Chapter 30, Problem 46AP

(a)

To determine

The energy produced.

Answer to Problem 46AP

The energy produced is $3.61\times {10}^{30}\text{J}$.

Explanation of Solution

Given info:

The mass of ocean is $1.32\times {10}^{21}\text{kg}$

The mass per mol is $18.0\times {10}^{-3}\text{kg/mol}$

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

The energy per atom $3.27\text{MeV}$

The formula to calculate the number of water molecules is,

$N_{H_{2}O}=\left(\frac{m_{ocean}}{M_{mol}}\right)N_A$

• $N_{H_{2}O}$ is the number of water molecules
• $m_{ocean}$ is mass of ocean water
• $N_A$ is Avogadro number
• $M_{mol}$ is the mass per mole

The formula to calculate the number of deuterons is,

$\begin{array}{c}{N}_D=2\left(1.56\times {10}^{-4}\right)N_{H_{2}0}\\ =2\left(1.56\times {10}^{-4}\right)\left(\frac{m_{ocean}}{M_{mol}}\right)N_A\end{array}$

• $N_D$ is the deuterons

The formula to calculate the number of events is,

$\begin{array}{c}{N}_{events}=\frac{N_D}{2}=2\left(1.56\times {10}^{-4}\right)N_{H_{2}0}\\ =\left(1.56\times {10}^{-4}\right)\left(\frac{m_{ocean}}{M_{mol}}\right)N_A\end{array}$

• $N_{events}$ is the number of events

The formula to calculate the energy produced is,

$\begin{array}{c}E=N_{events}{E}_0\\ =\left(1.56\times {10}^{-4}\left(\frac{m_{ocean}}{M_{mol}}\right)N_A\right)E_0\end{array}$

• $E$ is the energy
• $E_0$ is the energy of event

The mass of ocean is $1.32\times {10}^{21}\text{kg}$

The mass per mol is $18.0\times {10}^{-3}\text{kg/mol}$

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

The energy per atom $3.27\text{MeV}$

Substitute $1.32\times {10}^{21}\text{kg}$ for $m_{ocean}$, $18.0\times {10}^{-3}\text{kg/mol}$ for $M_{mol}$, $6.02\times {10}^{23}\text{atoms/mol}$ for $N_A$ and $3.27\text{MeV}$ for $E_0$ to find $E$.

$\begin{array}{c}E=\left(1.56\times {10}^{-4}\left(\frac{1.32\times {10}^{21}\text{kg}}{18.0\times {10}^{-3}\text{kg/mol}}\right)\left(6.02\times {10}^{23}\text{atoms/mol}\right)\right)\left(3.27\text{MeV}\right)\left(\frac{1.60\times {10}^{-13}\text{J}}{1\text{MeV}}\right)\\ =3.61\times {10}^{30}\text{J}\end{array}$

Thus, the energy produced is $3.61\times {10}^{30}\text{J}$.

Conclusion:

The energy produced is $3.61\times {10}^{30}\text{J}$.

(b)

To determine

The time the energy lasts.

Answer to Problem 46AP

The time the energy lasts is $1.63\times {10}^8\text{yr}$.

Explanation of Solution

Given info:

The rate of energy consumption is $7.00\times {10}^{14}\text{J/s}$

The formula to calculate the time is,

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

• $t$ is the time
• $P$ is the energy consumption rate

Substitute $3.61\times {10}^{30}\text{J}$ for $E$ and $7.00\times {10}^{14}\text{J/s}$ for $P$ to find $t$.

$\begin{array}{c}t=\frac{3.61\times {10}^{30}\text{J}}{\left(7.00\times {10}^{14}\text{J/s}\right)\left(\frac{3.156\times {10}^7\text{s}}{1\text{yr}}\right)}\\ =1.63\times {10}^8\text{yr}\end{array}$

Thus, the time the energy lasts is $1.63\times {10}^8\text{yr}$.

Conclusion:

The time the energy lasts is $1.63\times {10}^8\text{yr}$.