A black rectangular surface of area A emits e | vedclass

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(D) According to the Stefan-Boltzmann law,the energy emitted per second by a black body is given by $E = \sigma A T^4$. Initially,$E = \sigma A T_1^4$

Vedclass · Accepted Answer

$\frac{16 E}{9}$

Vedclass · Answer

(D) According to the Stefan-Boltzmann law,the energy emitted per second by a black body is given by $E = \sigma A T^4$. Initially,$E = \sigma A T_1^4$,where $T_1 = 27 + 273 = 300 \ K$. 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 $T_2 = 327 + 273 = 600 \ K$. The new energy emitted is $E' = \sigma A' T_2^4$. Taking the ratio: $\frac{E'}{E} = \frac{A'}{A} 	imes \left(\frac{T_2}{T_1}ight)^4$. Substituting the values: $\frac{E'}{E} = \frac{1}{9} 	imes \left(\frac{600}{300}ight)^4$. $\frac{E'}{E} = \frac{1}{9} 	imes (2)^4 = \frac{16}{9}$. 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 $1/3$ of their initial values and the temperature is raised to $327^{\circ} C$,then the energy emitted per second becomes:

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