Question

Suppose that you intercept$\mathbf{5}\mathbf{.0}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$ of the energy radiated by a hot sphere that has a radius of$\mathbf{0}\mathbf{.020}\mathbf{}\mathbf{\text{m}}$ , an emissivity of$\mathbf{0}\mathbf{.80}$ , and a surface temperature of$\mathbf{500}\mathbf{}\mathbf{\text{K}}$ . How much energy do you intercept in $\mathbf{2}\mathbf{.0}\mathbf{}\mathit{m}\mathit{i}\mathit{n}$?

Short answer

Energy interprets in 2 min is $8.6\text{J}$

Step-by-step solution

Identification of given data

1. Radius of hot sphere,$r=0.020\text{\hspace{0.17em}m}$
2. Emissivity of the surface,$\sigma =0.80$
3. Temperature of sphere,$T=500\text{\hspace{0.17em}K}$
4. Time, $t=2.0\text{minor}120\text{\hspace{0.17em}sec}.$

Significance of Stefan-Boltzmann's law

The total energy released or radiated by a blackbody per unit surface area across all wavelengths in a given period of time is inversely proportional to its thermodynamic temperature to the fourth power.

Using the values given in the problem and the formulas for energy in terms of power as well as Stefan-Boltzmann's law, we can get the answer.

Formula:

The energy radiated by the body in the given time, $E=Pt$ …(i)

The power radiated per area of the surface using Stefan-Boltzmann Law,

$P=\sigma \epsilon A{T}^4$ …(ii)

The total surface area of the sphere, $A=4\pi r^2$ …(iii)

Determining the energy

Using equation (iii), the total area of the surface is given as:

$\begin{array}{c}A=4\left(3.14\right){\left(0.020\text{\hspace{0.17em}m}\right)}^2\\ =5.024\times {10}^{-3}{\text{m}}^2\end{array}$

Using equation (ii) and the given values, we can get the power radiated by the surface as given:

$\begin{array}{c}P=\left(0.80\right)(5.67\times {10}^{-8}{\text{\hspace{0.17em}W/m}}^{\text{2}}{\text{K}}^{\text{4}})(5.024\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{2}}){\left(500\text{\hspace{0.17em}K}\right)}^4\\ =14.24\text{W}.\end{array}$

Now, the energy radiated by the surface body using equation (i) can be given as:

$\begin{array}{c}E=\left(0.005\right)\left(14.24\text{\hspace{0.17em}W}\right)\left(120\text{\hspace{0.17em}s}\right)\\ =8.54\text{J}\\ \approx 8.6\text{J}\end{array}$

Hence, the value of the energy is $8.6\text{J}$