\text{mol}$ number of atoms in reserve is $\text{7}\text{.8}\times {\text{10}}^{31}\text{atoms}$.

Explanation of Solution

Given Info:

The mass per one mole is $235\text{g}/\text{mol}$.

The Avogadro number is $6.02\times {10}^{23}\text{atoms}/\text{mol}$.

Formula to calculate the number of moles is,

$n=\frac{m_{{235}_U}}{M_{mol}}$

• $n$ is the number of moles
• $M_{mol}$ is mass per mol

Substitute $235\text{g}/\text{mol}$ for $M_{mol}$ and $3.1\times {10}^{10}\text{g}$ for $m_{{235}_U}$ to find n.

$\begin{array}{c}n=\frac{3.1\times {10}^{10}\text{g}}{235\text{g}/\text{mol}}\\ =1.3\times {10}^8\text{mol}\end{array}$

Thus, the number of moles is $1.3\times {10}^8\text{mol}$.

Formula to calculate the number of atoms is,

$N=n{N}_A$

• $N$ is the number of moles
• $N_A$ is the Avogadro number

Substitute $1.3\times {10}^8\text{mol}$ for n and $1.3\times {10}^8\text{mol}$ for $N_A$ to find N.

$\begin{array}{c}\text{N}=\left(1.3\times {10}^8\text{mol}\right)\left(6.02\times {10}^{23}\text{atoms}/\text{mol}\right)\\ =7.8\times {10}^{31}\text{atoms}\end{array}$

Thus, the number of atoms is $\text{7}\text{.8}\times {\text{10}}^{31}\text{atoms}$.

Conclusion:

The number of moles in reserve is $1.3\times {10}^8\text{mol}$ number of atoms in reserve is $\text{7}\text{.8}\times {\text{10}}^{31}\text{atoms}$.

(c)

To determine

The total energy available.

Answer to Problem 10P

The total energy available is $2.6\times {10}^{21}\text{J}$.

Explanation of Solution

Given Info:

The energy per atom is $208\text{MeV}/\text{atom}$.

Formula to calculate the total energy available is,

$E=N{E}_0$

• $E$ is the energy
• $E_0$ is the energy per atom

Substitute $208\text{MeV}/\text{atom}$ for $E_0$ and $7.87\times {10}^{31}\text{atom}$ for N to find $E$.

$\begin{array}{c}E=\left(208\text{MeV}/\text{atom}\right)\left(7.87\times {10}^{31}\text{atom}\right)\left(\frac{1.60\times {10}^{-13}\text{J}}{1\text{MeV}}\right)\\ =2.6\times {10}^{21}\text{J}\end{array}$

Thus, the total energy available is $2.6\times {10}^{21}\text{J}$.

Conclusion:

The total energy available is $2.6\times {10}^{21}\text{J}$.

(d)

To determine

The maximum time energy supply lasts.

Answer to Problem 10P

The maximum time energy supply lasts is $5.5\text{yr}$.

Explanation of Solution

Given Info:

The energy consumption rate is $1.5\times {10}^{13}\text{J}/\text{s}$.

Formula to calculate maximum time energy supply lasts is,

$t=\frac{E}{P}$

• $t$ is maximum time energy supply lasts
• $P$ is the energy consumption rate

Substitute $2.6\times {10}^{21}\text{J}$ for $E$ and $1.5\times {10}^{13}\text{J}/\text{s}$ for $P$ to find $t$.

$\begin{array}{c}t=\left(\frac{2.6\times {10}^{21}\text{J}}{1.5\times {10}^{13}\text{J}/\text{s}}\right)\left(\frac{1\text{yr}}{3.156\times {10}^7\text{s}}\right)\\ =5.5\text{yr}\end{array}$

Thus, the maximum time energy supply lasts is $5.5\text{yr}$.

Conclusion:

The maximum time energy supply lasts is $5.5\text{yr}$

(e)

To determine

The conclusion of the results.

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.