The earth receives solar radiation at a rate of 8.2 J cm^-2 min^-1. If the sun radiates as the black bodies, the temperature at the surface of the sun will be (the angle subtended by sun on the earth in suppose 0.53° and Stefan constant is σ = 5.67x 10^-6 Wm³ K⁴

5800 K

6700 K

8000 K

7800 K

Solution

5800 K

Let diameter of the sun is d while radius of the earth is r then

$\frac{D}{2} ≃ 0.53 \times \frac{π}{180}$

= 0.00925

⇒ $\frac{D}{2}$ = 9.25 × 10^-3

The radiation emitted per unit area of the sun is

$4 π {(\frac{d}{2})}^{2} {σT}^{4} = {πD}^{2} σ T^{4}$

The radiation received at the earth's surface per unit time per unit area is

$\frac{πσd T^{4}}{4 {πr}^{2}} = \frac{σ T^{4}}{4} {(\frac{d}{R})}^{2}$

⇒ $\frac{{σT}^{4}}{4} {(\frac{D}{R})}^{2}$ = 8.2 Jcm^-2 min^-1

⇒ $\frac{1}{4} \times 5.67 \times 10^{- 8} \times T^{4} \times {(9.25 \times 10^{- 3})}^{2}$ = $\frac{8.2}{10^{- 4} \times 60} {Wm}^{- 2}$

⇒ T = 5794 K

T ≅ 5800 K

For Daily Free Study Material Join wiredfaculty