Chapter 8, Problem 1P

To determine

Find the power radiated by a woodstove.

Answer to Problem 1P

The power radiated by a woodstove is $\underset{¯}{3.7\text{kW}}$.

Given information:

The size of the woodstove is $65\text{cm}\times \text{55}\text{cm}\times \text{85}\text{cm}$.

The surface temperature (T) is $120°\text{C}$.

The emissivity $\left(\epsilon \right)$ is 1.

Calculation:

Find the surface area of the woodstove.

$\begin{array}{c}A=\left(2\times H\times D\right)+\left(2\times H\times B\right)+\left(2\times D\times B\right)\\ =\left(2\times 65\times 55\right)+\left(2\times 65\times 85\right)+\left(2\times 55\times 85\right)\\ =27,550{\text{cm}}^{\text{2}}\\ =2.755{\text{m}}^{\text{2}}\end{array}$

Convert the surface temperature from Degree Celsius to Kelvin:

$\begin{array}{c}120°\text{C}=\left(120°\text{C}+273\right)\text{K}\\ =393\text{K}\end{array}$

Consider the Stephen Boltzmann constant $\left(\sigma \right)$ is $5.67\times {10}^{-8}\text{W}\cdot {\text{m}}^{\text{2}}\cdot {\text{K}}^{-}{}^{\text{4}}$.

Find the surface radiation (P):

$\begin{array}{c}P=A\epsilon \sigma T^4\\ =2.755{\text{m}}^{\text{2}}\times 1\times 5.67\times {10}^{-8}\text{W}\cdot {\text{m}}^{\text{2}}\cdot {\text{K}}^{-}{}^{\text{4}}\times {\left(393\text{K}\right)}^4\\ =3,726\text{W}\\ \approx 3.7\text{kW}\end{array}$

Thus, the power radiated by a woodstove is $\underset{¯}{3.7\text{kW}}$.

Explanation of Solution

Given information:

The size of the woodstove is $65\text{cm}\times \text{55}\text{cm}\times \text{85}\text{cm}$.

The surface temperature (T) is $120°\text{C}$.

The emissivity $\left(\epsilon \right)$ is 1.

Calculation:

Find the surface area of the woodstove.

$\begin{array}{c}A=\left(2\times H\times D\right)+\left(2\times H\times B\right)+\left(2\times D\times B\right)\\ =\left(2\times 65\times 55\right)+\left(2\times 65\times 85\right)+\left(2\times 55\times 85\right)\\ =27,550{\text{cm}}^{\text{2}}\\ =2.755{\text{m}}^{\text{2}}\end{array}$

Convert the surface temperature from Degree Celsius to Kelvin:

$\begin{array}{c}120°\text{C}=\left(120°\text{C}+273\right)\text{K}\\ =393\text{K}\end{array}$

Consider the Stephen Boltzmann constant $\left(\sigma \right)$ is $5.67\times {10}^{-8}\text{W}\cdot {\text{m}}^{\text{2}}\cdot {\text{K}}^{-}{}^{\text{4}}$.

Find the surface radiation (P):

$\begin{array}{c}P=A\epsilon \sigma T^4\\ =2.755{\text{m}}^{\text{2}}\times 1\times 5.67\times {10}^{-8}\text{W}\cdot {\text{m}}^{\text{2}}\cdot {\text{K}}^{-}{}^{\text{4}}\times {\left(393\text{K}\right)}^4\\ =3,726\text{W}\\ \approx 3.7\text{kW}\end{array}$

Thus, the power radiated by a woodstove is $\underset{¯}{3.7\text{kW}}$.