Solved Problem on Black Body Radiation
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a) Determine the rest mass lost per second by the sun in the form of radiation. Data: sun surface temperature: 5700 K; sun diameter: 1.4 × 109 m; Stefan-Boltzmann constant: \( \sigma =5.67\times 10^{-8}\;\frac{\text{W}}{\text{m}^{2}.\text{T}^{4}} \); speed of light in vacuum: 3.0 × 108 m/s;
b) What is the fraction of the rest mass lost each year by the sun from the electromagnetic radiation. Solar mass: 2.0 × 1030 kg.
Problem data:
- Sun surface temperature: T = 5700 K;
- Sun diameter: D =1.4 × 109 m;
- Solar mass: M = 2.0 × 1030 kg;
- Stefan-Boltzmann constant: \( \sigma =5.67\times 10^{-8}\;\frac{\text{W}}{\text{m}^{2}.\text{T}^{4}} \);
- Speed of light in vacuum: v = 3.0 × 108 m/s.
a) Using the Stefan-Boltzmann Law, spectral radiance is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{R_{T}=\sigma T^{4}} \tag{I}
\end{gather}
\]
Total radiance is defined as the irradiated power by area unit, we can write
\[
\begin{gather}
R_{T}=\frac{P}{A} \tag{II}
\end{gather}
\]
equating the expressions (I) and (II)
\[
\begin{gather}
\sigma T^{4}=\frac{P}{A}\\[5pt]
P=A\;\sigma T^{4} \tag{III}
\end{gather}
\]
Assuming that the sun is spherical, the surface area will be
\[
\begin{gather}
A=4\pi r^{2} \tag{IV}
\end{gather}
\]
The radius of a sphere is half the diameter
\[
\begin{gather}
r=\frac{D}{2} \tag{V}
\end{gather}
\]
substituting the expression (V) into expression (IV) and then in expression (III)
\[
\begin{gather}
P=4\pi \left(\frac{D}{2}\right)^{2}\;\sigma T^{4}\\[5pt]
P=4\pi \frac{D^{2}}{4}\;\sigma T^{4}\\[5pt]
P=\pi \;D^{2}\;\sigma T^{4} \tag{VI}
\end{gather}
\]
The power is given by the energy differentiation relative to time
\[
\begin{gather}
P=\frac{dE}{dt} \tag{VII}
\end{gather}
\]
the rest energy is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{E=m_{0}c^{2}} \tag{VIII}
\end{gather}
\]
where m0 is the rest mass, substituting the expressions (VI) and (VIII) into expression
(VII)
\[
\begin{gather}
\pi D^{2}\sigma T^{4}=\frac{d\left(m_{0}c^{2}\right)}{dt}
\end{gather}
\]
as the speed of light is constant
\[
\begin{gather}
c^{2}\frac{dm_{0}}{dt}=\pi D^{2}\sigma T^{4}
\end{gather}
\]
thus the variation of the mass will be
\[
\begin{gather}
\frac{dm_{0}}{dt}=\frac{\pi D^{2}\sigma T^{4}}{c^{2}}
\end{gather}
\]
assuming π = 3.14 and substituting the data of the problem
\[
\begin{gather}
\frac{dm_{0}}{dt}=\frac{3.14\times (1.4\times 10^{9})^{2}\times (5.67\times 10^{-8})\times (5700)^{4}}{(3\times 10^{8})^{2}}\\[5pt]
\frac{dm_{0}}{dt}=\frac{3.14\times 1.96\times 10^{18}\times 5.67\times 10^{-8}\times 1.06\times 10^{15}}{9\times 10^{16}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\frac{dm_{0}}{dt}=4.1\times 10^{9}\ \frac{\text{kg}}{\text{s}}}
\end{gather}
\]
b) One year in seconds is equal to
\[
\begin{gather}
t=1\;\cancel{\text{ano}}\times \;\frac{365\;\cancel{\text{dias}}}{1\;\cancel{\text{ano}}}\times \frac{24\;\cancel{\text{horas}}}{1\;\cancel{\text{dia}}}\times \frac{60\;\cancel{\text{minutos}}}{1\;\cancel{\text{hora}}}\times \frac{60\;\text{segundos}}{1\;\cancel{\text{minuto}}}=31,536,000\;\approx\;3.15\times 10^{7}\text{s}
\end{gather}
\]
Separating the variables and integrating the expression of the previous item
\[
\begin{gather}
\int d{m{\text '}}_{0}=\int 4.1\times 10^{9}dt{\text '}
\end{gather}
\]
Note: In the expression above m' and t' are dummy variables of integration.
The limits of the integral of the mass is from 0 to m0 (the total lost mass), the integral in time is from 0 to 3.15×107 s (one year in seconds), and moving out the constant 4.1× 109 from the integral
\[
\begin{gather}
\int_{{0}}^{m_{0}}d{m{\text '}}_{0}=\int_{{0}}^{3.15\times 10^{7}}4.1\times 10^{9}dt{\text '}\\[5pt]
\left.m{\text '}_{0}\;\right|_{0}^{m_{0}}=\left.t{\text '}\;\right|_{0}^{3.15\times 10^{7}}\\[5pt]
m_{0}-0=4.1\times 10^{9}\times (3.15\times 10^{7}-0)\\[5pt]
m_{0}=4.1\times 10^{9}\times 3.15\times 10^{7}\\[5pt]
m_{0}=1.3\times 10^{17}\ \text{kg}
\end{gather}
\]
so the lost mass fraction each year will be
\[
\begin{gather}
\frac{m_{0}}{M}=\frac{1.3\times 10^{17}}{2.0\times 10^{30}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\frac{m_{0}}{M}=6.5\times 10^{-14}}
\end{gather}
\]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .