Mass of uranium in reserve.

Question

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Answer 1

Question

Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

(a)

Expert Solution

Answer to Problem 10P

The mass of uranium in reserve is

$3.1×1010 g$ .

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70%$ .

The amount of uranium reserved is $4.4×106$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m235U=Δpm$

$m235U$ is the difference in time between the cars to reach the destination
$Δp$ is the percentage of mass reserved m is the speed of the slower car
m is the mass of uranium in reserve

Substitute $0.70%$ for $Δp$ and $4.4×106$ for m to find $m235U$ .

$m235U=(0.70%(1100%))(4.4×106 metric ton)(103 kg1 ton)(103 g1 kg)=3.1×1010 kg$

Thus, the mass of uranium in reserve is $3.1×1010 g$ .

Conclusion:

The mass of uranium in reserve is $3.1×1010 g$ .

(b)

To determine

The number of moles and atoms in reserve.

(b)

Expert Solution

Answer to Problem 10P

The number of moles in reserve is

$1.3×108 mol$ number of atoms in reserve is

$7.8×1031 atoms$ .

Explanation of Solution

Given Info:

The mass per one mole is $235 g/mol$ .

The Avogadro number is $6.02×1023 atoms/mol$ .

Formula to calculate the number of moles is,

$n=m235UMmol$

$n$ is the number of moles
$Mmol$ is mass per mol

Substitute $235 g/mol$ for $Mmol$ and $3.1×1010 g$ for $m235U$ to find n.

$n=3.1×1010 g235 g/mol=1.3×108 mol$

Thus, the number of moles is $1.3×108 mol$ .

Formula to calculate the number of atoms is,

$N=nNA$

$N$ is the number of moles
$NA$ is the Avogadro number

Substitute $1.3×108 mol$ for n and $1.3×108 mol$ for $NA$ to find N.

$N=(1.3×108 mol)(6.02×1023 atoms/mol)=7.8×1031 atoms$

Thus, the number of atoms is $7.8×1031 atoms$ .

Conclusion:

The number of moles in reserve is $1.3×108 mol$ number of atoms in reserve is $7.8×1031 atoms$ .

(c)

To determine

The total energy available.

(c)

Expert Solution

Answer to Problem 10P

The total energy available is

$2.6×1021 J$ .

Explanation of Solution

Given Info:

The energy per atom is $208 MeV/atom$ .

Formula to calculate the total energy available is,

$E=NE0$

$E$ is the energy
$E0$ is the energy per atom

Substitute $208 MeV/atom$ for $E0$ and $7.87×1031 atom$ for N to find $E$ .

$E=(208 MeV/atom)(7.87×1031 atom)(1.60×10−13 J1 MeV)=2.6×1021 J$

Thus, the total energy available is $2.6×1021 J$ .

Conclusion:

The total energy available is $2.6×1021 J$ .

(d)

To determine

The maximum time energy supply lasts.

(d)

Expert Solution

Answer to Problem 10P

The maximum time energy supply lasts is

$5.5 yr$ .

Explanation of Solution

Given Info:

The energy consumption rate is $1.5×1013 J/s$ .

Formula to calculate maximum time energy supply lasts is,

$t=EP$

$t$ is maximum time energy supply lasts
$P$ is the energy consumption rate

Substitute $2.6×1021 J$ for $E$ and $1.5×1013 J/s$ for $P$ to find $t$ .

$t=(2.6×1021 J1.5×1013 J/s)(1 yr3.156×107 s)=5.5 yr$

Thus, the maximum time energy supply lasts is $5.5 yr$ .

Conclusion:

The maximum time energy supply lasts is $5.5 yr$

(e)

To determine

The conclusion of the results.

(e)

Expert Solution

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

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Answer 2

Question

Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

(a)

Expert Solution

Answer to Problem 10P

The mass of uranium in reserve is

$3.1×1010 g$ .

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70%$ .

The amount of uranium reserved is $4.4×106$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m235U=Δpm$

$m235U$ is the difference in time between the cars to reach the destination
$Δp$ is the percentage of mass reserved m is the speed of the slower car
m is the mass of uranium in reserve

Substitute $0.70%$ for $Δp$ and $4.4×106$ for m to find $m235U$ .

$m235U=(0.70%(1100%))(4.4×106 metric ton)(103 kg1 ton)(103 g1 kg)=3.1×1010 kg$

Thus, the mass of uranium in reserve is $3.1×1010 g$ .

Conclusion:

The mass of uranium in reserve is $3.1×1010 g$ .

(b)

To determine

The number of moles and atoms in reserve.

(b)

Expert Solution

Answer to Problem 10P

The number of moles in reserve is

$1.3×108 mol$ number of atoms in reserve is

$7.8×1031 atoms$ .

Explanation of Solution

Given Info:

The mass per one mole is $235 g/mol$ .

The Avogadro number is $6.02×1023 atoms/mol$ .

Formula to calculate the number of moles is,

$n=m235UMmol$

$n$ is the number of moles
$Mmol$ is mass per mol

Substitute $235 g/mol$ for $Mmol$ and $3.1×1010 g$ for $m235U$ to find n.

$n=3.1×1010 g235 g/mol=1.3×108 mol$

Thus, the number of moles is $1.3×108 mol$ .

Formula to calculate the number of atoms is,

$N=nNA$

$N$ is the number of moles
$NA$ is the Avogadro number

Substitute $1.3×108 mol$ for n and $1.3×108 mol$ for $NA$ to find N.

$N=(1.3×108 mol)(6.02×1023 atoms/mol)=7.8×1031 atoms$

Thus, the number of atoms is $7.8×1031 atoms$ .

Conclusion:

The number of moles in reserve is $1.3×108 mol$ number of atoms in reserve is $7.8×1031 atoms$ .

(c)

To determine

The total energy available.

(c)

Expert Solution

Answer to Problem 10P

The total energy available is

$2.6×1021 J$ .

Explanation of Solution

Given Info:

The energy per atom is $208 MeV/atom$ .

Formula to calculate the total energy available is,

$E=NE0$

$E$ is the energy
$E0$ is the energy per atom

Substitute $208 MeV/atom$ for $E0$ and $7.87×1031 atom$ for N to find $E$ .

$E=(208 MeV/atom)(7.87×1031 atom)(1.60×10−13 J1 MeV)=2.6×1021 J$

Thus, the total energy available is $2.6×1021 J$ .

Conclusion:

The total energy available is $2.6×1021 J$ .

(d)

To determine

The maximum time energy supply lasts.

(d)

Expert Solution

Answer to Problem 10P

The maximum time energy supply lasts is

$5.5 yr$ .

Explanation of Solution

Given Info:

The energy consumption rate is $1.5×1013 J/s$ .

Formula to calculate maximum time energy supply lasts is,

$t=EP$

$t$ is maximum time energy supply lasts
$P$ is the energy consumption rate

Substitute $2.6×1021 J$ for $E$ and $1.5×1013 J/s$ for $P$ to find $t$ .

$t=(2.6×1021 J1.5×1013 J/s)(1 yr3.156×107 s)=5.5 yr$

Thus, the maximum time energy supply lasts is $5.5 yr$ .

Conclusion:

The maximum time energy supply lasts is $5.5 yr$

(e)

To determine

The conclusion of the results.

(e)

Expert Solution

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

(a)

Expert Solution

Answer to Problem 10P

The mass of uranium in reserve is

$3.1×1010 g$ .

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70%$ .

The amount of uranium reserved is $4.4×106$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m235U=Δpm$

$m235U$ is the difference in time between the cars to reach the destination
$Δp$ is the percentage of mass reserved m is the speed of the slower car
m is the mass of uranium in reserve

Substitute $0.70%$ for $Δp$ and $4.4×106$ for m to find $m235U$ .

$m235U=(0.70%(1100%))(4.4×106 metric ton)(103 kg1 ton)(103 g1 kg)=3.1×1010 kg$

Thus, the mass of uranium in reserve is $3.1×1010 g$ .

Conclusion:

The mass of uranium in reserve is $3.1×1010 g$ .

(b)

To determine

The number of moles and atoms in reserve.

(b)

Expert Solution

Answer to Problem 10P

The number of moles in reserve is

$1.3×108 mol$ number of atoms in reserve is

$7.8×1031 atoms$ .

Explanation of Solution

Given Info:

The mass per one mole is $235 g/mol$ .

The Avogadro number is $6.02×1023 atoms/mol$ .

Formula to calculate the number of moles is,

$n=m235UMmol$

$n$ is the number of moles
$Mmol$ is mass per mol

Substitute $235 g/mol$ for $Mmol$ and $3.1×1010 g$ for $m235U$ to find n.

