A rectangular black body of temperature 127^{ | vedclass

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(A) According to Stefan-Boltzmann Law,the rate of radiation $E$ is given by $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is

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$8E$

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(A) According to Stefan-Boltzmann Law,the rate of radiation $E$ is given by $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,and $T$ is the absolute temperature in Kelvin.Initial state: $T_1 = 127 + 273 = 400 \ K$,$A_1 = 4 \ cm 	imes 2 \ cm = 8 \ cm^2$,$E_1 = E = \sigma A_1 T_1^4$.Final state: $T_2 = T_1 + 400 = 400 + 400 = 800 \ K$,$A_2 = A_1 / 2 = 4 \ cm^2$.The new rate of radiation $E_2 = \sigma A_2 T_2^4$.Taking the ratio: $E_2 / E_1 = (A_2 / A_1) 	imes (T_2 / T_1)^4$.$E_2 / E = (1/2) 	imes (800 / 400)^4 = (1/2) 	imes (2)^4 = 16 / 2 = 8$.Therefore,$E_2 = 8E$.

$A$ rectangular black body of temperature $127^{\circ} C$ has surface area $4 \ cm \times 2 \ cm$ and rate of radiation is $E$. If its temperature is increased by $400^{\circ} C$ and surface area is reduced to half of the initial value,then the rate of radiation is:

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