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et safari File Edit View History Bookmarks Develop Window Help 27% SunMay 18 10:52 AM Natasha Ell way a : Pacific Union College PHYS 113 - Spring14 Andrianarijaona: HW: Nuclear Energy 1 | || | + www.sa plinglearning.com /ibiscms/mod/ibis/view.php?id=1136125 Reader :: Sapling Lear. Instruction Mastering Biology Log Into Canvas Pacific Unio n Education Apple facebook Yahoo! Google Maps YouTube Wikipedia News Popular Selected Masses - Physindex Pacific Union College PHYS 113 Spring... Files: PHYS-113L-01 LAB (Andrianarijaona. graduation post card - Google Search Gradebook Assignment Information Available From: 5/16/2014 10:45 Due Date: Points Possible: 100 Grade Category: Graded Description: Policies: # Attempts Score Due Date: 5/19/2014 09:00 AM Print Calculator Periadic Table Question 4 of 4 5/19/2014 09:00 Map Map och 2 48 sapling learning The average energy released by the fission of a single atom of uranium-235 is approximately 205 MeV How much total energy would be released by the complete fission of 4.35 kg of 235U? Express your answer in units of Joules 4 Number You can check your answers. You can view solutions when you up on any question. You can keep trying to answer each you get it right or give up A typical small town might use about 28.5 MW of power, on average. How long, in days, would it take such a town to use the amount of energy calculated above? You lose 5% of the points available to in your question for each incorrect att answer. Number days Help With This Topic Web Help & Video:s Technical Support and Bug Re Previous Give Up & View Solution O Check Answer xt Exit Hint 18Explanation / Answer
Average energy released by the fission of a single atom of Uranium-235
is approximately 205 Mev= 205*106*1.6*10-19 J=0.328*10-10 J
( Here M= mega=106 and ev= electron volt =1.6*10-19 J)
4.35 Kg Uranium-235 is 4350/235=18.51 mole Uranium= 18.51*6.022*1023 number of atoms
1 atom gives 0.328 J
So, 18.51*6.022*1023 number of atoms gives 18.51*6.022*1023*0.328*10-10= 36.56*1013 J
28.5 MW power is using 28.5*106 J energy in 1 second.
Now, seconds taken to use the power = 36.56*1013 /(28.5*106)
So, days taken to use that power=36.56*1023 /(28.5*106) / 86400 =148.48 days
(as a day has 24*60*60 =86400 seconds )
So, rounding off to the next day, as it takes half day more to finish after the 148th day it takes actually 149 days to use that power.
So, that's the way we do this!