Using the Sun's luminosity what is the mass of hydrogen that is converted into helium during 1 second?

Answer 1

Somewhat more than 6 x 10^11 kilograms of hydrogen are converted into helium per second inside the Sun.

I complete gm of matter corresponds to c^2 ergs = 8.99 x 10^20 ergs.

The solar luminosity is 3.86 x 10^33 erg/s.

Therefore the amount of mass being converted into ENERGY is 4.29 x 10^12 gm/s.

But only 0.007 of the mass of hydrogen taking part in nuclear reactions is CONVERTED into energy; the remaining 0.993 becomes helium. So the amount of hydrogen being converted into helium per second is :

4.29 x 10^12 x 0.993 / 0.007 gm = 6.1 x 10^14 gm = 6.1 x 10^11 kg.

Note:

1. To get this number more accurately, greater significance would be necessary in ALL of the above inputs. In fact, even the "1" of "6.1" is suspect because of the the conversion figure 0.007 only being given to one significant figure.

2. This calculation also ignored the fact that neutrinos actually carry off a certain fraction of the energy produced, on the order of ten or more percent. That means that, by only using the solar luminosity, we slightly undercounted the energy being produced per second, and therefore underestimated the total amount of hydrogen being converted into helium per second.

The only thing that we're therefore justified in concluding is that somewhat more than 6 x 10^11 kilograms of hydrogen are converted into helium per second inside the Sun.

Live long and prosper.

POSTSCRIPT : Note that 'Charles' has calculated the amount of hydrogen "disappearing" into ENERGY per second, but that is NOT the amount being converted into HELIUM. That's where what I call the BOND factor (really the nuclear bond factor), 0.007 of course ("Licensed to convert"), comes in, and therefore the additional factor of 0.993/0.007.

Answer 2

Using the sun's luminosity: By measuring how much gets to the Earth (which is called the Solar Constant and is around 1353 watts/square meter), we can calculate how much energy must be leaving the surface of the Sun. It is about 3.86 * 10^26 watts or joules/second or 3.86 * 10^33 ergs/second. We know that one AMU (Atomic Mass Unit) is equal to 1.660 * 10^-24 grams, so the amount that "disappears" in each fusion sequence (which we know to be 0.028697 AMU) must be 4.76 * 10^-26 gram. We also know Einstein's famous E = m * c^2, the equation that converts mass into energy. c = 3 * 10^10 cm/sec. So, multiplying, we find that each fusion event releases 4.29 * 10^-5 erg of energy. If we divide, we now know that there must be (3.86 * 10^33 ergs/second) / (4.29 * 10^-5 erg) occurring every second, or about 9 * 10^37 fusions/second occurring in the Sun. Now then, we could multiply this number by the amount of mass that disappears in each fusion sequence and get 4.28 * 10^12 grams/second. This is 4.28 * 10^9 kilograms or over 4 million metric tons that disappears every second.

Answer 3

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