Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
5th Edition
ISBN: 9781305586871
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 30, Problem 58P

(a)

To determine

Energy produced by the fission of 1.00kg of 239Pu.

(a)

Expert Solution
Check Mark

Answer to Problem 58P

Energy produced by the fission of 1.00kg of 239Pu is 2.24×107kWh.

Explanation of Solution

Write the equation to find the number of nuclie in the given mass.

    n=NavaMPum

Here, n is the number of nuclei, Nava is the Avogadro number, MPu is the atomic mass of plutonium, and m is the given mass in gram.

Write the equation to find the total energy produced.

    Etotal=nE

Here, Etotal is the total energy produced and E is the energy produced per fission.

Conclusion:

Substitute 6.02×1023 nuclei/mol for Nava, 239.05g/mol for MPu, and 1.00kg for m in the equation for n.

    n=6.02×1023 nuclei/mol239.05g/mol(1.00kg(103g1.00kg))=2.52×1024nuclei

Substitute 2.52×1024nuclei for n and 200MeV for E in the equation for Etotal.

  Etotal=(2.52×1024nuclei(1 fissionnucleus))(200 MeV fission)=5.04×1026MeV(4.44×1020kWh1MeV)=2.24×107kWh

Therefore, the energy produced by the fission of 1.00kg of 239Pu is 2.24×107kWh.

(b)

To determine

Energy produced by the deuterium –tritium fusion reaction.

(b)

Expert Solution
Check Mark

Answer to Problem 58P

Energy produced by the deuterium –tritium fusion reaction is 17.6MeV/fission.

Explanation of Solution

Write the equation to find the Energy produced by the deuterium –tritium fusion reaction.

    E=Δmc2

Here, E is the energy produced in fusion, Δm is the mass defect, and c is the speed of light in vacuum.

Write the equation to find Δm.

    Δm=m2H+m3Hm4Hemneutron

Here, m2H is the mass of deuterium, m3H is the mass of tritium, m4He is the mass of helium nuclei, and mneutron is the mass of neutron.

Rewrite the expression for E by substituting the equation for Δm.

    E=(m3H+m2Hm4Hemneutron)c2

Conclusion:

Substitute 3.016049u for m3H, 2.014102 u for m2H, 4.002603u for m4He, 1.008665u for mneutron, and 931.5MeV/u for c2 in the above equation to find E.

    E=(3.016049u+2.014102u4.002603u1.008665u)(931.5MeV/u)2=17.6MeV/fission

Therefore, the energy produced by the deuterium –tritium fusion reaction is 17.6MeV/fission.

(c)

To determine

Energy produced by the deuterium –tritium fusion reaction for 1.00kg of deuterium.

(c)

Expert Solution
Check Mark

Answer to Problem 58P

Energy produced by the deuterium –tritium fusion reaction for 1.00kg of deuterium is 2.34×108kWh.

Explanation of Solution

Write the equation to find the energy produced by the deuterium –tritium fusion reaction for given mass of deuterium.

    Etotal=ndeE                                                                                                  (I)

Here, Etotal is the energy produced in fusion and nde is the total number of deuterium nuclie.

Write the equation to find the number of nuclie in the given mass.

    n=Navam2Hm

Here, n is the number of nuclie in the given mass.

Rewrite equation (I) by substituting the above relation for n.

  Etotal=(Navam2Hm)E

Conclusion:

Substitute 6.02×1023 nuclei/mol for Nava, 239.05g/mol for MPu, 1.00kg for m, and 17.6MeV/fission for E in the above equation.

  Etotal=(6.02×10232.014 g/mol(1.00kg(103g1.00kg)))(17.6MeV/fission(1kWh4.44×1020MeV))=2.34×108kWh

Therefore, the energy produced by the deuterium –tritium fusion reaction for 1.00kg of deuterium is 2.34×108kWh.

(d)

To determine

Energy produced by the combustion of 1.00kg of carbon.

(d)

Expert Solution
Check Mark

Answer to Problem 58P

Energy produced by the combustion of 1.00kg carbon is 9.36 kWh.

Explanation of Solution

Write the equation to find the energy produced by the deuterium –tritium fusion reaction for given mass of deuterium.

    Etotal=nCEcom                                                                                                   (I)

Here, Etotal is the energy produced in combustion, nC is the number of carbon atoms, and Ecom is the energy produced by the formation of one CO2 molecule.

Write the equation to find the number of nuclie in the given mass.

    n=NavaMCmc

Here, MC is the atomic mass of carbon and mc is the given mass in gram.

Rewrite the equation for Etotal by substituting the above relation for n.

  Etotal=(NavaMCmc)Ecom

Conclusion:

Substitute 6.02×1023 nuclei/mol for Nava, 12g/mol for MC, 1.00kg for m, and 4.20MeV for Ecom in the above equation.

