A black rectangular surface of area A emits e | vedclass

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(A) According to the Stefan-Boltzmann law, the energy emitted per unit time (power) $P$ by a black body is given by $P = \sigma A T^4$, where $\sigma$

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$E$

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(A) According to the Stefan-Boltzmann law, the energy emitted per unit time (power) $P$ by a black body is given by $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature in Kelvin.Initial state: $P_1 = E = \sigma A T_1^4$, where $T_1 = 27 + 273 = 300 	ext{ K}$.Final state: The length and breadth are reduced to $1/4$ of their initial values, so the new area $A' = (A/4) 	imes (1/4) = A/16$. The new temperature $T_2 = 327 + 273 = 600 	ext{ K}$.The new power $P_2 = \sigma A' T_2^4 = \sigma (A/16) (600)^4$.Taking the ratio: $P_2 / P_1 = [\sigma (A/16) (600)^4] / [\sigma A (300)^4] = (1/16) 	imes (600/300)^4 = (1/16) 	imes 2^4 = 16/16 = 1$.Therefore, $P_2 = E$.

$A$ black rectangular surface of area $A$ emits energy $E$ per unit time at $27^{\circ}C$. If length and breadth are reduced to $(1/4)^{th}$ of their initial values and the temperature is raised to $327^{\circ}C$, then the energy emitted per unit time becomes:

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