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Amount of energy generated from fusion

  1. Jul 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Assuming that we could generate Australia's electricity from a fusion reactor, that converted hydrogen to iron and turned the energy into electricity with 100% efficiency, what mass of fuel would it use per year? You may assume that this fusion reaction converts 1% of the mass into energy. Australia uses 225 billion kilo-watt hours of electricity per year which is 8.2*10^17 Joules.

    2. Relevant equations
    E=mc^2

    3. The attempt at a solution
    m = (8.2*10^17)/c^2
    => m = 10 kg (approx.)
    Since the fusion converts only 1% mass into energy, I multiply the result by 100 to get 1000 kg.

    While my answer is within the margin of error (the real answer is 900 kg.), my way of going about solving the problem might be wrong as the answer given is:
    "Divide the energy by 1% of the speed of light squared, to get the mass needed."

    What I don't understand is: why would we wanna divide it by 1% the speed of light? What's the logic behind that? We obviously are not changing the speed of light, just reducing the efficiency, so, wouldn't increasing the total energy required (as I did) be a more logical approach?
     
  2. jcsd
  3. Jul 20, 2016 #2

    haruspex

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    It is strangely worded, but if we read it as "divide by 1% of ((speed of light)2), that is the same as dividing by c2 to get the mass lost, and then multiplying by 100 to get the total mass of fuel, as you did.
    One source of discrepancy is that the conversion from kWh to J that they provided is not quite right, but it's only a 1.25% error. But you seem to have made a larger error in your division by c2.
     
  4. Jul 20, 2016 #3
    My error comes from dividing 8.2 by 3 * 3. I divided them both to get one, while now that I actually do the division, my answer is coming to 911 which is much closer to 900 than my previous answer of 1000 was. And from your comment, doesn't my way of solving seem more intuitive since we apply that 1% error in the entire energy and not in the speed of light?
     
  5. Jul 20, 2016 #4

    haruspex

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    If you ignore the given 8.2 and start from the 225 bn kWh you will get even closer to 900.
    It's hard to say what their way of solving it was. They specify an arithmetic process, but not the logic behind it.
     
  6. Jul 20, 2016 #5
    Exactly! So I suppose I'll stick to my logic and understand the solution that way, cool?
     
  7. Jul 20, 2016 #6

    haruspex

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    Sure.
     
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