
In the Sun, about 4 billion kg of matter is converted into energy each second. The power output of the Sun in watt is.
Answer
507.9k+ views
Hint: Here, Mass is given which is 4 billion kg, that is the quantity of matter which is converting, and it is said that we have to find the energy, that is, the capacity for doing work. It may exist in different forms like potential, kinetic, heat, etc which are formed in the sun per second. In the case of the sun, energy is in heat or you can say it as thermal energy and we get the output power of the sun by using the Einstein relation.
Formula Used: We can calculate energy by using Einstein's equation which is:- \[E=m{{c}^{2}}\]
Complete step-by-step solution:
Converted Mass, m=\[4\times {{10}^{9}}\]kg
Using Einstein equation :- \[E=m{{c}^{2}}\]
here, C is the speed of light that is $3\times {{10}^{8}}$m/s
Therefore, E = \[4\times {{10}^{9}}\times {{\left( 3\times {{10}^{8}} \right)}^{2}}\]\[kg{{m}^{2}}/{{s}^{2}}\]
E = \[3.6\times {{10}^{26}}\]J
Since, the mass is getting converted per second. So, the energy output is also in per second that is, E = \[3.6\times {{10}^{26}}\]J/s
Hence, the power output of the Sun in watt is \[3.6\times {{10}^{26}}\]W. This means that In the Sun \[3.6\times {{10}^{26}}\] W amount of power is required to convert about 4 billion kg of matter into energy each second.
Additional Information: Power is then defined as the amount of energy transferred or converted or you can say required in a simple way per unit time. According to the international system of units, the unit of power is the Watt represented by the symbol ‘W’ that is equal to one Joule per second or you can write in the symbol form which is 1J/s. In older works, power is sometimes called activity.
Note: Don't get confused while solving an equation. Always prefer solving calculations in a stepwise manner to get rid of confusion as well as problems. Also, remember the Einstein formula by heart as it has several usages in studies.
Formula Used: We can calculate energy by using Einstein's equation which is:- \[E=m{{c}^{2}}\]
Complete step-by-step solution:
Converted Mass, m=\[4\times {{10}^{9}}\]kg
Using Einstein equation :- \[E=m{{c}^{2}}\]
here, C is the speed of light that is $3\times {{10}^{8}}$m/s
Therefore, E = \[4\times {{10}^{9}}\times {{\left( 3\times {{10}^{8}} \right)}^{2}}\]\[kg{{m}^{2}}/{{s}^{2}}\]
E = \[3.6\times {{10}^{26}}\]J
Since, the mass is getting converted per second. So, the energy output is also in per second that is, E = \[3.6\times {{10}^{26}}\]J/s
Hence, the power output of the Sun in watt is \[3.6\times {{10}^{26}}\]W. This means that In the Sun \[3.6\times {{10}^{26}}\] W amount of power is required to convert about 4 billion kg of matter into energy each second.
Additional Information: Power is then defined as the amount of energy transferred or converted or you can say required in a simple way per unit time. According to the international system of units, the unit of power is the Watt represented by the symbol ‘W’ that is equal to one Joule per second or you can write in the symbol form which is 1J/s. In older works, power is sometimes called activity.
Note: Don't get confused while solving an equation. Always prefer solving calculations in a stepwise manner to get rid of confusion as well as problems. Also, remember the Einstein formula by heart as it has several usages in studies.
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