Answer:
83.2 W/m^2
Explanation:
The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:
[tex]P=\sigma A T^4[/tex]
where
[tex]\sigma[/tex] is the Stefan-Boltzmann constant
A is the surface area
T is the surface temperature
So, we see that the radiation per unit area is proportional to the fourth power of the temperature:
[tex]I \propto T^4[/tex]
So in our problem we can write:
[tex]I_1 : T_1^4 = I_2 : T_2^4[/tex]
where
[tex]I_1 = 1400 W/m^2[/tex] is the power per unit area of the present sun
[tex]T_1 = 5800 K[/tex] is the temperature of the sun
[tex]I_2[/tex] is the power per unit area of sun X
[tex]T_2 = 2864 K[/tex] is the temperature of sun X
Solving for I2, we find
[tex]I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2[/tex]
