Chapter 20, Problem 75AP

(a)

To determine

The power radiating from the sun at uniform temperature.

Answer to Problem 75AP

The power radiating from the sun at uniform temperature is $3.16\times {10}^{22}\text{W}$.

Explanation of Solution

Given info: The area of photosphere is $5.14\times {10}^{14}{\text{m}}^2$, The coefficient of emissivity for sun is $0.965$, and the uniform temperature is $5800\text{K}$.

The expression for the power radiated from the surface of the photosphere is,

$P=eA\sigma T^4$

Here,

$P$ is the power radiated from the surface.

$e$ is the coefficient of emissivity.

$A$ is the area of photosphere.

$\sigma$ is the Stefan’s constant.

$T$ is the temperature of the surface.

Substitute $0.965$ for $e$, $5.10\times {10}^{14}{\text{m}}^2$ for $A$, $5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}$ for $\sigma$ and $5800\text{K}$ for $T$.

$\begin{array}{c}P=0.965\times 5.10\times {10}^{14}{\text{m}}^2\times 5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}\times {\left(5800\text{K}\right)}^4\\ =3.1576\times {10}^{\text{22}}\text{W}\\ \approx \text{3}\text{.16}\times {\text{10}}^{22}\text{W}\end{array}$

Conclusion:

Therefore, the power radiating from the sun at uniform temperature is $3.16\times {10}^{\text{22}}\text{W}$.

(b)

To determine

The power output of the patch from the sun at non uniform temperature.

Answer to Problem 75AP

The power output of the patch from the sun at non uniform temperature is $3.17\times {10}^{22}\text{W}$.

Explanation of Solution

Given info: The $10\%$ area of the patch is at $4800\text{K}$ and $90\%$ area of the patch is at $5890\text{K}$.

The expression for the power radiated from the surface of the photosphere is,

$P=e{A}_1\sigma T_1^4+e{A}_2\sigma T_2^4$

Here,

$P$ is the power radiated from the surface.

$e$ is the coefficient of emissivity.

$A_1$ is the area of patch at having temperature $4800\text{K}$.

$\sigma$ is the Stefan’s constant.

$T_1$ is the temperature of the 10% area of patch.

$A_2$ is the area of patch at having temperature $5890\text{K}$.

$T_2$ is the temperature of the 90% area of patch.

Substitute $0.965$ for $e$, $0.1\times 5.10\times {10}^{14}{\text{m}}^2$ for $A_1$, $0.9\times 5.10\times {10}^{14}{\text{m}}^2$ for $A_2$ $5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}$ for $\sigma$, $4800\text{K}$ for $T_1$ and $5890\text{K}$ for $T_2$.

$\begin{array}{c}{P}^{\prime }=\left[\begin{array}{c}0.965\times 5.10\times {10}^{14}{\text{m}}^2\times \left(0.1\times 5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^4\right)\times {\left(4800\text{K}\right)}^4\\ +0.965\times 5.10\times {10}^{14}{\text{m}}^2\times \left(0.9\times 5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^4\right)\times {\left(5890\text{K}\right)}^4\end{array}\right]\\ =1.4812\times {10}^{21}+3.0224\times {10}^{22}\\ =3.17\times {10}^{22}\text{W}\end{array}$

Conclusion:

Therefore, The power output of the patch from the sun at non uniform temperature is $3.17\times {10}^{22}\text{W}$.

(c)

To determine

The comparison of the answers of part (a) and (b).

Answer to Problem 75AP

The answer of part (b) is $0.408\%$ greater than the answer of part (a).

Explanation of Solution

The expression for comparing the answers is,

$r=\frac{\left(P^{\text{'}}-P\right)}{P}\times 100$

Here,

$r$ is the percentage increment in the answer of part (b) with respect to part (a).

$P$ is the power output from patch at uniform temperature.

$P^{\text{'}}$ is the power output from patch at non uniform temperature.

Substitute $3.17\times {10}^{22}\text{W}$ for $P^{\text{'}}$ and $3.1576\times {10}^{\text{22}}\text{W}$ for $P$ in above equation.

$\begin{array}{c}r=\frac{\left(3.17\times {10}^{22}\text{W}-3.1576\times {10}^{\text{22}}\text{W}\right)}{3.1576\times {10}^{\text{22}}\text{W}}\times 100\\ =0.408\%\end{array}$

Conclusion:

Therefore, the answer of part (b) is $0.408\%$ greater than the answer of part (a).

(d)

To determine

The average temperature of the patch.

Answer to Problem 75AP

The average temperature of the patch is $5.78\times {10}^3\text{K}$

Explanation of Solution

Let the average temperature is $T\text{'}$.

The Stefan’s law is,

$P\text{'}=eA\sigma {\left(T\text{'}\right)}^4$

Here,

$P\text{'}$ is the power radiated from the surface.

$e$ is the coefficient of emissivity.

$A$ is the area of photosphere.

$\sigma$ is the Stefan’s constant.

$T\text{'}$ is the average temperature of the surface.

Substitute $0.965$ for $e$, $5.10\times {10}^{14}{\text{m}}^2$ for $A$, $5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^4$ for $\sigma$ and $3.17\times {10}^{22}\text{W}$ for $P\text{'}$ in above equation.

$\begin{array}{c}3.17\times {10}^{22}\text{W}=0.965\times 5.10\times {10}^{14}{\text{m}}^2\times 5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^4\times {\left(T\text{'}\right)}^4\\ {\left(T\text{'}\right)}^4=\frac{3.17\times {10}^{22}\text{W}}{0.965\times 5.10\times {10}^{14}{\text{m}}^2\times 5.6696\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}}\\ T\text{'}=5.78\times {10}^3\text{K}\end{array}$

Conclusion:

Therefore, the average temperature of the patch is $5.78\times {10}^3\text{K}$.