Question 43.8: Deuterium versus Fossil Fuel In seawater there is about 1 de......
Deuterium versus Fossil Fuel
In seawater there is about 1 deuterium atom for every 6400 hydrogen atoms. If the deuterium from 1 gal of such water were used in the fusion reaction given in Equation 43.21,
{ }_1^2 H +{ }_1^3 H \rightarrow{ }_2^4 He +{ }_0^1 n \quad \quad (43.21)estimate how much energy would be released? Don’t worry about details such as obtaining the tritium. In the Check and Think step compare your answer to energy released by the same volume of fossil fuel such as coal or gasoline. Hints: See Table 9.1 (page 265) and Table 15.1 (page 420); also 1 m³ = 264 gal. Report your final answer to two significant figures.
| TABLE 9.1 Energy content in joules. | |||
| Astronomical sources | Joules (J) | Human needs and everyday phenomena | Joules (J) |
| Typical supernova (exploding star) | 10^{43} | 1000-MW power station in 1 year | 10^{16} |
| Milky Way’s radiation in 1 s | 10^{38} | 1 lb of uranium-235 | 3.7 \times 10^{13} |
| Sun’s radiation in 1 year | 10^{34} | Energy to put space shuttle into orbit | 10^{13} |
| Crossing Atlantic Ocean in a jet plane | 10^{12} | ||
| 1 U.S. gallon of gasoline | 1.3 \times 10^8 | ||
| 1 lb of coal | 1.6 \times 10^7 | ||
| Two-ton truck traveling at highway speed | 10^6 | ||
| Terrestrial sources | Joules (J) | 1 lb of TNT | 10^6 |
| Earth’s rotation | 10^{29} | 1 can lighter fluid | 10^6 |
| Volcanic detonation | 10^{19} | 1 candy bar, order of french fries, or slice of pizza | 10^6 |
| Largest recorded earthquake | 10^{18} | 1 tablespoon sugar, 1 apple, or 5 crackers | 10^5 |
| San Francisco earthquake of 1906 | 10^{17} | 1 AA battery | 10^3 |
| Tornado or thunderstorm | 10^{15} | Major league pitch | 10^2 |
| Lightning flash | 10^{10} | Striking a typewriter key | 10^{-2} |
| Adapted from Howard Keller, “Energy Yield of Various Sources,” Physics Teacher 30:455 (1992). | |||
| Table 15.1 Typical density of common fluids. | |
| Fluid | Density \rho( kg/m^3)^a |
| Acetone | 784.58 |
| Air | 1.29 |
| Ethyl alcohol | 785.06 |
| Gasoline | 737.22 |
| Helium gas | 0.18 |
| Hydrogen gas | 0.089 |
| Mercury | 13,600 |
| Sunflower oil | 920 |
| Seawater | 1025.18 |
| Water (at 4°C) | 1000.00 |
| Water | 998.2 |
| ^a Densities are at room temperature (15°–25°C) except where indicated. | |
INTERPRET and ANTICIPATE
We need to find the number of deuterium atoms in a gallon of seawater. We don’t need to worry about the details such as ionizing and heating the deuterium. We’ll assume that the number of deuterium nuclei available for the fusion reaction equals the number of deuterium atoms. Finally, we’ll use the fact that each fusion reaction releases energy equal to 17.59 MeV.
SOLVE
First, we need the mass of 1 gallon of seawater. The density of seawater is 1025.18 kg/m³ (Table 15.1).
Next we need the number of water molecules in the gallon. The molar mass of H_2O is 18 g, which means 1 mole of water has a mass of 18 g. (Of course, heavy water is slightly heavier, but we are making an estimate to two significant figures, and we don’t need to be concerned about such details.)
\begin{aligned} & N_{\text {molecules }}=m_{\text {water }} \frac{N_{ A }}{18 g } \\ & N_{\text {molecules }}=m_{\text {water }} \frac{6.02 \times 10^{23} \text { molecules }}{18 g } \\& N_{\text {molecules }}=\left(3.88 \times 10^3 g \right) \frac{6.02 \times 10^{23} \text { nuclei }}{18 g } \\& N_{\text {molecules }}=1.3 \times 10^{26} \text { molecules }\end{aligned}There are two hydrogen atoms in each molecule of water. So we’ll multiply the number of molecules by 2 to find the number of hydrogen atoms.
N_{ H }=2 N_{\text {molecules }}=2.6 \times 10^{26}We need the number of deuterium atoms. Because there is 1 deuterium atom for every 6400 hydrogen atoms, we divide by 6400. As described, this is also the number of nuclei that undergo fusion.
N_D=\frac{N_{ H }}{6400}=\frac{2.6 \times 10^{26}}{6400}=4.1 \times 10^{22} \text { nuclei }Assume deuterium nuclei undergo a fusion reaction and that each reaction releases 17.59 MeV of energy.
\begin{aligned}& E_{\text {released }}=(17.59 MeV / \text { nuclei }) N_{ D } \\& E_{\text {released }}=(17.59 MeV / \text { nuclei })\left(4.1 \times 10^{22} \text { nuclei }\right) \\& E_{\text {released }}=7.1 \times 10^{23} MeV =1.1 \times 10^{11} J\end{aligned}CHECK and THINK
According to Table 9.1, 1 gal of gasoline contains 1.3 \times 10^8 J. Fusing the deuterium in 1 gal of seawater releases about a thousand times more energy than burning 1 gal of gasoline, and fusion doesn’t produce greenhouse gases.