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Calculating energy and power from fusion

(a) Calculate the energy released by the fusion of a 1.00-kg mixture of deuterium and tritium, which produces helium. There are equal numbers of deuterium and tritium nuclei in the mixture.

(b) If this takes place continuously over a period of a year, what is the average power output?

Strategy

According to 2 H + 3 H 4 He + n , the energy per reaction is 17.59 MeV. To find the total energy released, we must find the number of deuterium and tritium atoms in a kilogram. Deuterium has an atomic mass of about 2 and tritium has an atomic mass of about 3, for a total of about 5 g per mole of reactants or about 200 mol in 1.00 kg. To get a more precise figure, we will use the atomic masses from Appendix A. The power output is best expressed in watts, and so the energy output needs to be calculated in joules and then divided by the number of seconds in a year.

Solution for (a)

The atomic mass of deuterium ( 2 H ) is 2.014102 u, while that of tritium ( 3 H ) is 3.016049 u, for a total of 5.032151 u per reaction. So a mole of reactants has a mass of 5.03 g, and in 1.00 kg there are ( 1000 g ) / ( 5.03 g/mol ) =198 . 8 mol of reactants . The number of reactions that take place is therefore

( 198.8 mol ) 6 . 02 × 10 23 mol 1 = 1.20 × 10 26 reactions .

The total energy output is the number of reactions times the energy per reaction:

E = 1.20 × 10 26 reactions ( 17.59 MeV/reaction ) 1.602 × 10 13 J/MeV = 3 . 37 × 10 14 J . alignl { stack { size 12{E= left (1 "." "20" times "10" rSup { size 8{"26"} } `"reactions" right ) \( "17" "." "59"`"MeV/reaction" \) left (1 "." 6 times "10" rSup { size 8{ - "13"} } `"J/MeV" right )} {} #" "= 3 "." "37" times "10" rSup { size 8{"14"} } `J "." {} } } {}

Solution for (b)

Power is energy per unit time. One year has 3 . 16 × 10 7 s size 12{3 "." "16" times "10" rSup { size 8{7} } `s} {} , so

P = E t = 3 . 37 × 10 14 J 3 . 16 × 10 7 s = 1 . 07 × 10 7 W = 10 . 7 MW . alignl { stack { size 12{P= { {E} over {t} } = { {3 "." "37" times "10" rSup { size 8{"14"} } `J} over {3 "." "16" times "10" rSup { size 8{7} } `s} } } {} #" "=1 "." "07" times "10" rSup { size 8{7} } `W="10" "." 7`"MW" "." {} } } {}

Discussion

By now we expect nuclear processes to yield large amounts of energy, and we are not disappointed here. The energy output of 3 . 37 × 10 14 J size 12{3 "." "37" times "10" rSup { size 8{"14"} } `J} {} from fusing 1.00 kg of deuterium and tritium is equivalent to 2.6 million gallons of gasoline and about eight times the energy output of the bomb that destroyed Hiroshima. Yet the average backyard swimming pool has about 6 kg of deuterium in it, so that fuel is plentiful if it can be utilized in a controlled manner. The average power output over a year is more than 10 MW, impressive but a bit small for a commercial power plant. About 32 times this power output would allow generation of 100 MW of electricity, assuming an efficiency of one-third in converting the fusion energy to electrical energy.

Test prep for ap courses

The figure is a graph with an arrow pointing up for the y-, vertical axis and labeled PEtot. The x-, horizontal axis is labeled with arrow r. A red line indicating the plot starts horizontally below the x-axis and then quickly rises toward the horizontal line marked 0 and is labeled Attractive nuclear below the 0 line. There is a light nuclei dot on the slope of the line above the x-axis with an arrow pointing down to the left along the slope and is labeled Pulled together. After the plot reaches a maximum, the red line of the plot moves down and at less of slope. This portion of the plot is labeled Repulsive Coulomb. On this portion is another light nuclei dot labeled Repelled with a line pointing down and to the right along the slope of the plot.

This figure shows a graph of the potential energy between two light nuclei as a function of the distance between them. Fusion can occur between the nuclei if the distance is

  1. large so that kinetic energy is low.
  2. large so that potential energy is low.
  3. small so that nuclear attractive force can overcome Coulomb’s repulsion.
  4. small so that nuclear attractive force cannot overcome Coulomb’s repulsion.

(c)

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In a nuclear fusion reaction, 2 g of hydrogen is converted into 1.985 g of helium. What is the energy released?

  1. 4.5 × 10 3 J
  2. 4.5 × 10 6 J
  3. 1.35 × 10 12 J
  4. 1.35 × 10 15 J
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When deuterium and tritium nuclei fuse to produce helium, what else is produced?

  1. positron
  2. proton
  3. α-particle
  4. neutron

(d)

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Suppose two deuterium nuclei are fused to produce helium.

  1. Write the equation for the fusion reaction.
  2. Calculate the difference between the masses of reactants and products.
  3. Using the result calculated in (b), find the energy produced in the fusion reaction.

Assume that the mass of deuterium is 2.014102 u, the mass of helium is 4.002603 u and 1 u = 1.66 × 10 -27 kg.

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Practice Key Terms 6

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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