A rectangular block of surface area A emits e | vedclass

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

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$4 E$

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(B) According to Stefan-Boltzmann Law,the energy emitted per second (power) $P$ 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 $T_1 = 27 + 273 = 300 \ K$.Final state: Length and breadth are halved,so the new area $A' = (L/2) 	imes (B/2) = A/4$. The new temperature $T_2 = 327 + 273 = 600 \ K$.The new power $P_2 = \sigma A' T_2^4 = \sigma (A/4) (600)^4$.Taking the ratio: $P_2 / P_1 = [\sigma (A/4) (600)^4] / [\sigma A (300)^4] = (1/4) 	imes (600/300)^4 = (1/4) 	imes 2^4 = (1/4) 	imes 16 = 4$.Therefore,$P_2 = 4 E$.

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

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