If the radius of a star is R and it acts as a | vedclass

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(B) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $H = \sigma A T^4$,where $A$ is the surface area of the body.

Vedclass · Accepted Answer

$\left(\frac{Q}{4 \pi R^2 \sigma}\right)^{1 / 4}$

Vedclass · Answer

(B) According to the Stefan-Boltzmann law,the power radiated by a black body is given by $H = \sigma A T^4$,where $A$ is the surface area of the body.For a star of radius $R$,the surface area is $A = 4 \pi R^2$.Given that the rate of energy production is $Q$,we have $Q = \sigma (4 \pi R^2) T^4$.Rearranging the equation to solve for the temperature $T$:$T^4 = \frac{Q}{4 \pi R^2 \sigma}$.Taking the fourth root on both sides:$T = \left( \frac{Q}{4 \pi R^2 \sigma} \right)^{1/4}$.

$X = x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

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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