7. Nuclear energy question: (show your work) a. If thorium (molecular weight = 2

Question

7. Nuclear energy question: (show your work)

a. If thorium (molecular weight = 229 grams per mol) is being used as fuel in a nuclear power plant and the amount of thorium fuel needed is 0.5 grams every second (the whole fuel, not the mass lost), what is the amount of energy produced in one month (30 days)? It is known that 0.03 g are lost when 1 mol of thorium fuel is consumed. Answer: 1.53x1016J

b. The spent thorium fuel is highly radioactive, but some estimates show that it can decay into a non-radioactive material. The half-life of the spent fuel is estimated to be at 560 years. The facility storing the radioactive waste needs to know for how long they need to worry about radioactive material. How long will it take for 98% of the thorium fuel consumed to decay to non-radioactive material? Answer: 3154.9yr

Explanation / Answer

a) given

time = 30 days

time = 30 x 24 hr

time = 30 x 24 x 60 min

time = 30 x 24 x 60 x 60 sec

time = 2.592 x 10^6 sec

given

0.5 grams of thorium is required per second

now

amount of thorium required for 30 days = 0.5 x 2.592 x 10^6 g

amount of thorium required for 30 days = 1.296 x 10^6 grams

now

we know that

moles = mass / molar mass

molar mass of thorium = 229 g/mol

so

moles of thorium required for 30 days = 1.296 x 10^6 / 229

moles of thorium required for 30 days = 5659.4

now

0.03 of mass is lost per 1 mol of thorium used

so

amount of mass lost for 30 days = 5659.4 x 0.03 g

amount of mass lost for 30 days = 169.78 g

amount of mass lost for 30 days = 0.16978 kg

amount of mass lost for 30 days = 0.17 kg

now

we know that

E = dm x c2

here

c = speed of light = 3 x 10^8

so

E = 0.17 x (3 x 10^8)^2

E = 1.53 x 10^16 J

so

the energy produced is 1.53 x 10^16 J

b)

we know that

for radioactive decay

N= No x e^(-kt)

No/N = e^(kt)

ln (No/N) = kt

now

we know that

decay constant (k) = ln2 / half life

given

half life = 560 yr

so

k = ln2 / 560

k = 1.23776 x 10-3

k = 1.24 x 10-3 yr-1

now

given

98% of the thorium is coverted to non radioactive material

so

2 % of the thorium is left

now

consider initial amount of thorium (No) = 100

then final amount of thorium (N) = 2

now

ln (No/N) = kt

ln (100/2) = 1.24 x 10-3 x t

t = 3154.857

t = 3154.9

so

the time taken is 3154.9 years

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