Answer: [tex]4.1602(10)^{21} W[/tex]
Explanation:
The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":
[tex]P=\sigma A T^{4}[/tex] (1)
Where:
[tex]P[/tex] is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing [tex]1W=\frac{1Joule}{second}=1\frac{J}{s}[/tex]
[tex]\sigma=5.6703(10)^{-18}\frac{W}{m^{2} K^{4}}[/tex] is the Stefan-Boltzmann's constant.
[tex]A[/tex] is the Surface of the body
[tex]T=12000K[/tex] is the effective temperature of the body (its surface absolute temperature) in Kelvin .
However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.
Therefore, for the case of the star Rigel, we will use the Stefan-Boltzmann law for real radiator bodies:
[tex]P=\sigma A \epsilon T^{4}[/tex] (2)
Where [tex]\epsilon=0.955[/tex] is the star's emissivity
Now, firstly we need to find [tex]A[/tex], in the case of Rigel, its surface area can be approximated to a sphere, so:
[tex]A_{Rigel}=4 \pi r^{2}[/tex] (3)
[tex]A_{Rigel}=4 \pi (5.43(10)^{10}m)^{2}[/tex]
[tex]A_{Rigel}=3.705(10)^{22}m^{2}[/tex] (4)
Knowing this value, let's substitute it in (2):
[tex]P=(5.6703(10)^{-18}\frac{W}{m^{2} K^{4}})(3.705(10)^{22}m^{2})(0.955)(12000K)^{4}[/tex] (5)
[tex]P=4.1602(10)^{21}W[/tex] (6) This is the total energy radiated by Rigel each second.
