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If the radius of a star is $R$ and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is $Q$ ?
( $\sigma$ stands for Stefan's constant.)
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( $\sigma$ stands for Stefan's constant.)
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Verified Answer
The correct answer is:
$\left(Q / 4 \pi R^2 \sigma\right)^{1 / 4}$
From Stefan' law,
$E=\sigma T^4$
So, the rate energy production
$\begin{aligned}
& Q=E \times A \\
& Q=\sigma T^4 \times 4 \pi R^2
\end{aligned}$
Temperature of star
$T=\left(\frac{Q}{4 \pi R^2 \sigma}\right)^{1 / 4}$
$E=\sigma T^4$
So, the rate energy production
$\begin{aligned}
& Q=E \times A \\
& Q=\sigma T^4 \times 4 \pi R^2
\end{aligned}$
Temperature of star
$T=\left(\frac{Q}{4 \pi R^2 \sigma}\right)^{1 / 4}$
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