
#1
Mar3107, 09:47 PM

P: 52

Hello, I'm having trouble with this question:
mass of proton= (1.674*10^27)kg mass of neutron= (1.675*10^27)kg mass of 4He= (6.648*10^27)kg How many Kg of Hydrogen does the sun convert into Helium per year? Now my main problem is that I don't have the rate of conversion. I know that each 4 protons (H atoms) for one Helium atom, but at what rate? I multiplied mass of one proton by 4 to get (6.696*10^27)kg but I'm not really sure how to convert this into Kg per year since I dont have a rate. Any help is greatly appreciated, thanks 



#2
Mar3107, 09:48 PM

P: 52

Hello, I'm having trouble with this question:
mass of proton= (1.674*10^27)kg mass of neutron= (1.675*10^27)kg mass of 4He= (6.648*10^27)kg How many Kg of Hydrogen does the sun convert into Helium per year? Now my main problem is that I don't have the rate of conversion. I know that each 4 protons (H atoms) for one Helium atom, but at what rate? I multiplied mass of one proton by 4 to get (6.696*10^27)kg but I'm not really sure how to convert this into Kg per year since I dont have a rate. Any help is greatly appreciated, thanks 



#3
Mar3107, 10:12 PM

Sci Advisor
HW Helper
P: 2,886





#4
Mar3107, 10:31 PM

P: 137

Hydrogen to Helium
From a scientific video I remember the sun loses 4 million tons of mass every second.




#5
Mar3107, 10:33 PM

P: 52

i have Luminosity= (3.9*10^26)joules/sec , I don't understand how to convert this to Kg of Hydrogen...




#6
Mar3107, 10:34 PM

P: 52

I have the luminosity as (3.9*10^26) joules/sec. I am not sure how to convert this to Kg of Hydrogen though...




#7
Mar3107, 10:38 PM

P: 52

I have the luminosity as (3.9*10^26) joules/sec. I am not sure how to convert this to Kg of Hydrogen though...




#8
Mar3107, 10:42 PM

Sci Advisor
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P: 2,886





#9
Mar3107, 10:43 PM

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P: 2,886





#10
Mar3107, 11:05 PM

P: 1,351

well the conversion of hydrogen to He, does not exactly balance mass wise; you understand that part right. So you take the inital mass less the final mass to figure energy per fusion rxnE=delta(mass)*c^2. The you just need to compute number of protons per Kg and divide by 4 then multiply by the energy liberated in a single rxn. This help, or is it the last part of the question?




#11
Apr107, 01:24 AM

P: 691

==> m=E/c^2. With E=3.9e26 ; c=3.0e8 you get m=4.33e9kg/sec, quite similar to the value given by Haiha 



#12
Apr107, 01:40 AM

P: 1,351

funny, i thought i hinted at same in a vanishing post several hours ago.




#13
Apr107, 05:57 PM

P: 32

I don't think you can just use hydrogen atoms, the fusion process involves deuterium and tritium.




#14
Apr107, 10:40 PM

P: 1,351

although these are indeed intermediaries, I think the overall stoichiometry of the process can be represented as above...Good point though.




#15
Apr207, 10:25 AM

P: 52

thanks guys



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