
In the radius of a star $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is $Q$? ($\sigma$ for Stefan’s constant)
\[\begin{align}
& A.{{\left( \dfrac{4\pi {{R}^{2}}Q}{\sigma } \right)}^{\dfrac{1}{4}}} \\
& B.{{\left( \dfrac{Q}{4\pi {{R}^{2}}\sigma } \right)}^{\dfrac{1}{4}}} \\
& C.\left( \dfrac{Q}{4\pi {{R}^{2}}\sigma } \right) \\
& D.{{\left( \dfrac{Q}{4\pi {{R}^{2}}\sigma } \right)}^{\dfrac{-1}{4}}} \\
\end{align}\]
Answer
588k+ views
Hint: We know that the rate at which a body absorbs heat is given by the Stefan–Boltzmann law. Here to find the temperature of the star, we must equate the Stefan–Boltzmann law to the energy produced in the star, as the start is a black body.
Formula used:
$\sigma AT^{4}=E_{ab}$
Complete step-by-step answer:
Stefan-Boltzmann constant or the Stefan’s constant is denoted by $\sigma$ , it is a constant of proportionality, used in the Stefan–Boltzmann law of blackbody radiation; which is stated as the total intensity radiated over the wavelength, it is proportional the temperature. It is given as $E_{ab}=\sigma AT^{4}$ where $E_{ab}$ is the energy absorbed by the body here the star, whose surface area is $A$ and temperature is $T$.
Let us assume the star to be spherical, then the surface area of the star is given as $A=4\pi R^{2}$ where $R$ is the radius of the star.
Also given that the rate of energy production or the energy radiated by the star is $Q$.
Since it is given that the body is a black body, we know that the black body absorbs and radiates all heat. This means that there is no loss in energy during the transmission of heat, then we can say that, $E_{abs}=Q$
Then, $Q=\sigma AT^{4}$
Or, we get $T^{4}=\dfrac{Q}{\sigma A}$
Or $T^{4}=\dfrac{Q}{\sigma 4\pi R^{2}}$
Then we get that the temperature $T=\left(\dfrac{Q}{\sigma 4\pi R^{2}}\right)^{\dfrac{1}{4}}$
Hence the answer is \[B.{{\left( \dfrac{Q}{4\pi {{R}^{2}}\sigma } \right)}^{\dfrac{1}{4}}}\]
So, the correct answer is “Option B”.
Note: Stefan-Boltzmann constant is $5.673\times10^{-8} \dfrac{kg}{sec^{3}kelvin^{4}}$, where SI units are $\dfrac{kg}{sec^{3}kelvin^{4}}$, this can be used to reduce dimension of the constant. The value of the Stefan–Boltzmann constant is derivable and experimentally tested as well. It is derived from the Boltzmann constant.
Formula used:
$\sigma AT^{4}=E_{ab}$
Complete step-by-step answer:
Stefan-Boltzmann constant or the Stefan’s constant is denoted by $\sigma$ , it is a constant of proportionality, used in the Stefan–Boltzmann law of blackbody radiation; which is stated as the total intensity radiated over the wavelength, it is proportional the temperature. It is given as $E_{ab}=\sigma AT^{4}$ where $E_{ab}$ is the energy absorbed by the body here the star, whose surface area is $A$ and temperature is $T$.
Let us assume the star to be spherical, then the surface area of the star is given as $A=4\pi R^{2}$ where $R$ is the radius of the star.
Also given that the rate of energy production or the energy radiated by the star is $Q$.
Since it is given that the body is a black body, we know that the black body absorbs and radiates all heat. This means that there is no loss in energy during the transmission of heat, then we can say that, $E_{abs}=Q$
Then, $Q=\sigma AT^{4}$
Or, we get $T^{4}=\dfrac{Q}{\sigma A}$
Or $T^{4}=\dfrac{Q}{\sigma 4\pi R^{2}}$
Then we get that the temperature $T=\left(\dfrac{Q}{\sigma 4\pi R^{2}}\right)^{\dfrac{1}{4}}$
Hence the answer is \[B.{{\left( \dfrac{Q}{4\pi {{R}^{2}}\sigma } \right)}^{\dfrac{1}{4}}}\]
So, the correct answer is “Option B”.
Note: Stefan-Boltzmann constant is $5.673\times10^{-8} \dfrac{kg}{sec^{3}kelvin^{4}}$, where SI units are $\dfrac{kg}{sec^{3}kelvin^{4}}$, this can be used to reduce dimension of the constant. The value of the Stefan–Boltzmann constant is derivable and experimentally tested as well. It is derived from the Boltzmann constant.
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