Difference between revisions of "Giancoli Physics (5th ed) Chapter 31"

Revision as of 06:56, 12 March 2009 (view source) Cmurray (talk | contribs) (/* 20. Suppose that the average power consumption, day and night, in a typical house is 300 W. What initial mass of U-235 would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fissi) ← Older edit		Revision as of 16:21, 9 March 2010 (view source) Cmurray (talk | contribs) (/* 20. Suppose that the average power consumption, day and night, in a typical house is 300 W. What initial mass of U-235 would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fissi) Newer edit →
Line 136:		Line 136:
	To find the number of fissions, we use:<br><br>		To find the number of fissions, we use:<br><br>
−	P = E / t<br>	+	Power = Energy / time    (1 Watt = 1 J/s)<br>
	300 W = E/(3.16E7 s/yr)<br />		300 W = E/(3.16E7 s/yr)<br />
	E = 9.48E9 J<br>		E = 9.48E9 J<br>

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Revision as of 16:21, 9 March 2010

Main Page > Giancoli Physics (5th ed) Solutions > Giancoli Physics (5th ed) Chapter 31

Problems

1. Question

²⁸/₁₃Al
B- emitter
²⁸/₁₄Si

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2. Question

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3. Question

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4. Question

17.35 MeV is released

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5. Question

+ 18.000953 u

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6. Question

5.700 MeV is released

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17. Calculate the energy released in the fission reaction n + 92-U-235 -> 38-Sr-88 + 54-Xe-136 + 12n. Use Appendix F; assume that the initial KE of the neutron is small.

To calculate the released energy, we use the formula:

Q = [m(²³⁵U) + m(n) - 12m(n) - m(⁸⁸Sr) - m(¹³⁶Xe)]c²

Q = [(235.043924 u) - 11(1.008665 u) - (87.905618 u) - (135.90721 u)]c²(931.5 MeV / uc²) = 126.5 MeV

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18. Question

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20. Suppose that the average power consumption, day and night, in a typical house is 300 W. What initial mass of U-235 would have to undergo fission to supply the electrical needs of such a house for a year? (Assume 200 MeV is released per fission)

To find the number of fissions, we use:

Power = Energy / time (1 Watt = 1 J/s)
300 W = E/(3.16E7 s/yr)
E = 9.48E9 J
so the number of fissions you need is:
N = (9.48E9 J)/[(200E6 eV)(1.602E-19 J/eV)]

N = 2.96E20 fissions

Using this number, knowing each fission requires one uranium atom, we can set up the equation:

m = [(2.96E20 atoms) / (6.02E23 atoms/mol)](235 g/mol) = .116 g

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