Question 970274: One gram of Pu-238 generates 0.56 watts of thermal power and has a half life of 87.7 years. The generator that converts thermal power to to electrical power for the probe's instruments operates at 6% efficiency(i.e. 1 watt of thermal power generates 0.06 watts of electrical power). The instruments on the spacecraft require 85 watts of electrical power and need to operate for 19 years to complete the mission.
How many whole grams of Pu-238 are required to complete the mission?
I assumed I should be using either the half life A=Ao(1/2)^(t/h)
But is the A grams or watts?
Or should I be using the interest formula?
I am having trouble setting up the problem. Thanks!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! One gram of Pu-238 generates 0.56 watts of thermal power and has a half life of 87.7 years. The generator that converts thermal power to to electrical power for the probe's instruments operates at 6% efficiency(i.e. 1 watt of thermal power generates 0.06 watts of electrical power). The instruments on the spacecraft require 85 watts of electrical power and need to operate for 19 years to complete the mission.
How many whole grams of Pu-238 are required to complete the mission?
:
Since the half-life is 87 yrs and only has to operate for 19 yrs, that part of it does not come into the problem
:
let g = no. of grams required to generate 85 watts
.56 * .06 * g = 85
.0336g = 85
g = 85/.0336
g = 2530 grams or 2.53 kg required
:
Thought about this one this morning, perhaps they want the initial amt of P-238 so they would still have 2530 grams at the end of 19 yrs. Let that amt = Ao
Ao*2^(-19/87.7) = 2530
use your calc to find 2^(-19/87.7)
Ao*.86056 = 2530
Ao = 2530/.86056
Ao = 2940 grams required initially, Sorry, should have realized this yesterday
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