$n=3.1×1010 g235 g/mol=1.3×108 mol$

Thus, the number of moles is $1.3×108 mol$ .

Formula to calculate the number of atoms is,

$N=nNA$

$N$ is the number of moles
$NA$ is the Avogadro number

Substitute $1.3×108 mol$ for n and $1.3×108 mol$ for $NA$ to find N.

$N=(1.3×108 mol)(6.02×1023 atoms/mol)=7.8×1031 atoms$

Thus, the number of atoms is $7.8×1031 atoms$ .

Conclusion:

The number of moles in reserve is $1.3×108 mol$ number of atoms in reserve is $7.8×1031 atoms$ .

(c)

To determine

The total energy available.

(c)

Expert Solution

Answer to Problem 10P

The total energy available is

$2.6×1021 J$ .

Explanation of Solution

Given Info:

The energy per atom is $208 MeV/atom$ .

Formula to calculate the total energy available is,

$E=NE0$

$E$ is the energy
$E0$ is the energy per atom

Substitute $208 MeV/atom$ for $E0$ and $7.87×1031 atom$ for N to find $E$ .

$E=(208 MeV/atom)(7.87×1031 atom)(1.60×10−13 J1 MeV)=2.6×1021 J$

Thus, the total energy available is $2.6×1021 J$ .

Conclusion:

The total energy available is $2.6×1021 J$ .

(d)

To determine

The maximum time energy supply lasts.

(d)

Expert Solution

Answer to Problem 10P

The maximum time energy supply lasts is

$5.5 yr$ .

Explanation of Solution

Given Info:

The energy consumption rate is $1.5×1013 J/s$ .

Formula to calculate maximum time energy supply lasts is,

$t=EP$

$t$ is maximum time energy supply lasts
$P$ is the energy consumption rate

Substitute $2.6×1021 J$ for $E$ and $1.5×1013 J/s$ for $P$ to find $t$ .

$t=(2.6×1021 J1.5×1013 J/s)(1 yr3.156×107 s)=5.5 yr$

Thus, the maximum time energy supply lasts is $5.5 yr$ .

Conclusion:

The maximum time energy supply lasts is $5.5 yr$

(e)

To determine

The conclusion of the results.

(e)

Expert Solution

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

(a)

Expert Solution

Answer to Problem 10P

The mass of uranium in reserve is

$3.1×1010 g$ .

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70%$ .

The amount of uranium reserved is $4.4×106$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m235U=Δpm$

$m235U$ is the difference in time between the cars to reach the destination
$Δp$ is the percentage of mass reserved m is the speed of the slower car
m is the mass of uranium in reserve

Substitute $0.70%$ for $Δp$ and $4.4×106$ for m to find $m235U$ .

$m235U=(0.70%(1100%))(4.4×106 metric ton)(103 kg1 ton)(103 g1 kg)=3.1×1010 kg$

Thus, the mass of uranium in reserve is $3.1×1010 g$ .

Conclusion:

The mass of uranium in reserve is $3.1×1010 g$ .

(b)

To determine

The number of moles and atoms in reserve.

(b)

Expert Solution

Answer to Problem 10P

The number of moles in reserve is

$1.3×108 mol$ number of atoms in reserve is

$7.8×1031 atoms$ .

Explanation of Solution

Given Info:

The mass per one mole is $235 g/mol$ .

The Avogadro number is $6.02×1023 atoms/mol$ .

Formula to calculate the number of moles is,

$n=m235UMmol$

$n$ is the number of moles
$Mmol$ is mass per mol

Substitute $235 g/mol$ for $Mmol$ and $3.1×1010 g$ for $m235U$ to find n.

$n=3.1×1010 g235 g/mol=1.3×108 mol$

Thus, the number of moles is $1.3×108 mol$ .

Formula to calculate the number of atoms is,

$N=nNA$

$N$ is the number of moles
$NA$ is the Avogadro number

Substitute $1.3×108 mol$ for n and $1.3×108 mol$ for $NA$ to find N.

$N=(1.3×108 mol)(6.02×1023 atoms/mol)=7.8×1031 atoms$

Thus, the number of atoms is $7.8×1031 atoms$ .

Conclusion:

The number of moles in reserve is $1.3×108 mol$ number of atoms in reserve is $7.8×1031 atoms$ .

