Calculating Hydrogen to Helium Conversion in the Sun

[b_o3]

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 don't have a rate.

Any help is greatly appreciated,
thanks

 

[b_o3]

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 don't have a rate.

Any help is greatly appreciated,
thanks

 

[nrqed]

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 don't have a rate.

Any help is greatly appreciated,
thanks

You have to look up the luminosity of the Sun to know how much energy is produced by the Sun in one year (L something of order 10^26 W so multiplying this by the number of seconds per year will give you the energy produced per year...Check the number, I am quoting from memory and I know it's not exactly 10^26)

 

[haiha]

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

 

[b_o3]

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

 

[b_o3]

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

 

[b_o3]

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

 

[nrqed]

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

You have to find the energy released by each fusion reaction of hydrogen. How many Joules are produced by each reaction (hint: what is the change of mass? What equation does one use to then get the energy released?) Once you have the energy released by each reaction, take the total amount of energy produced by the Sun in one year and divide it by the energy produced per reaction. That will tell you how many reactions occur per year. Once you have that, multiply that number by the mass of hydrogen converted to helium in each reaction.

 

[nrqed]

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

Yes, that's the correct number...where "ton" refers to metric tons, so 4 billion kg per second.

 

[denverdoc]

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 rxn--E=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?

 

[pixel01]

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

You just apply the equation : E=mc^2
==> 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

 

[denverdoc]

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

 

[wk1989]

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

 

[denverdoc]

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

 

[b_o3]

thanks guys