Question

The Sun radiates like a perfect black body with an emissivity of exactly 1. (a) Calculate the surface temperature of the Sun, given that it is a sphere with a\({\bf{7}}{\bf{.00 \times 1}}{{\bf{0}}^{\bf{8}}}\;{\bf{m}}\)radius that radiates\({\bf{3}}{\bf{.80 \times 1}}{{\bf{0}}^{{\bf{26}}}}{\bf{ W}}\)into 3-K space. (b) How much power does the Sun radiate per square meter of its surface? (c) How much power in watts per square meter is that value at the distance of Earth,\({\bf{1}}{\bf{.50 \times 1}}{{\bf{0}}^{{\bf{11}}}}{\bf{ m}}\)away? (This number is called the solar constant.)

Short answer

(a) The surface temperature of Sun is\({\bf{5}}{\bf{.74 \times 1}}{{\bf{0}}^{\bf{3}}}\;{\bf{K}}\).

(b) The power radiated by sun in unit area is\({\bf{6}}{\bf{.17 \times 1}}{{\bf{0}}^{\bf{7}}}\;{\bf{W/}}{{\bf{m}}^{\bf{2}}}\).

(c) The power radiated by sun in unit area is \({\bf{1}}{\bf{.34 \times 1}}{{\bf{0}}^{\bf{3}}}\;{\bf{W/}}{{\bf{m}}^{\bf{2}}}\).

Step-by-step solution

Determination of surface temperature of Sun

(a)

Given Data:

The radius of sun is \(R = 7 \times {10^8}\;{\rm{m}}\)

The energy radiated by sun is \(Q = 3.80 \times {10^{26}}\;{\rm{W}}\)

The distance of Earth from Sun is \(d = 1.50 \times {10^{11}}\;{\rm{m}}\)

The rate of heat transfer by radiation from skin of body is calculated by using the Stefan’s law. The rate of increase in body temperature is found by heat balance equation.

The surface area of the Sun is given as:

\(\begin{aligned}{}A& = 4\pi {R^2}\\A &= 4\pi {\left( {7 \times {{10}^8}\;{\rm{m}}} \right)^2}\\A &= 6.157 \times {10^{18}}\;{{\rm{m}}^2}\end{aligned}\)

The surface temperature of Sun is given as:

\(Q = \varepsilon \sigma AT_s^4\)

Here,\(\sigma \)is Stefan Boltzmann constant and its value is\(5.67 \times {10^{ - 8}}\;{\rm{W}}/{{\rm{m}}^2} \cdot {{\rm{K}}^4}\)

Substitute all the values in the above equation.

\(\begin{aligned}{}3.80 \times {10^{26}}\;{\rm{W}} &= \left( 1 \right)\left( {5.67 \times {{10}^{ - 8}}\;{\rm{W}}/{{\rm{m}}^2} \cdot {{\rm{K}}^4}} \right)\left( {6.157 \times {{10}^{18}}\;{{\rm{m}}^2}} \right)T_s^4\\{T_s} &= 5.74 \times {10^3}\;{\rm{K}}\end{aligned}\)

Therefore, the surface temperature of Sun is \(5.74 \times {10^3}\;{\rm{K}}\).

Determination of power radiated by sun in unit area

(b)

The power radiated by sun in unit area is given as:

\(P = \frac{Q}{A}\)

Substitute all the values in the above equation.

\(\begin{aligned}P &= \frac{{3.80 \times {{10}^{26}}\;{\rm{W}}}}{{6.157 \times {{10}^{18}}\;{{\rm{m}}^2}}}\\P &= 6.17 \times {10^7}\;{\rm{W}}/{{\rm{m}}^2}\end{aligned}\)

Therefore, the power radiated by sun in unit area is \(6.17 \times {10^7}\;{\rm{W}}/{{\rm{m}}^2}\).

Determination of power radiated by sun on Earth

(c)

The power radiated by sun on Earth is given as:

\(P = \frac{Q}{{4\pi {d^2}}}\)

Substitute all the values in the above equation.

\(\begin{aligned}P &= \frac{{3.80 \times {{10}^{26}}\;{\rm{W}}}}{{4\pi {{\left( {1.50 \times {{10}^{11}}\;{\rm{m}}} \right)}^2}}}\\P &= 1.34 \times {10^3}\;{\rm{W}}/{{\rm{m}}^2}\end{aligned}\)

Therefore, the power radiated by sun in unit area is \(1.34 \times {10^3}\;{\rm{W}}/{{\rm{m}}^2}\).