Question

The earth receives at its surface radiation from the sun at the rate of $1400w/m^2$. The distance of the centre of the sun from the surface of the earth is $1.5\times {10}^{11}m$ and the radius of the sun is $7\times {10}^{8}m$. Treating sun as a black body, it follows from the above data that its surface temperature is

• A. $5801$ K
• B. ${10}^6$ K
• C. $50.1$ K
• D. $5801$${}^0$C

Answer 1

Answer: (A)

$5801$ K

Let $R$ be the radius of Sun and $d$ be the distance of earth from centre of Sun.

Power radiated by Sun $=\sigma A{T}^4=\sigma \left(4\pi R^2\right)T^4$

The power radiates out over a spherical surface. If we consider a sphere with radius $=$ distance between Sun and earth with centre at centre of Sun, the power radiated by Sun is distributed over the surface of sphere.

So power per unit area (intensity) at location of earth's surface $=\frac{\sigma 4\pi R^2{T}^4}{4\pi d^2}=1400w/m^2\Rightarrow T^4=\frac{1400\times 4\pi d^2}{\sigma 4\pi R^2}\Rightarrow T^4=\frac{1400\times d^2}{\sigma R^2}\Rightarrow T^4=\frac{1400\times {(1.5\times {10}^{11})}^2}{{(5.67\times {10}^{-8})}^2{(7\times {10}^8)}^2}\Rightarrow T={(1.1337\times {10}^{15})}^{\frac{1}{4}}\Rightarrow T=5801K$