Question

The luminosity of the Sun is 4 x 1026 W, which means that the sun emits 4 x 1026 J of energy every second. How much mass does it change into energy every second? Express your answer in kilograms.

In fusion reactions, only a small fraction of the hydrogen mass is converted into energy. 99.3% of the hydrogen is converted into helium, and just 0.7% of the hydrogen is converted into energy. Based on this information, how much hydrogen undergoes fusion in the Sun every second? Express your answer in kilograms.

Answer 1

The sun emits =
`$4 * 10^(26) $` J of energy per second

=
`$4 * 10^(26) \ kg m^2 s^(-3) $`

We know,
`$1 \ J =1 \ kg \ m^2 / s^2 $`

`$E=mC^2$` , where C =
`$3 * 10^8 \ m/s$`

`$E=mC^2$`

`$J=M(M/s)^2$`

Dividing both sides by 1 second

`$(J)/(s)=(M * m^2 s^(-2))/(sec)$`

`$(J)/(s)=M * m^2 s^(-3)$`

Then,
`$4 * 10^(26) \ J/s = M * m^2 s^(-3) * (3 * 10^8)^2$`

`$M = (4 * 10^(26))/(9 *10^(16))$`

`$M=4.44 * 10^9 \ kg$`

Now according to the information, 99.3% hydrogen.

If 0.7 % of hydrogen produce =
`$4 * 10^(26) $` J of energy per second

Then 1% of hydrogen will produce =
`$(4 * 10^(26))/(0.7)$` J energy per second

So, 100% of hydrogen will produce =
`$(4 * 10^(26))/(0.7) * 100$` J energy per second

=
`$5.7143 * 10^(28 )$` J energy per second

Mass of hydrogen undergo fusion in sun per second

`$E=mC^2$`

Similarly,
`$(J)/(s)=M * m^2 s^(-3)$`

`$5.714 * 10^(28) \ J/s = M * (3 * 10^8)^2 \ m^2 \ s^(-3)$`

`$M = (5.7143 * 10^(28))/(9 * 10^(16))$` kg

`$M= 6.349 * 10^(11)$` kg