Heat released from the first-order, radioactive decay of plutonium-238 (238Pu) has been proposed as a steady..? - steady heat tech deck
Heat is released from the first order, was the radioactive decay of plutonium-238 (238Pu) as a constant source of heat for proposed long-term applications. 238Pu half-life of 86 years. What fraction of the heat flow of the origin would be a 238Pu-powered device to produce during recovery from a remote location 43 years after the implementation?
Tuesday, February 2, 2010
Steady Heat Tech Deck Heat Released From The First-order, Radioactive Decay Of Plutonium-238 (238Pu) Has Been Proposed As A Steady..?
1 comment:
This is the equation of a first-order reaction:
[A l] = [A] o * e ^ (-kt)
[I] n this case, the heat flow after a certain time
[A] o is the heat flow of
k is the rate
t is the time
To find k, use the information you provided on the half-life of 238Pu. Use the following equation:
t1 / 2 = ln (2) / k. The half-life (t1 / 2) is 86 years, so
86 = ln (2) / k me -----> k-value of 0.00806.
Now plug it in, just what you know in the rate equation of first order:
[A l] = [A] o * e ^ (-0.00806 * 43)
The percentage of heat flow, the home would be the expression [A t] / [A] o, because if something breaks theorhetically over time, the less and less of it than the par value.
How to solve for the [A t] / [A] o.
[A l] / [A] o = e ^ (-0.00806 * 43) = 0.7071 = 70.71%. This is the answer.
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