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 34P

(a)

To determine

The amount of energy released in the nuclear fusion.

(a)

Expert Solution
Check Mark

Answer to Problem 34P

The amount of energy released in the nuclear fusion is 2.53×1031J_.

Explanation of Solution

Write the expression for the mass of water.

    mH2O=ρV        (I)

Here, mH2O is the mass of water, ρ is the density of water, V is the volume of water.

Write the expression for mass of hydrogen.

    mH2=(MH2MH2O)mH2O        (II)

Here, mH2 is the mass of hydrogen, MH2 is the molar mass of hydrogen, MH2O is the Molar mass of water.

Write the expression for the mass of deuterium.

    mdeuterium=(0.030%)mH2        (III)

Here, mdeuterium is the mass of deuterium.

Write the expression for the number of deuterium nuclei in the mass mdeuterium.

    N=mdeuteriummdeuteron        (IV)

Here, N is the number of deuterium nuclei, mdeuteron is the mass of deuteron.

Two deuterium nuclei are used per fusion. The number of events take place in the fusion process is N2.

Write the expression for the energy released per event.

    Q=[MH12+MH12MHe24](931.5MeV/u)        (V)

Here, MH12 is the mass of the deuteron, MHe24 is the mass of the helium.

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

    E=(N2)Q        (VI)

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

Conclusion:

Substitute 103kg/m3 for ρ, 317million mi3 for V in equation (I) to find mH2O.

    mH2O=(103kg/m3)[317million mi3×(1609m1mi)3](103kg/m3)[317×106 mi3×(1609m1mi)3]=(103kg/m3)(1.32×1018m3)=1.32×1021kg

Substitute 1.32×1021kg for mH2O, 2.016u for MH2, 18.015u for MH2O in equation (II) to find mH2.

    mH2=(2.016u18.015u)(1.32×1021kg)=1.48×1020kg

Substitute 1.48×1020kg for mH2 in equation (III) to find mdeterium.

    mdeuterium=(0.030×102)(1.48×1020kg)=4.43×1016kg

Substitute 4.43×1016kg for mdeuterium, 2.014u for mdeuteron in equation (IV) to find N.

    N=4.43×1016kg(2.014u×1.67×1027kg1u)=1.33×1043nuclei

Substitute 2.014102u for MH2, 4.002603u for MHe24 in equation (V) to find Q.

    Q=[2(2.014102u)4.002603u](931.5MeV/u)=23.8MeV

Substitute 23.8MeV for Q, 1.33×1043nuclei for N in equation (VI) to find E.

    E=1.33×1043nuclei2(23.8MeV×1.69×1013J1MeV)=2.53×1031J

Therefore, the amount of energy released in the nuclear fusion is 2.53×1031J_.

(b)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 34P

The time required to last the available energy in the fusion reaction is 5.34×108yr_.

Explanation of Solution

Write the expression for the power consumption in the fusion.

    P=EΔt        (VII)

Here, P is the power consumed, E is the energy released, Δt is the time interval.

Use equation (VII) to solve for Δt.

    Δt=EP        (VIII)

Conclusion:

Substitute 2.53×1031J for E, 100(1.50×1013W) for P in equation (VIII) to find Δt.

    Δt=2.53×1031J100(1.50×1013W)=(1.69×1016s)Δt(inyr)=(1.69×1016s)(1yr3.16×107s)=5.34×108yr

Therefore, the time required to last the available energy in the fusion reaction is 5.34×108yr_.

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Chapter 30 Solutions

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

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