How Does Uranium Decay Impact the Temperature of Surrounding Lead?

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SUMMARY

The discussion focuses on the thermal impact of uranium-238 decay on surrounding lead, specifically examining the temperature change of 120.0 kg of lead after 3.8 million years due to the decay of 1.00 g of uranium. The decay process converts a mass of 6.83x10-8 kg to energy, calculated using Einstein's equation e=mc2. An initial calculation yielded an implausible temperature of 394,035 degrees Celsius, prompting a request for clarification on the methodology used to arrive at this result.

PREREQUISITES
  • Understanding of nuclear decay processes, specifically uranium-238.
  • Familiarity with Einstein's mass-energy equivalence formula, e=mc2.
  • Knowledge of half-life calculations, particularly for uranium-238 (4.7 billion years).
  • Basic principles of thermodynamics related to heat transfer and temperature change.
NEXT STEPS
  • Review the principles of nuclear decay and its implications on surrounding materials.
  • Study the application of the mass-energy equivalence formula in practical scenarios.
  • Learn about thermal conductivity and heat transfer in solid materials like lead.
  • Explore advanced decay calculations using different isotopes and their thermal effects.
USEFUL FOR

Students in physics or engineering, nuclear scientists, and anyone interested in the thermal effects of radioactive decay on surrounding materials.

Zoey Brown
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1. The
Lead is the final decay product of uranium-238 (half life = 4.7 billion years), so often the

uranium is embedded in lead. The decay of 1.00 g of uranium to thorium converts

6.83x10-8 kg of mass to energy. Assuming the uranium absorbs none of the heat, what

would be the temperature change of 120.0 kg of lead surrounding 1.00 g of uranium after

3.8 million years?

Homework Equations


e=mc^2
mf=mi(0.5)^t/h

The Attempt at a Solution


i got a really unreasonable answer; 394 035 degrees celsius
 
Physics news on Phys.org
Zoey Brown said:
i got a really unreasonable answer; 394 035 degrees celsius
Be helpful if you showed HOW you got that result.
 

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