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

83.2 W/m^2

Step-by-step 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 = (I_1 T_2^4)/(T_1^4)=((1400 W/m^2)(2864 K)^4)/((5800 K)^4)=83.2 W/m^2`