Chapter 14, Problem 85P

(a)

To determine

The temperature of the leaf.

Answer to Problem 85P

The temperature of the leaf is $\text{39}\text{°C}$.

Explanation of Solution

Write the expression for Stefan’s law of radiation.

$p=e\sigma A{T}^4$

Here, $\sigma$ is the Stefan constant, $e$ is the emissivity, $A$ is the area, $T$ is the temperature, $p$ is the power absorbed.

Re-arrange the expression to find the temperature.

$T={\left(\frac{p}{e\sigma A}\right)}^{\frac{1}{4}}$

Substitute $I_{\text{top}}A+e\sigma A{T}_s{}^4$ for $p$ to find the temperature of the leaf.

$T={\left(\frac{I_{\text{top}}A+e\sigma A{T}_s{}^4}{e\sigma A_b}\right)}^{\frac{1}{4}}$

Here, $I_{\text{top}}$ is the intensity at the top surface of the leaf, $T_s$ is the temperature of the surroundings, $A_b$ area of both side leaf.

Conclusion:

Substitute $0.700\left(9.00\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)$ for $I_{\text{top}}$, $5.00\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A$, $1$ for $e$, $5.670\times {10}^{-8}\frac{\text{W}}{{\text{m}}^2\cdot {\text{K}}^{\text{4}}}$ for $\sigma$, $25.\text{0}\text{°C}$ for $T_s$ and $2\left(5.00\times {10}^{-3}{\text{m}}^{\text{2}}\right)$ for $A_b$ to find the temperature of the leaf.

$\begin{array}{c}T={\left(\frac{\begin{array}{l}\left(0.700\left(9.00\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)\right)\left(5.00\times {10}^{-3}{\text{m}}^{\text{2}}\right)+\\ \left(1\right)\left(5.670\times {10}^{-8}\frac{\text{W}}{{\text{m}}^2\cdot {\text{K}}^{\text{4}}}\right)\left(5.00\times {10}^{-3}{\text{m}}^{\text{2}}\right){\left(\left(25.\text{0}\text{°C}+273\text{K}\right)\text{K}\right)}^4\end{array}}{\left(1\right)\left(5.670\times {10}^{-8}\frac{\text{W}}{{\text{m}}^2\cdot {\text{K}}^{\text{4}}}\right)\left(2\left(5.00\times {10}^{-3}{\text{m}}^{\text{2}}\right)\right)}\right)}^{\frac{1}{4}}\\ ={\left(\frac{5.39\text{W}}{\left(1\right)\left(5.670\times {10}^{-8}\frac{\text{W}}{{\text{m}}^2\cdot {\text{K}}^{\text{4}}}\right)\left(2\left(5.00\times {10}^{-3}{\text{m}}^{\text{2}}\right)\right)}\right)}^{\frac{1}{4}}\\ =312\text{K}-273\text{K}\\ =3\text{9}\text{°C}\end{array}$

(b)

To determine

The power per unit area need to be lost by other methods to keep the temperature at $25.\text{0}\text{°C}$.

Answer to Problem 85P

The power per unit area need to be lost by other methods to keep the temperature at $25.\text{0}\text{°C}$ is $182{\text{W/m}}^{\text{2}}$.

Explanation of Solution

The one side of the leaf, the bottom side can be neglected because it absorb and emit in same rate.

Write the expression to find the power per unit area need to be lost.

$\begin{array}{c}\frac{p_{\text{ab,sun}}}{A}=\frac{p_{\text{ra}}}{A}+\frac{p_{\text{othr}}}{A}\\ =e\sigma T^4+\frac{p_{\text{othr}}}{A}\end{array}$

Here, $\frac{p_{\text{ab,sun}}}{A}$ is the radiant energy absorbed by the top surface of the leaf, $\frac{p_{\text{othr}}}{A}$ is the power per unit area need to be lost.

Re-arrange the expression to find the power per unit area need to be lost.

$\frac{p_{\text{othr}}}{A}=\frac{p_{\text{ab,sun}}}{A}-e\sigma T^4$

Conclusion:

Substitute $0.700\left(9.00\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)$ for $\frac{p_{\text{ab,sun}}}{A}$, $1$ for $e$, $5.670\times {10}^{-8}\frac{\text{W}}{{\text{m}}^2\cdot {\text{K}}^{\text{4}}}$ for $\sigma$ and $25.\text{0}\text{°C}$ for $T_s$ to find the t power per unit area need to be lost.

$\begin{array}{l}\frac{p_{\text{othr}}}{A}=0.700\left(9.00\times {10}^2\frac{\text{W}}{{\text{m}}^{\text{2}}}\right)-\left(1\right)\left(5.670\times {10}^{-8}\frac{\text{W}}{{\text{m}}^2\cdot {\text{K}}^{\text{4}}}\right){\left(\left(25.\text{0}\text{°C}+273\text{K}\right)\text{K}\right)}^4\\ =182{\text{W/m}}^{\text{2}}\end{array}$