(c)

To determine

The total energy available.

(c)

Expert Solution

Answer to Problem 10P

The total energy available is

$2.6×1021 J$ .

Explanation of Solution

Given Info:

The energy per atom is $208 MeV/atom$ .

Formula to calculate the total energy available is,

$E=NE0$

$E$ is the energy
$E0$ is the energy per atom

Substitute $208 MeV/atom$ for $E0$ and $7.87×1031 atom$ for N to find $E$ .

$E=(208 MeV/atom)(7.87×1031 atom)(1.60×10−13 J1 MeV)=2.6×1021 J$

Thus, the total energy available is $2.6×1021 J$ .

Conclusion:

The total energy available is $2.6×1021 J$ .

(d)

To determine

The maximum time energy supply lasts.

(d)

Expert Solution

Answer to Problem 10P

The maximum time energy supply lasts is

$5.5 yr$ .

Explanation of Solution

Given Info:

The energy consumption rate is $1.5×1013 J/s$ .

Formula to calculate maximum time energy supply lasts is,

$t=EP$

$t$ is maximum time energy supply lasts
$P$ is the energy consumption rate

Substitute $2.6×1021 J$ for $E$ and $1.5×1013 J/s$ for $P$ to find $t$ .

$t=(2.6×1021 J1.5×1013 J/s)(1 yr3.156×107 s)=5.5 yr$

Thus, the maximum time energy supply lasts is $5.5 yr$ .

Conclusion:

The maximum time energy supply lasts is $5.5 yr$

(e)

To determine

The conclusion of the results.

(e)

Expert Solution

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

(a)

Expert Solution

Answer to Problem 10P

The mass of uranium in reserve is

$3.1×1010 g$ .

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70%$ .

The amount of uranium reserved is $4.4×106$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m235U=Δpm$

$m235U$ is the difference in time between the cars to reach the destination
$Δp$ is the percentage of mass reserved m is the speed of the slower car
m is the mass of uranium in reserve

Substitute $0.70%$ for $Δp$ and $4.4×106$ for m to find $m235U$ .

$m235U=(0.70%(1100%))(4.4×106 metric ton)(103 kg1 ton)(103 g1 kg)=3.1×1010 kg$

Thus, the mass of uranium in reserve is $3.1×1010 g$ .

Conclusion:

The mass of uranium in reserve is $3.1×1010 g$ .

(b)

To determine

The number of moles and atoms in reserve.

(b)

Expert Solution

Answer to Problem 10P

The number of moles in reserve is

$1.3×108 mol$ number of atoms in reserve is

$7.8×1031 atoms$ .

Explanation of Solution

Given Info:

The mass per one mole is $235 g/mol$ .

The Avogadro number is $6.02×1023 atoms/mol$ .

Formula to calculate the number of moles is,

$n=m235UMmol$

$n$ is the number of moles
$Mmol$ is mass per mol

Substitute $235 g/mol$ for $Mmol$ and $3.1×1010 g$ for $m235U$ to find n.

$n=3.1×1010 g235 g/mol=1.3×108 mol$

Thus, the number of moles is $1.3×108 mol$ .

Formula to calculate the number of atoms is,

$N=nNA$

$N$ is the number of moles
$NA$ is the Avogadro number

Substitute $1.3×108 mol$ for n and $1.3×108 mol$ for $NA$ to find N.

$N=(1.3×108 mol)(6.02×1023 atoms/mol)=7.8×1031 atoms$

Thus, the number of atoms is $7.8×1031 atoms$ .

Conclusion:

The number of moles in reserve is $1.3×108 mol$ number of atoms in reserve is $7.8×1031 atoms$ .

(c)

To determine

The total energy available.

(c)

Expert Solution

Answer to Problem 10P

The total energy available is

$2.6×1021 J$ .

Explanation of Solution

Given Info:

The energy per atom is $208 MeV/atom$ .

Formula to calculate the total energy available is,

$E=NE0$

$E$ is the energy
$E0$ is the energy per atom

Substitute $208 MeV/atom$ for $E0$ and $7.87×1031 atom$ for N to find $E$ .

$E=(208 MeV/atom)(7.87×1031 atom)(1.60×10−13 J1 MeV)=2.6×1021 J$

Thus, the total energy available is $2.6×1021 J$ .

Conclusion:

The total energy available is $2.6×1021 J$ .

