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- Attachment 655378The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

The mark scheme is a shambles so i really need help in understanding what on earth i have to do please

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(Original post by

what have you done so far ?

**Uni12345678**)what have you done so far ?

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#4

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ive only changed the 1600 MW to Joules but the mark scheme says that the 1600 MW is what is does at 100% efficiency then proceeds to multiply ot by 4 and call the answer 25% efficiency ? or ive completely misunderstood it :/ either way im stumped

**UFOEEK**)ive only changed the 1600 MW to Joules but the mark scheme says that the 1600 MW is what is does at 100% efficiency then proceeds to multiply ot by 4 and call the answer 25% efficiency ? or ive completely misunderstood it :/ either way im stumped

so that gives 400 MJ... then, I got 1.26 x 10^6 J released in one year

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#5

(Original post by

**UFOEEK**)- The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

The mark scheme is a shambles so i really need help in understanding what on earth i have to do please

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#6

I'm not entirely sure if this is how you're meant to go about this, but I'll try.

Converting the 200MeV to joules gives us 3.2043546e-11J. So dividing 1600000000J by that gives us about 5*10^19. This is the number of uranium-238 nuclei required per second. To convert this to years, multiply by 60*60*24*365 to get 1.6*10^27.

Now to convert this from 3% to 100%. Divide by 3 and multiply by 100 to get 5.25*10^28. And because the efficiency is only 25%, multiply by a final 4 to get 2.1*10^29 atoms of uranium.

Divide this by Avogadro's constant to get the number of moles. This gives you 348637.4 moles of uranium. Divide by the relative atomic mass to get 1464g, or 1.464kg per year.

Converting the 200MeV to joules gives us 3.2043546e-11J. So dividing 1600000000J by that gives us about 5*10^19. This is the number of uranium-238 nuclei required per second. To convert this to years, multiply by 60*60*24*365 to get 1.6*10^27.

Now to convert this from 3% to 100%. Divide by 3 and multiply by 100 to get 5.25*10^28. And because the efficiency is only 25%, multiply by a final 4 to get 2.1*10^29 atoms of uranium.

Divide this by Avogadro's constant to get the number of moles. This gives you 348637.4 moles of uranium. Divide by the relative atomic mass to get 1464g, or 1.464kg per year.

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#7

**UFOEEK**)

- The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

The mark scheme is a shambles so i really need help in understanding what on earth i have to do please

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(Original post by

what paper is this?

**localwhiteasian**)what paper is this?

im assuming theyre all past paper qus from different specs in some cases ?

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could you post a photo of the mark scheme?

**Uni12345678**)could you post a photo of the mark scheme?

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#13

(Original post by

i think i already did but ill do so again

**UFOEEK**)i think i already did but ill do so again

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(Original post by

Oh you multiply by 4 because 1600 MW is the output power not the input

**Uni12345678**)Oh you multiply by 4 because 1600 MW is the output power not the input

but then what

i kind of get what theyve done with the rest of it but i dont get how they got the energy from 1 kilo of fuel

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#15

(Original post by

**UFOEEK**)- Attachment 655378The question is when a Uranium-235 nucleus undergoes fission ,approximetly 200 MeVof energy is released . Estimate the total mass of original fuel required per year in a 1600 MW nuclear reactor that uses enriched fuel containing 3% uranium -235 and 97% uranium-238 and operates at an efficiency of 25% .

The mark scheme is a shambles so i really need help in understanding what on earth i have to do please

So we know that 1 U-235 Nucleus releases 200MeV. I converted this into Joules

200 MeV = 200 x 1.6 x 10^-13 =

**3.2 x 10^-11 J <= 1 U-235 nucleus**

So to produce 1600 MW for 1 year ==> P = 1600 x 10^6 W and t = 365 x 24 x 60 x60

using E = P x t

E = (1600 x 10^6) x (365 x 24 x 60 x 60) =

**5.04576 x 10^16 J of energy required**

so we need

**5.04576 x 10^16 / 3.2 x 10^-11 =**1.5768 x 10^27 U-235 nuclei

to produce 1600MW in 1 year

**so now we can calculate the mass of this amount by calculating the number of moles using avogadro's constant.**

1.5768 x 10^27 / 6.02 x 10^23 = 2619 moles of U-235

1 mole of U-235 is 0.235kg

there 2619 moles = 2619 x 0.235 =

**615.5kg of U-235**

**but since the fuel rods contain 3% of this U-235 then the total amount = 615.5 / 0.03 = 20520 kg**

**Now since its 25% effcient then we need 4 times this amount to achieve this output**

**20520 x 4 = 8 x 10^4**

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(Original post by

So let me show you how I did it. I done it slightly different to the mark scheme

So we know that 1 U-235 Nucleus releases 200MeV. I converted this into Joules

200 MeV = 200 x 1.6 x 10^-13 =

So to produce 1600 MW for 1 year ==> P = 1600 x 10^6 W and t = 365 x 24 x 60 x60

using E = P x t

E = (1600 x 10^6) x (365 x 24 x 60 x 60) =

so we need

to produce 1600MW in 1 year

1.5768 x 10^27 / 6.02 x 10^23 = 2619 moles of U-235

1 mole of U-235 is 0.235kg

there 2619 moles = 2619 x 0.235 =

**PawanAviator**)So let me show you how I did it. I done it slightly different to the mark scheme

So we know that 1 U-235 Nucleus releases 200MeV. I converted this into Joules

200 MeV = 200 x 1.6 x 10^-13 =

**3.2 x 10^-11 J <= 1 U-235 nucleus**So to produce 1600 MW for 1 year ==> P = 1600 x 10^6 W and t = 365 x 24 x 60 x60

using E = P x t

E = (1600 x 10^6) x (365 x 24 x 60 x 60) =

**5.04576 x 10^16 J of energy required**so we need

**5.04576 x 10^16 / 3.2 x 10^-11 =**1.5768 x 10^27 U-235 nucleito produce 1600MW in 1 year

**so now we can calculate the mass of this amount by calculating the number of moles using avogadro's constant.**1.5768 x 10^27 / 6.02 x 10^23 = 2619 moles of U-235

1 mole of U-235 is 0.235kg

there 2619 moles = 2619 x 0.235 =

**615.5kg of U-235****but since the fuel rods contain 3% of this U-235 then the total amount = 615.5 / 0.03 = 20520 kg****Now since its 25% effcient then we need 4 times this amount to achieve this output****20520 x 4 = 8 x 10^4**
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#17

(Original post by

i think i already did but ill do so again

**UFOEEK**)i think i already did but ill do so again

200 Ã— 1.6 Ã— 10

^{-13}is converting 200 MeV to the unit of joules.

(200 Ã— 10

^{6}eV) Ã— (1.6 Ã— 10

^{-19}J/eV) is simplified to 200 Ã— 1.6 Ã— 10

^{-13}

1/0.238 Ã— 0.03 is to find out the number of moles Uranium-235 in 1 kg of fuel. The need of times 0.03 is because you are told the following:

...enriched fuel containing 3% uranium -235 and 97% uranium-238 ....

Thus in 1 kg fuel, only 0.03 kg is uranium -235.

1/0.238 Ã— 0.03 Ã— 6.0 Ã— 10

^{23}is to find out how many uranium -235 nuclei in 1 kg of the fuel.

Hope it helps.

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