91Ó°ÊÓ

(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 \(5.74 \times {10^3}\;{\rm{K}}\).

(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 \(6.17 \times {10^7}\;{\rm{W}}/{{\rm{m}}^2}\).

(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 \(1.34 \times {10^3}\;{\rm{W}}/{{\rm{m}}^2}\).