Chapter 11, Problem 35P

(a)

To determine

Find the amount of radiation emitted from a hot pavement in Arizona.

Answer to Problem 35P

In SI units, The amount of radiation emitted from a hot pavement in Arizona is $493.7\text{W}$ and in U.S customary units, it is $1684.7\frac{\text{Btu}}{\text{h}}$.

Explanation of Solution

Given data:

Area of the surface, $A=1{\text{m}}^{\text{2}}$.

Emissivity of the surface, $\epsilon =0.8$.

Temperature of the surface, $T_s=50°\text{C}$.

Formula used:

The relationship between degree Celsius $\left(°\text{C}\right)$ and Kelvin is,

$T\left(\text{K}\right)=T\left(\text{C}\right)+273$ (1)

Here,

$T\left(°\text{C}\right)$ is the temperature in degree Celsius,

$T\left(\text{K}\right)$ is the temperature in Kelvin.

The formula for the amount of radiant energy emitted by a surface is,

$q=\epsilon \sigma A{T}_s{}^4$ (2)

Here,

$\epsilon$ is the emissivity of the surface,

$\sigma$ is the Boltzmann constant,

$A$ is the area of the surface,

$T_s$ is the temperature of the surface.

Calculation:

Substitute $50°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1) to find the surface temperature in Kelvin,

$\begin{array}{c}{T}_s\left(\text{K}\right)=50+273\\ =323\text{K}\end{array}$

Substitute $0.8$ for $\epsilon$, $5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}}$ for $\sigma$, $1{\text{m}}^{\text{2}}$ for $A$, and $323\text{K}$ for $T_s$ in equation (2) to find $q$,

$\begin{array}{c}q=\left(0.8\right)\left(5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}}\right)\left(1{\text{m}}^{\text{2}}\right){\left(323\text{K}\right)}^4\\ =0.8\times 5.67\times {10}^{-8}\times {\left(323\right)}^4\text{W}\\ =493.\text{7}\text{W}\end{array}$

Do the unit conversion in above result,

$\begin{array}{c}q=493.7\text{W}\left[\therefore 1\text{W}=3.4123\frac{\text{Btu}}{\text{h}}\right]\\ =\left(493.7\text{W}\right)\left(\frac{3.4123\frac{\text{Btu}}{\text{h}}}{1\text{W}}\right)\\ =1684.7\frac{\text{Btu}}{\text{h}}\end{array}$

Therefore, in SI units, The amount of radiation emitted from a hot pavement in Arizona is $493.7\text{W}$ and in U.S customary units, it is $1684.7\frac{\text{Btu}}{\text{h}}$.

Conclusion:

Hence, in SI units, The amount of radiation emitted from a hot pavement in Arizona is $493.7\text{W}$ and in U.S customary units, it is $1684.7\frac{\text{Btu}}{\text{h}}$.

(b)

To determine

Find the amount of radiated emitted from a hood of a car.

Answer to Problem 35P

In SI units, The amount of radiation emitted from a hood of a car is $489.8\text{W}$ and in U.S customary units, it is $1671.2\frac{\text{Btu}}{\text{h}}$.

Explanation of Solution

Given data:

Area of the surface, $A=1{\text{m}}^{\text{2}}$.

Emissivity of the surface, $\epsilon =0.9$.

Temperature of the surface, $T_s=40°\text{C}$.

Calculation:

Substitute $40°\text{C}$ for $T\left(\text{C}\right)$ in equation (1) to find the surface temperature in Kelvin,

$\begin{array}{c}{T}_s\left(\text{K}\right)=40+273\\ =313\text{K}\end{array}$

Substitute $0.9$ for $\epsilon$, $5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}}$ for $\sigma$, $1{\text{m}}^{\text{2}}$ for $A$, and $313\text{K}$ for $T_s$ in equation (2) to find $q$,

$\begin{array}{c}q=\left(0.9\right)\left(5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}}\right)\left(1{\text{m}}^{\text{2}}\right){\left(313\text{K}\right)}^4\\ =0.8\times 5.67\times {10}^{-8}\times {\left(313\right)}^4\text{W}\\ =489.8\text{W}\end{array}$

Do the unit conversion in above result,

$\begin{array}{c}q=498.8\text{W}\left[\therefore 1\text{W}=3.4123\frac{\text{Btu}}{\text{h}}\right]\\ =\left(498.8\text{W}\right)\left(\frac{3.4123\frac{\text{Btu}}{\text{h}}}{1\text{W}}\right)\\ =1671.2\frac{\text{Btu}}{\text{h}}\end{array}$

Therefore, in SI units, the amount of radiation emitted from a hood of a car is $489.8\text{W}$ and in U.S customary units, it is $1671.2\frac{\text{Btu}}{\text{h}}$.

Conclusion:

Hence, in SI units, the amount of radiation emitted from a hood of a car is $489.8\text{W}$ and in U.S customary units, it is $1671.2\frac{\text{Btu}}{\text{h}}$.

(c)

To determine

Find the amount of radiated emitted from a sunbather.

Answer to Problem 35P

In SI units, The amount of radiation emitted from a sunbather is $477.4\text{W}$ and in U.S customary units, it is $1629.0\frac{\text{Btu}}{\text{h}}$.

Explanation of Solution

Given data:

Area of the surface, $A=1{\text{m}}^{\text{2}}$.

Emissivity of the surface, $\epsilon =0.9$.

Temperature of the surface, $T_s=38°\text{C}$.

Calculation:

Substitute $38°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1) to find the surface temperature in Kelvin,

$\begin{array}{c}{T}_s\left(\text{K}\right)=38+273\\ =311\text{K}\end{array}$

Substitute $0.9$ for $\epsilon$, $5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}}$ for $\sigma$, $1{\text{m}}^{\text{2}}$ for $A$, and $311\text{K}$ for $T_s$ in equation (2) to find $q$,

$\begin{array}{c}q=\left(0.9\right)\left(5.67\times {10}^{-8}\frac{\text{W}}{{\text{m}}^{\text{2}}\cdot {\text{K}}^{\text{4}}}\right)\left(1{\text{m}}^{\text{2}}\right){\left(311\text{K}\right)}^4\\ =0.9\times 5.67\times {10}^{-8}\times {\left(311\right)}^4\text{W}\\ =477.4\text{W}\end{array}$

Do the unit conversion in above result,

$\begin{array}{c}q=477.4\text{W}\left[\therefore 1\text{W}=3.4123\frac{\text{Btu}}{\text{h}}\right]\\ =\left(477.4\text{W}\right)\left(\frac{3.4123\frac{\text{Btu}}{\text{h}}}{1\text{W}}\right)\\ =1629.0\frac{\text{Btu}}{\text{h}}\end{array}$

Therefore, in SI units, the amount of radiation emitted from a sunbather is $477.4\text{W}$ and in U.S customary units, it is $1629.0\frac{\text{Btu}}{\text{h}}$.

Conclusion:

Hence, in SI units, the amount of radiation emitted from a sunbather is $477.4\text{W}$ and in U.S customary units, it is $1629.0\frac{\text{Btu}}{\text{h}}$.