Assuming perfect efficient transfer of energy,how long would it take 2.6Kg of 239PU to turn 4.2Kg of ice at -10 celsius into 4.2Kg of H20 vapour with a temperature of 128 celsius?How much mass is lost each time one of these atoms decay?  -primary isotopes involved in nuclear decay are 239Pu --> 235U

Expert Answers

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First let us compute how much energy one needs to turn the ice into vapour.

`Q = m_(ice)*c_(ice)*10 +m_(ice)*lambda_(ice) +m_(water)*c_(water)*100 +m_(water)*lambda_(water) +m_(vapour)*c_(vapour)*28`

`Q =4.2*2110*10 +4.2*334*10^3+4.2*4180*100+4.2*2260*10^3+4.2*2080*28=`

`=1.298*10^7 J =13 MJ `

Second let us write the fission equation of (239)Pu

`(239)Pu = (235)U + (4)He`

The decay heat per mass unit for this reaction is known

`Q/(m*t) =P/m = 1.9 W/(kg)`

Therefore for a total mass `m =2.6 kg` of fissionable material we get

`Q/t = P= 2.6*1.9 =4.94 W`

Now the total time that is needed to transform ice into vapour comes from

`P*t = Q`

`t=Q/P =(13*10^6)/4.94 =2.63*10^6 seconds=731 hours =30.46 days`

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