    Etotal=(6.02×1023 nuclei/mol12g/mol(1.00kg))(4.20MeV(1kWh4.44×1020MeV))=9.36 kWh

Therefore, the energy produced by the combustion of 1.00kg carbon is 9.36 kWh.

(e)

To determine

The pros and corns of energy production by fission, fusion, and combustion.

(e)

Expert Solution
Check Mark

Explanation of Solution

To produce energy by combustion, coal is used. Coal is abundant form of fossil fuel and it is very cheap. The disadvantage is the high carbon emission and thereby acts as a major contributor in global warming. Energy production by nuclear fission cannot produce carbon and the working of reactor cannot cause any pollution. But the disposal of radioactive waste materials from the reactor is a very big challenge.

Nuclear fusion is better than fission since it does not produce radioactive by-products. But researches on nuclear fusion reactor are in developing stage only and it requires extreme high temperature for its working that cannot be achieved by man at present. Plutonium is a very risky material to handle in fission process.

Therefore, combustion is cheaper, but carbon emission is high, fission cannot cause global warming but produce radioactive bye products, and nuclear fusion cannot produce radioactive pollution but requires very large temperature that cannot be achieved in a laboratory at present.

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Chapter 30 Solutions

Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)

Ch. 30 - Prob. 5OQCh. 30 - Prob. 6OQCh. 30 - Prob. 7OQCh. 30 - Prob. 8OQCh. 30 - Prob. 9OQCh. 30 - Prob. 10OQCh. 30 - Which of the following quantities represents the...Ch. 30 - Prob. 12OQCh. 30 - Prob. 1CQCh. 30 - Prob. 2CQCh. 30 - Prob. 3CQCh. 30 - Prob. 4CQCh. 30 - Prob. 5CQCh. 30 - Prob. 6CQCh. 30 - Prob. 7CQCh. 30 - If no more people were to be born, the law of...Ch. 30 - Prob. 9CQCh. 30 - Prob. 10CQCh. 30 - Prob. 11CQCh. 30 - What fraction of a radioactive sample has decayed...Ch. 30 - Prob. 13CQCh. 30 - Prob. 14CQCh. 30 - Prob. 15CQCh. 30 - Prob. 16CQCh. 30 - Prob. 17CQCh. 30 - Prob. 1PCh. 30 - Prob. 2PCh. 30 - Prob. 3PCh. 30 - Prob. 4PCh. 30 - Prob. 5PCh. 30 - Prob. 6PCh. 30 - Prob. 7PCh. 30 - Prob. 8PCh. 30 - Prob. 9PCh. 30 - Prob. 10PCh. 30 - Prob. 11PCh. 30 - Prob. 12PCh. 30 - Prob. 13PCh. 30 - Prob. 14PCh. 30 - Prob. 16PCh. 30 - Prob. 17PCh. 30 - Prob. 18PCh. 30 - What time interval elapses while 90.0% of the...Ch. 30 - Prob. 20PCh. 30 - Prob. 21PCh. 30 - Prob. 22PCh. 30 - Prob. 23PCh. 30 - Prob. 24PCh. 30 - Prob. 25PCh. 30 - Prob. 26PCh. 30 - Prob. 27PCh. 30 - Prob. 28PCh. 30 - Prob. 29PCh. 30 - Prob. 30PCh. 30 - Prob. 31PCh. 30 - Prob. 32PCh. 30 - Prob. 33PCh. 30 - Prob. 34PCh. 30 - Prob. 35PCh. 30 - Prob. 36PCh. 30 - Prob. 37PCh. 30 - Prob. 38PCh. 30 - Prob. 39PCh. 30 - Prob. 41PCh. 30 - Prob. 42PCh. 30 - Prob. 43PCh. 30 - Prob. 45PCh. 30 - Prob. 46PCh. 30 - Prob. 47PCh. 30 - Prob. 48PCh. 30 - Prob. 49PCh. 30 - Prob. 50PCh. 30 - Prob. 51PCh. 30 - Prob. 52PCh. 30 - Prob. 53PCh. 30 - Prob. 54PCh. 30 - Prob. 55PCh. 30 - Prob. 56PCh. 30 - Prob. 57PCh. 30 - Prob. 58PCh. 30 - Prob. 59PCh. 30 - Prob. 60PCh. 30 - Prob. 61PCh. 30 - Prob. 62PCh. 30 - Prob. 63PCh. 30 - Prob. 64PCh. 30 - Prob. 65PCh. 30 - Prob. 66PCh. 30 - Prob. 67PCh. 30 - Prob. 68PCh. 30 - Prob. 69PCh. 30 - Prob. 70P
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