A black rectangular surface of area 'A' emits | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{16 E}{9}$

Vedclass · Answer

(A) According to Stefan-Boltzmann Law,the energy emitted per second is given by $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the area,and $T$ is the absolute temperature in Kelvin.Initial state: $E = \sigma A T_1^4$,where $T_1 = 27 + 273 = 300 \ K$.Final state: $E' = \sigma A' T_2^4$,where $T_2 = 327 + 273 = 600 \ K$.The area $A = \ell 	imes b$. If length and breadth are reduced to $\frac{1}{3}$ of their initial values,the new area $A' = (\frac{\ell}{3}) 	imes (\frac{b}{3}) = \frac{A}{9}$.Taking the ratio: $\frac{E'}{E} = \frac{A'}{A} 	imes (\frac{T_2}{T_1})^4$.Substituting the values: $\frac{E'}{E} = \frac{1}{9} 	imes (\frac{600}{300})^4 = \frac{1}{9} 	imes (2)^4 = \frac{16}{9}$.Therefore,$E' = \frac{16 E}{9}$.

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

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