Chapter 45, Problem 27P

(a)

To determine

Find the amount of energy would be released when all the mass of deuterium in ocean are fused to ${}_2^4\text{He}$.

Answer to Problem 27P

The amount of energy that would be released when all the mass of deuterium in ocean are fused to ${}_2^4\text{He}$ is $2.53\times {10}^{31}\text{ J}$.

Explanation of Solution

The volume of the oceans is $317{\text{ million mi}}^3$ and $0.0300\%$ of all the hydrogen in the ocean is deuterium. The mass of hydrogen is $1.0079\text{ u}$, oxygen is $15.9994\text{ u}$ and the water molecule is $18.0152\text{ u}$. The water density of the ocean is ${10}^3\text{kg}/{\text{m}}^{\text{3}}$.

Write the formula for power

    $N=\frac{m}{M}$                                                                           (I)

Write the formula for energy released

  $Q=\left(\Delta m\right)c^2$                                                                     (II)

Write the formula for energy

    $E=NQ$                                                                            (III)

Write the formula for mass

    $m=\rho V$                                                                              (IV)

Where, E is the total energy converted, $N$ is the number of fissionable isotope, $\rho$ is density, $V$ is the volume, $Q$ is the energy released per reaction, $m$ is the mass, $M$ is the molar mass, $Q$ is the energy released, $\Delta m$ is the mass defect and $c$ is the speed of light (the value of $c^2$ in $\text{MeV}/\text{u}$ is $931.494\text{MeV}/\text{u}$).

Conclusion:

Converting miles into meter: Volume of the oceans if $V=317\times {10}^6{\text{ mi}}^3{\left(\frac{1609\text{ m}}{1\text{ mi}}\right)}^3=1.32\times {10}^8{\text{ m}}^3$.

Substitute ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $1.32\times {10}^8{\text{ m}}^3$ for $V$ in equation (IV) to find the value of $m_{water}$

$\begin{array}{c}{m}_{water}=\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\times 1.32\times {10}^8{\text{ m}}^3\\ =1.32\times {10}^{21}\text{ kg}\end{array}$

The mass of the water molecule in ocean is $1.32\times {10}^{21}\text{ kg}$.

To find the mass of hydrogen in ocean,

$\begin{array}{c}{m}_{H_2}=\left(\frac{M_{H_2}}{M_{H_{2}O}}\right)m_{H_{2}O}\\ \text{=}\left(\frac{2.016\text{ u}}{18.015\text{ u}}\right)1.32\times {10}^{21}\text{ kg}\\ \text{=1}\text{.48}\times {\text{10}}^{20}\text{ kg}\end{array}$

The mass of hydrogen in ocean is $\text{1}\text{.48}\times {\text{10}}^{20}\text{ kg}$.

As $0.0300\%$ of the mass of hydrogen in ocean is deuterium. The mass of deuterium in ocean is

$\begin{array}{c}{m}_D=\left(0.0300\%\right)m_{H_2}\\ =\left(0.0300\times {10}^{-2}\right)\times \text{1}\text{.48}\times {\text{10}}^{20}\text{ kg}\\ \text{=4}\text{.43}\times {\text{10}}^{16}\text{ kg}\end{array}$

The mass of deuterium in ocean is $\text{4}\text{.43}\times {\text{10}}^{16}\text{ kg}$.

Substitute $\text{4}\text{.43}\times {\text{10}}^{16}\text{ kg}$ for $m$ and $2.014\text{ u}$ for $M$ in equation (I) to find the value of N. [Note: $1\text{ u}=1.66\times {10}^{-27}\text{ kg}$

$\begin{array}{c}{N}_D=\frac{\text{4}\text{.43}\times {\text{10}}^{16}\text{ kg}}{2.014\text{ u}\times \left(1.66\times {10}^{-27}\text{kg}/\text{u}\right)}\\ =1.33\times {10}^{43}\end{array}$

The number of deuterium nuclei in the ocean is $1.33\times {10}^{43}$.

Since, all the mass of deuterium in ocean are fused to ${}_2^4\text{He}$. The number of ${}_2^4\text{He}$ nuclei is $\frac{N_D}{2}=6.63\times {10}^{42}$.

Substitute $2.04102\text{ u}$ for $M_{2_H}$, $4.002603\text{ u}$ for $M_{4_{He}}$ and $931.494\text{MeV}/\text{u}$ for $c^2$ in equation (II) to find the value of $Q$

$\begin{array}{c}Q=\left[M_{2_H}+M_{2_H}-M_{4_{He}}\right]c^2\\ =\left[2\times \left(2.04102\text{ u}\right)-4.002603\text{ u}\right]\times 931.494\text{MeV}/\text{u}\\ =23.8\text{ MeV}\end{array}$

The energy released per fusion reaction is $23.8\text{ MeV}$ .

Substitute $23.8\text{ MeV}$ for $Q$ and $6.63\times {10}^{42}$ for $N$  in equation (III) to find the value of $E$

$\begin{array}{c}E=\left(6.63\times {10}^{42}\right)\times 23.8\text{ MeV}\\ \text{=}\left(6.63\times {10}^{42}\right)\times 23.8\text{ MeV}\times \left(\frac{1.60\times {10}^{-13}\text{ J}}{1\text{ MeV}}\right)\\ =2.53\times {10}^{31}\text{ J}\end{array}$

Thus, the amount of energy that would be released when all the mass of deuterium in ocean are fused to ${}_2^4\text{He}$ is $2.53\times {10}^{31}\text{ J}$.

(b)

To determine

How many years would the energy calculated in part (a) last.

Answer to Problem 27P

The energy calculated in part (a) would last for $5.34\times {10}^8\text{ yr}$.

Explanation of Solution

The world power consumption is $1.50\times {10}^{13}\text{ W}$ and from part (a), the amount of energy that would be released when all the mass of deuterium in ocean are fused to ${}_2^4\text{He}$ is $2.53\times {10}^{31}\text{ J}$. If the power consumption is $100\text{ times}$ the current date, $P=100\times 1.50\times {10}^{13}\text{ W}=1.50\times {10}^{15}\text{ W}$.

Write the formula for power

    $P=E/\Delta t$                                                                           (V)

Where, E is the total energy converted, $P$ is the power and $\Delta t$ is the time interval.

Conclusion:

Substitute $1.50\times {10}^{15}\text{ W}$ for $P$ and $2.53\times {10}^{31}\text{ J}$ for $E$ in equation (V) to find the value of $\Delta t$

$\begin{array}{c}\Delta t=\frac{2.53\times {10}^{31}\text{ J}}{1.50\times {10}^{15}\text{J}/\text{s}}\\ =\left(1.69\times {10}^{16}\text{ s}\right)\left(\frac{1\text{ yr}}{3.16\times {10}^7\text{ s}}\right)\\ =5.34\times {10}^8\text{ yr}\end{array}$

Thus, the energy calculated in part (a) would last for $5.34\times {10}^8\text{ yr}$.