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

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(C) According to Stefan-Boltzmann Law,the energy emitted per second by a black body is given by $E = \sigma A T^4$.Initially,$E = \sigma A T^4$,where

Vedclass · Accepted Answer

$\frac{16 E}{9}$

Vedclass · Answer

(C) According to Stefan-Boltzmann Law,the energy emitted per second by a black body is given by $E = \sigma A T^4$.Initially,$E = \sigma A T^4$,where $T = 27^{\circ}C = 300 \ K$.When the length and breadth are reduced to $1/3$ of their initial values,the new area $A' = (L/3) 	imes (B/3) = A/9$.The new temperature $T' = 327^{\circ}C = 600 \ K$.The new energy emitted per second is $E' = \sigma A' (T')^4$.Taking the ratio: $\frac{E'}{E} = \frac{A'}{A} 	imes \left(\frac{T'}{T}ight)^4$.Substituting the values: $\frac{E'}{E} = \frac{1}{9} 	imes \left(\frac{600}{300}ight)^4 = \frac{1}{9} 	imes (2)^4 = \frac{16}{9}$.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 $(1/3)^{rd}$ of their initial values and temperature is raised to $327^{\circ}C$,then the energy emitted per second becomes:

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