Question

The radiation per unit area from the Sun reaching the earth is 1400 W/m2 , approximately the amount of radiative power per unit area reaching a sun bather on the banks of Barton Springs at noon on a clear day in June. The temperature of the Sun is 5800 K. Now suppose instead of the present Sun we received radiation from sun X at temperature 2864 K, located at the same position as our Sun. How much radiative power per unit area would reach the sun bather from the new sun X?

Answer 1

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:

$P=\sigma A T^4$

where

$\sigma$ 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:

$I \propto T^4$

So in our problem we can write:

$I_1 : T_1^4 = I_2 : T_2^4$

where

$I_1 = 1400 W/m^2$ is the power per unit area of the present sun

$T_1 = 5800 K$ is the temperature of the sun

$I_2$ is the power per unit area of sun X

$T_2 = 2864 K$ is the temperature of sun X

Solving for I2, we find

$I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2$