Chapter 30, Problem 34P

(a)

To determine

The amount of energy released in the nuclear fusion.

Answer to Problem 34P

The amount of energy released in the nuclear fusion is $\underset{\_}{2.53\times {10}^{31}\text{J}}$.

Explanation of Solution

Write the expression for the mass of water.

    $m_{{\text{H}}_{\text{2}}\text{O}}=\rho V$        (I)

Here, $m_{{\text{H}}_{\text{2}}\text{O}}$ is the mass of water, $\rho$ is the density of water, $V$ is the volume of water.

Write the expression for mass of hydrogen.

    $m_{{\text{H}}_2}=\left(\frac{M_{{\text{H}}_2}}{M_{{\text{H}}_2\text{O}}}\right)m_{{\text{H}}_2\text{O}}$        (II)

Here, $m_{{\text{H}}_2}$ is the mass of hydrogen, $M_{{\text{H}}_2}$ is the molar mass of hydrogen, $M_{{\text{H}}_{\text{2}}\text{O}}$ is the Molar mass of water.

Write the expression for the mass of deuterium.

    $m_{\text{deuterium}}=\left(0.030\%\right)m_{{\text{H}}_2}$        (III)

Here, $m_{\text{deuterium}}$ is the mass of deuterium.

Write the expression for the number of deuterium nuclei in the mass $m_{\text{deuterium}}$.

    $N=\frac{m_{\text{deuterium}}}{m_{\text{deuteron}}}$        (IV)

Here, $N$ is the number of deuterium nuclei, $m_{\text{deuteron}}$ is the mass of deuteron.

Two deuterium nuclei are used per fusion. The number of events take place in the fusion process is $\frac{N}{2}$.

Write the expression for the energy released per event.

    $Q=\left[M_{{\text{H}}_1^2}+M_{{\text{H}}_1^2}-M_{{\text{He}}_2^4}\right]\left(931.5\text{MeV}/\text{u}\right)$        (V)

Here, $M_{{\text{H}}_1^2}$ is the mass of the deuteron, $M_{{\text{He}}_2^4}$ is the mass of the helium.

Write the expression for the total energy in the fusion reaction.

    $E=\left(\frac{N}{2}\right)Q$        (VI)

Here, $E$ is the total energy available in the fusion reaction.

Conclusion:

Substitute ${10}^3\text{kg}/{\text{m}}^3$ for $\rho$, $317{\text{million mi}}^3$ for $V$ in equation (I) to find $m_{{\text{H}}_2\text{O}}$.

    $\begin{array}{c}{m}_{{\text{H}}_2\text{O}}=\left({10}^3\text{kg}/{\text{m}}^3\right)\left[317{\text{million mi}}^3\times {\left(\frac{1609\text{m}}{1\text{mi}}\right)}^3\right]\\ \left({10}^3\text{kg}/{\text{m}}^3\right)\left[317\times {10}^6{\text{ mi}}^3\times {\left(\frac{1609\text{m}}{1\text{mi}}\right)}^3\right]\\ =\left({10}^3\text{kg}/{\text{m}}^3\right)\left(1.32\times {10}^{18}{\text{m}}^3\right)\\ =1.32\times {10}^{21}\text{kg}\end{array}$

Substitute $1.32\times {10}^{21}\text{kg}$ for $m_{{\text{H}}_{\text{2}}\text{O}}$, $2.016\text{u}$ for $M_{{\text{H}}_2}$, $18.015\text{u}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ in equation (II) to find $m_{{\text{H}}_{\text{2}}}$.

    $\begin{array}{c}{m}_{{\text{H}}_2}=\left(\frac{2.016\text{u}}{18.015\text{u}}\right)\left(1.32\times {10}^{21}\text{kg}\right)\\ =1.48\times {10}^{20}\text{kg}\end{array}$

Substitute $1.48\times {10}^{20}\text{kg}$ for $m_{{\text{H}}_2}$ in equation (III) to find $m_{\text{deterium}}$.

    $\begin{array}{c}{m}_{\text{deuterium}}=\left(0.030\times {10}^{-2}\right)\left(1.48\times {10}^{20}\text{kg}\right)\\ =4.43\times {10}^{16}\text{kg}\end{array}$

Substitute $4.43\times {10}^{16}\text{kg}$ for $m_{\text{deuterium}}$, $2.014\text{u}$ for $m_{\text{deuteron}}$ in equation (IV) to find $N$.

    $\begin{array}{c}N=\frac{4.43\times {10}^{16}\text{kg}}{\left(2.014\text{u}\times \frac{1.67\times {10}^{-27}\text{kg}}{1\text{u}}\right)}\\ =1.33\times {10}^{43}\text{nuclei}\end{array}$

Substitute $2.014102\text{u}$ for $M_{{\text{H}}_2}$, $4.002603\text{u}$ for $M_{{\text{He}}_2^4}$ in equation (V) to find $Q$.

    $\begin{array}{c}Q=\left[2\left(2.014102\text{u}\right)-4.002603\text{u}\right]\left(931.5\text{MeV}/\text{u}\right)\\ =23.8\text{MeV}\end{array}$

Substitute $23.8\text{MeV}$ for $Q$, $1.33\times {10}^{43}\text{nuclei}$ for $N$ in equation (VI) to find $E$.

    $\begin{array}{c}E=\frac{1.33\times {10}^{43}\text{nuclei}}{2}\left(23.8\text{MeV}\times \frac{1.69\times {10}^{-13}\text{J}}{1\text{MeV}}\right)\\ =2.53\times {10}^{31}\text{J}\end{array}$

Therefore, the amount of energy released in the nuclear fusion is $\underset{\_}{2.53\times {10}^{31}\text{J}}$.

(b)

To determine

The time required to last the available energy in the fusion reaction.

Answer to Problem 34P

The time required to last the available energy in the fusion reaction is $\underset{\_}{5.34\times {10}^8\text{yr}}$.

Explanation of Solution

Write the expression for the power consumption in the fusion.

    $P=\frac{E}{\Delta t}$        (VII)

Here, $P$ is the power consumed, $E$ is the energy released, $\Delta t$ is the time interval.

Use equation (VII) to solve for $\Delta t$.

    $\Delta t=\frac{E}{P}$        (VIII)

Conclusion:

Substitute $2.53\times {10}^{31}\text{J}$ for $E$, $100\left(1.50\times {10}^{13}\text{W}\right)$ for $P$ in equation (VIII) to find $\Delta t$.

    $\begin{array}{c}\Delta t=\frac{2.53\times {10}^{31}\text{J}}{100\left(1.50\times {10}^{13}\text{W}\right)}\\ =\left(1.69\times {10}^{16}\text{s}\right)\\ \Delta t\left(\text{in}\text{yr}\right)\text{=}\left(1.69\times {10}^{16}\text{s}\right)\left(\frac{1\text{yr}}{3.16\times {10}^7\text{s}}\right)\\ =5.34\times {10}^8\text{yr}\end{array}$

Therefore, the time required to last the available energy in the fusion reaction is $\underset{\_}{5.34\times {10}^8\text{yr}}$.