A black rectangular surface of area A emits e | vedclass

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(C) According to the Stefan-Boltzmann law,the energy emitted per second is given by $E = \sigma A T^4$.Here,$\sigma$ is the Stefan-Boltzmann constant,

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$16 E / 9$

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(C) According to the Stefan-Boltzmann law,the energy emitted per second is given by $E = \sigma A T^4$.Here,$\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature in Kelvin.Let the initial state be $E_1 = \sigma A_1 T_1^4$ and the final state be $E_2 = \sigma A_2 T_2^4$.Given that the length and breadth are reduced to $1/3$ of their initial values,the new area $A_2 = (L/3) 	imes (B/3) = A_1 / 9$.Initial temperature $T_1 = 27 + 273 = 300 \ K$.Final temperature $T_2 = 327 + 273 = 600 \ K$.Taking the ratio: $\frac{E_2}{E_1} = \frac{A_2}{A_1} 	imes \left(\frac{T_2}{T_1}ight)^4$.Substituting the values: $\frac{E_2}{E_1} = \frac{1}{9} 	imes \left(\frac{600}{300}ight)^4 = \frac{1}{9} 	imes (2)^4 = \frac{16}{9}$.Therefore,$E_2 = \frac{16}{9} E$.

$A$ black rectangular surface of area $A$ emits energy $E$ per second at $27^{\circ} C$. If length and breadth are reduced to $1/3$ of their initial values and the temperature is raised to $327^{\circ} C$,then the energy emitted per second becomes:

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