Chapter 30, Problem 12P

(a)

To determine

The amount of uranium dissolved in the ocean.

Answer to Problem 12P

The amount of uranium dissolved in the ocean is $4\times 0^{15}\text{g}$.

Explanation of Solution

Given Info:

The density is $3\times {10}^{-3}{\text{g/m}}^3$.

The deapth is $4\times {10}^3\text{m}$.

The radius is $6.38\times {10}^6\text{m}$.

Formula to calculate the amount of uranium dissolved in the ocean is,

$\begin{array}{c}{m}_U=cV\\ c\left(\frac{2}{3}A\right)h_{av}\\ =c\left(\frac{2}{3}\left(4\pi R_E^2\right)\right)h_{av}\\ =\frac{8c{h}_{av}\pi R_E^2}{3}\end{array}$

• $m_U$ is the amount of uranium dissolved
• $c$ is the density
• $V$ is the volume
• $A$ is the area
• $h_{av}$ is the depth
• $R_E$ is the radius of Earth

Substitute $3\times {10}^{-3}{\text{g/m}}^3$ for $c$, $4\times {10}^3\text{m}$ for $h_{av}$ and $6.38\times {10}^6\text{m}$ for  $R_E$ to find $m_U$.

$\begin{array}{c}{m}_U=\frac{8\left(3\times {10}^{-3}{\text{g/m}}^3\right)\left(4\times {10}^3\text{m}\right)\left(3.14\text{rad}\right){\left(6.38\times {10}^6\text{m}\right)}^2}{3}\\ =4\times {10}^{15}\text{g}\end{array}$

Thus, the amount of uranium dissolved in the ocean is $4\times 0^{15}\text{g}$.

Conclusion:

The amount of uranium dissolved in the ocean is $4\times 0^{15}\text{g}$.

(b)

To determine

The time the uranium lasts.

Answer to Problem 12P

The time the uranium lasts is $5\times {10}^3\text{year}$.

Explanation of Solution

Given Info:

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

The percentage of isotope present is $0.70\%$

The energy per event $200\text{MeV/atoms}$

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

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

Formula to calculate the amount of uranium isotope is,

$m_{{}^{235}U}=\Delta p{m}_U$

• $m_{{}^{235}U}$ is the mass of isotope
• $\Delta p$ is the percentage of isotope

Formula to calculate the number of atoms is,

$\begin{array}{c}N=\frac{N_A}{M_{mol}}{m}_{{}^{235}U}\\ =\frac{N_A}{M_{mol}}\Delta p{m}_U\end{array}$

• $N_A$ is the Avogadro number
• $M_{mol}$ is the mass per mol
• $N$ is the number of atoms

Formula to calculate the energy released is,

$\begin{array}{c}E=N{E}_0\\ =\frac{N_A}{M_{mol}}\Delta p{m}_U{E}_0\end{array}$

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

Formula to calculate the time is,

$\begin{array}{c}t=\frac{E}{P}\\ =\frac{\frac{N_A}{M_{mol}}\Delta p{m}_U{E}_0}{P}\\ =\frac{N_A}{P{M}_{mol}}\Delta p{m}_U{E}_0\end{array}$

• $t$ is the energy released
• $P$ is the energy released per atom

Substitute $6.02\times {10}^{23}\text{atoms/mol}$ for $N_A$, $0.70\%$ for $\Delta p$, $200\text{MeV/atoms}$ for $E$, $1.5\times {10}^{13}\text{J/s}$ for $P$, $4\times {10}^{15}\text{g}$ for $m_U$ and $235\text{g/mol}$ for  $M_{mol}$ to find $t$.

$\begin{array}{c}t=\frac{\left(6.02\times {10}^{23}\text{atoms/mol}\right)}{\left(1.5\times {10}^{13}\text{J/s}\right)\left(235\text{g/mol}\right)}\left(0.70\%\right)\left(4\times {10}^{15}\text{g}\right)\left(200\text{MeV/atoms}\right)\\ =5\times {10}^3\text{year}\end{array}$

Thus, the time the uranium lasts is $5\times {10}^3\text{year}$.

Conclusion:

The time the uranium lasts is $5\times {10}^3\text{year}$.

(c)

To determine

The source of uranium and the possibility of renewal of the energy source.

Explanation of Solution

The uranium comes from natural resources as rocks and minerals.

The uranium comes from dissolving rock and minerals. Rivers carry such solutes into the oceans, however, the Earth’s supply of uranium is not renewable. However, if breeder reactors are used, the current ocean supply can last about a half- million years.

Conclusion:

The sources of uranium are natural sources as rocks and minerals which are not renewable.