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Energy emitted from Sun

  1. May 18, 2003 #1
    Hi
    I have the following question
    4Hydrogen->Helium+2electrons+2neutrino+gamma

    In the sun, Hydrogen is converted into helium(see equation above) and that the reaction releases about 26.7MeV of energy. the sun radiates energy at the rate of 3.9*10^26W and has a mass about 1.99*10^30 kg of which 75% is hydrogen. Find out how long it will take the sun to convert 12% of its hydrogen into helium.

    I'm abit blocked. anyhelp is appreciated
    I think i have the right ideas.
    For Four hydrogen into helium is 26.7Mev or 4.272*10^-12Joules.
    now a hydrogen is about 1.008 or 1.0u or 1* (1.66*10^-27)kg.
    so I do, (1.99*10^30) *.75=1.425*10^30kg and
    i do answer * .12 to get those 12% of hydrogen to helium
    1.42*10^30 * .12 = 1.79*10^29kg.
    so i can do (1.79*10^29)/(1.66*10^-27) to get the number of Hydrogen
    so i get 1.08*10^56 hydrogen.
    since 4 Hydrogen into helium i divide by the total number of hydrogen to get
    (1.08*10^56)/4=2.7*10^55
    i multiply 2.7*10^55 by 26.7MeV to get the total energy output.
    so i get
    (2.7*10^55)*(4.272*10^-12)=1.15344*10^44

    and since
    watts=J/S, so i do 1/(3.9*10^26) * (1.15344*10^44) and i get
    2.96*10^17 seconds.
    I could get it into years by dividing the answer by (60*60*24*365) and i get about
    9.4*10^9 years.

    Any help is appreciated.
     
  2. jcsd
  3. May 18, 2003 #2

    marcus

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    Hello ant, you say you are blocked and want help but it seems to me that you are doing well and that you have gotten a reasonable answer namely about 10 billion years.

    Perhaps at the end of this time the sun will go red giant and
    start burning helium?

    Would it help if one of us would do a parallel calculation and see if we get the same 10 billion years, or 9.4 billion or whatever?

     
  4. May 18, 2003 #3

    marcus

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    For fun, I will do the calculation in natural units (c=G=hbar=k=1).

    the sun's mass is 93E36

    12 percent of 75 percent is 9 percent

    9 percent of 93E36 is 8.37E36 (the Planck masses of H to consume)

    The mass of H is 1/13E18, so the number of H to consume is
    13E18 x 8.37E36 = 1.1E56.

    But each H consumed supplies 6.675 MeV, which is 5.5E-22

    therefore 6E34 Planck energy units must be released!

    [this figure for the total energy release seems to be critical]

    Now in natural units, the assumed power of 3.9E26 watts is equal to 1.07E-26

    So one simply asks how long does it take to deliver an amount of energy 6E34 at the rate 1.07E-26

    The answer is 5.6E60. This is planck units of time and may be converted to 3.0E17 seconds or to 9.5 billion years.

    I have generally be using a little more accuracy than I have been bothering to write down. I am fairly confident of the figure and see that it agrees substantially with your 9.4 billion years.

    Is this satisfactory?
     
  5. May 19, 2003 #4
    Hi

    Thank you, the funny thing is that i was really blocked!! before writting the question on the forum, then i wrote the step to show my working, and I was thinking about it, and suddently i got the correct answer just by writting to this forum.
    Very strange

    Thank you
     
  6. May 19, 2003 #5

    marcus

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    perhaps not so strange or even so unusual

    it may help one to think if one has (even an imaginary)
    someone to talk to and explain things to.

    here is a bit of solar trivia for you:

    in a cubic kilometer of space, at this distance from the sun, how many photons of sunlight are there?

    I am in northern california and you are in france. probably we often walk outside and look at the sunlight. Think about the occupation of space by sunlight quanta----how many are there in the volume around you?

    it would be rather easy to calculate for you, I think.
    perhaps it is a "general culture" number which one should know!
    I happened to calculate it in cubic miles rather than cubic kilometers and there are 1023 photons in a cubic mile.
     
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