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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 $Q$ ?
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Verified Answer
The correct answer is:
$\left(\mathrm{Q} / 4 \pi R^{2} \sigma\right)^{1 / 4}$
Stefan's law for black bodyradiation $Q=\sigma e A T^{4}$
$$
T=\left[\frac{Q}{\sigma\left(4 \pi R^{2}\right)}\right]^{1 / 4}
$$
Here $e=1$
$$
A=4 \pi R^{2}
$$
$$
T=\left[\frac{Q}{\sigma\left(4 \pi R^{2}\right)}\right]^{1 / 4}
$$
Here $e=1$
$$
A=4 \pi R^{2}
$$
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