(d)

To determine

The maximum time energy supply lasts.

(d)

Expert Solution

Answer to Problem 10P

The maximum time energy supply lasts is

$5.5 yr$ .

Explanation of Solution

Given Info:

The energy consumption rate is $1.5×1013 J/s$ .

Formula to calculate maximum time energy supply lasts is,

$t=EP$

$t$ is maximum time energy supply lasts
$P$ is the energy consumption rate

Substitute $2.6×1021 J$ for $E$ and $1.5×1013 J/s$ for $P$ to find $t$ .

$t=(2.6×1021 J1.5×1013 J/s)(1 yr3.156×107 s)=5.5 yr$

Thus, the maximum time energy supply lasts is $5.5 yr$ .

Conclusion:

The maximum time energy supply lasts is $5.5 yr$

(e)

To determine

The conclusion of the results.

(e)

Expert Solution

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Answer 6

Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

(a)

Expert Solution

Answer to Problem 10P

The mass of uranium in reserve is

$3.1×1010 g$ .

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70%$ .

The amount of uranium reserved is $4.4×106$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m235U=Δpm$

$m235U$ is the difference in time between the cars to reach the destination
$Δp$ is the percentage of mass reserved m is the speed of the slower car
m is the mass of uranium in reserve

Substitute $0.70%$ for $Δp$ and $4.4×106$ for m to find $m235U$ .

$m235U=(0.70%(1100%))(4.4×106 metric ton)(103 kg1 ton)(103 g1 kg)=3.1×1010 kg$

Thus, the mass of uranium in reserve is $3.1×1010 g$ .

Conclusion:

The mass of uranium in reserve is $3.1×1010 g$ .

Answer 7

Chapter 30, Problem 10P

(a)

To determine

Mass of uranium in reserve.

Answer 8

(a)

Expert Solution

Answer to Problem 10P

The mass of uranium in reserve is

$3.1×1010 g$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given Info:

The percentage of mass reserved is $0.70%$ .

The amount of uranium reserved is $4.4×106$ metric ton.

Formula used to calculate the mass of uranium reserved is,

$m235U=Δpm$

$m235U$ is the difference in time between the cars to reach the destination
$Δp$ is the percentage of mass reserved m is the speed of the slower car
m is the mass of uranium in reserve

Substitute $0.70%$ for $Δp$ and $4.4×106$ for m to find $m235U$ .

$m235U=(0.70%(1100%))(4.4×106 metric ton)(103 kg1 ton)(103 g1 kg)=3.1×1010 kg$

Thus, the mass of uranium in reserve is $3.1×1010 g$ .

Conclusion:

The mass of uranium in reserve is $3.1×1010 g$ .

Answer 13

(b)

To determine

The number of moles and atoms in reserve.

(b)

Expert Solution

Answer to Problem 10P

The number of moles in reserve is

$1.3×108 mol$ number of atoms in reserve is

$7.8×1031 atoms$ .

Explanation of Solution

Given Info:

The mass per one mole is $235 g/mol$ .

The Avogadro number is $6.02×1023 atoms/mol$ .

Formula to calculate the number of moles is,

$n=m235UMmol$

$n$ is the number of moles
$Mmol$ is mass per mol

Substitute $235 g/mol$ for $Mmol$ and $3.1×1010 g$ for $m235U$ to find n.

$n=3.1×1010 g235 g/mol=1.3×108 mol$

Thus, the number of moles is $1.3×108 mol$ .

Formula to calculate the number of atoms is,

$N=nNA$

$N$ is the number of moles
$NA$ is the Avogadro number

Substitute $1.3×108 mol$ for n and $1.3×108 mol$ for $NA$ to find N.

$N=(1.3×108 mol)(6.02×1023 atoms/mol)=7.8×1031 atoms$

Thus, the number of atoms is $7.8×1031 atoms$ .

Conclusion:

The number of moles in reserve is $1.3×108 mol$ number of atoms in reserve is $7.8×1031 atoms$ .

Answer 14

(b)

To determine

The number of moles and atoms in reserve.

Answer 15

(b)

Expert Solution

Answer to Problem 10P

The number of moles in reserve is

$1.3×108 mol$ number of atoms in reserve is

$7.8×1031 atoms$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given Info:

The mass per one mole is $235 g/mol$ .

The Avogadro number is $6.02×1023 atoms/mol$ .

