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

Solution :

The sun emits = $4 \times 10^{26}$ J of energy per second

                        = $4 \times 10^{26} \ kg m^2 s^{-3}$

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

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

$E=mC^2$

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

Dividing both sides by 1 second

$\frac{J}{s}=\frac{M \times m^2 s^{-2}}{sec}$

$\frac{J}{s}=M \times m^2 s^{-3}$

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

      $M = \frac{4 \times 10^{26}}{9 \times10^{16}}$

      $M=4.44 \times 10^9 \ kg$

Now according to the information, 99.3% hydrogen.

If 0.7 % of hydrogen produce = $4 \times 10^{26}$ J of energy per second

Then 1% of hydrogen will produce = $\frac{4 \times 10^{26}}{0.7}$   J energy per second

So, 100% of hydrogen will produce = $\frac{4 \times 10^{26}}{0.7} \times 100$  J energy per second

                                                          = $5.7143 \times 10^{28 }$  J energy per second

Mass of hydrogen undergo fusion in sun per second

$E=mC^2$

Similarly, $\frac{J}{s}=M \times m^2 s^{-3}$

$5.714 \times 10^{28} \ J/s = M \times (3 \times 10^8)^2 \ m^2 \ s^{-3}$

$M = \frac{5.7143 \times 10^{28}}{9 \times 10^{16}}$   kg

$M= 6.349 \times 10^{11}$ kg