Two spheres S_{1} and S_{2} have the same rad | vedclass

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(A) The emissive power $P$ of a body is given by the Stefan-Boltzmann law: $P = \sigma e T^{4}$,where $\sigma$ is the Stefan-Boltzmann constant,$e$ is

Vedclass · Accepted Answer

$\sqrt{2}: 1$

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(A) The emissive power $P$ of a body is given by the Stefan-Boltzmann law: $P = \sigma e T^{4}$,where $\sigma$ is the Stefan-Boltzmann constant,$e$ is the emissivity,and $T$ is the absolute temperature.Given that the emissive power of both spheres is the same,we have $P_{1} = P_{2}$.Therefore,$\sigma e_{1} T_{1}^{4} = \sigma e_{2} T_{2}^{4}$.This simplifies to $\frac{T_{1}^{4}}{T_{2}^{4}} = \frac{e_{2}}{e_{1}}$.Given the ratio of emissivity $e_{1} : e_{2} = 1 : 4$,we have $\frac{e_{2}}{e_{1}} = \frac{4}{1} = 4$.Substituting this into the equation: $\frac{T_{1}^{4}}{T_{2}^{4}} = 4$.Taking the fourth root on both sides: $\frac{T_{1}}{T_{2}} = (4)^{1/4} = (2^{2})^{1/4} = 2^{1/2} = \sqrt{2}$.Thus,the ratio $T_{1} : T_{2} = \sqrt{2} : 1$.

Two spheres $S_{1}$ and $S_{2}$ have the same radii but temperatures $T_{1}$ and $T_{2}$ respectively. Their emissive power is the same and their emissivity is in the ratio $1: 4$. Then the ratio of $T_{1}$ to $T_{2}$ is:

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