Formula to calculate the number of moles is,

$n=m235UMmol$

$n$ is the number of moles
$Mmol$ is mass per mol

Substitute $235 g/mol$ for $Mmol$ and $3.1×1010 g$ for $m235U$ to find n.

$n=3.1×1010 g235 g/mol=1.3×108 mol$

Thus, the number of moles is $1.3×108 mol$ .

Formula to calculate the number of atoms is,

$N=nNA$

$N$ is the number of moles
$NA$ is the Avogadro number

Substitute $1.3×108 mol$ for n and $1.3×108 mol$ for $NA$ to find N.

$N=(1.3×108 mol)(6.02×1023 atoms/mol)=7.8×1031 atoms$

Thus, the number of atoms is $7.8×1031 atoms$ .

Conclusion:

The number of moles in reserve is $1.3×108 mol$ number of atoms in reserve is $7.8×1031 atoms$ .

Answer 20

(c)

To determine

The total energy available.

(c)

Expert Solution

Answer to Problem 10P

The total energy available is

$2.6×1021 J$ .

Explanation of Solution

Given Info:

The energy per atom is $208 MeV/atom$ .

Formula to calculate the total energy available is,

$E=NE0$

$E$ is the energy
$E0$ is the energy per atom

Substitute $208 MeV/atom$ for $E0$ and $7.87×1031 atom$ for N to find $E$ .

$E=(208 MeV/atom)(7.87×1031 atom)(1.60×10−13 J1 MeV)=2.6×1021 J$

Thus, the total energy available is $2.6×1021 J$ .

Conclusion:

The total energy available is $2.6×1021 J$ .

Answer 21

(c)

To determine

The total energy available.

Answer 22

(c)

Expert Solution

Answer to Problem 10P

The total energy available is

$2.6×1021 J$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given Info:

The energy per atom is $208 MeV/atom$ .

Formula to calculate the total energy available is,

$E=NE0$

$E$ is the energy
$E0$ is the energy per atom

Substitute $208 MeV/atom$ for $E0$ and $7.87×1031 atom$ for N to find $E$ .

$E=(208 MeV/atom)(7.87×1031 atom)(1.60×10−13 J1 MeV)=2.6×1021 J$

Thus, the total energy available is $2.6×1021 J$ .

Conclusion:

The total energy available is $2.6×1021 J$ .

Answer 27

(d)

To determine

The maximum time energy supply lasts.

(d)

Expert Solution

Answer to Problem 10P

The maximum time energy supply lasts is

$5.5 yr$ .

Explanation of Solution

Given Info:

The energy consumption rate is $1.5×1013 J/s$ .

Formula to calculate maximum time energy supply lasts is,

$t=EP$

$t$ is maximum time energy supply lasts
$P$ is the energy consumption rate

Substitute $2.6×1021 J$ for $E$ and $1.5×1013 J/s$ for $P$ to find $t$ .

$t=(2.6×1021 J1.5×1013 J/s)(1 yr3.156×107 s)=5.5 yr$

Thus, the maximum time energy supply lasts is $5.5 yr$ .

Conclusion:

The maximum time energy supply lasts is $5.5 yr$

Answer 28

(d)

To determine

The maximum time energy supply lasts.

Answer 29

(d)

Expert Solution

Answer to Problem 10P

The maximum time energy supply lasts is

$5.5 yr$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given Info:

The energy consumption rate is $1.5×1013 J/s$ .

Formula to calculate maximum time energy supply lasts is,

$t=EP$

$t$ is maximum time energy supply lasts
$P$ is the energy consumption rate

Substitute $2.6×1021 J$ for $E$ and $1.5×1013 J/s$ for $P$ to find $t$ .

$t=(2.6×1021 J1.5×1013 J/s)(1 yr3.156×107 s)=5.5 yr$

Thus, the maximum time energy supply lasts is $5.5 yr$ .

Conclusion:

The maximum time energy supply lasts is $5.5 yr$

Answer 34

(e)

To determine

The conclusion of the results.

(e)

Expert Solution

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Answer 35

(e)

To determine

The conclusion of the results.

Answer 36

(e)

Expert Solution

Answer to Problem 10P

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium since the energy produced will last only 5.5 years for a large amount of uranium.

Conclusion:

Fission alone cannot meet the world’s energy needs at a price of $130 or less per kilogram of uranium.