How Long Will the Uranium Supply Last?

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SUMMARY

The discussion focuses on the estimation of how long the Earth's supply of natural uranium, specifically 235U, will last if utilized for energy production. The total economically mineable uranium is estimated at 1.0 x 10^9 tons, which translates to approximately 343,000 tons of usable 235U. The energy output from fission events is calculated to be 5.33 x 10^26 eV/kg, leading to a total energy of 2.66 x 10^23 J. However, the initial calculations contained errors in unit conversions, particularly between grams and kilograms, affecting the final estimate of 1204 years.

PREREQUISITES
  • Understanding of nuclear fission and energy calculations
  • Familiarity with Avogadro's number and its application in chemistry
  • Knowledge of energy unit conversions, specifically eV to Joules
  • Basic grasp of isotopes, particularly uranium-235
NEXT STEPS
  • Review nuclear fission energy calculations and unit conversions
  • Study the properties and abundance of uranium isotopes, focusing on 235U
  • Learn about the implications of uranium supply on energy policy and sustainability
  • Explore advanced topics in nuclear physics, including reactor design and efficiency
USEFUL FOR

This discussion is beneficial for physics students, nuclear energy researchers, and environmental scientists interested in the sustainability of uranium resources and energy production methods.

proudtobeavol
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I'm seriously lost!

Here's the problem:
It has been estimated that the Earth contains 1.0 x10^9 tons of natural uranium that can be mined economically. If all the world's energy needs (7.0 x10^12 J/s) were supplied by 235(U) fission, how long would this supply last? Assume that the average energy released in a fission event is 208 MeV. (Hint: See Appendix B for the percent abundance of _92^235(U).)

My (pathetic attempt at a) solution

OK, I've converted to 343000 tons of usable 235U and that this is equal to 3.11x10^12g which, using avogadro's number equals 7.97x10^33 nuclei available for a reaction. this converts to 1.65x10^36MeV which converts to 5.33x10^32 ev/kg

Now I need to get from here to how many years this would last. I tried multiplying the number of kg on hand (3111643658) by 5.33x10^32 ev/kg which gives me 1.66x10^42ev and since there are 6.24x10^18 J in an ev this gives me 2.66x10^23J

BUT I tried to go from there to 1s/7x10^12J=3.80x10^10s which converted to 1204 years... but it's wrong.

Does anyone see where I went wrong?

Much appreciation!
Shanna
(Who is hugely grateful that this is the LAST physics problem I will ever have to work!)
 
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You intersperse grams with kilograms...plus you may have transposed some digits...so you're off by factors of 10.

Consider, you get 5.33x10^32 ev/kg ... but it is really 5.33x10^23 ev/g, which is 5.33x10^26 ev/kg.
 

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