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(C) According to Stefan-Boltzmann Law,the rate of energy emission per unit area $E$ is proportional to the fourth power of the absolute temperature $T

Vedclass · Accepted Answer

$527$

Vedclass · Answer

(C) According to Stefan-Boltzmann Law,the rate of energy emission per unit area $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.Given:$T_1 = 127^{\circ}C = 127 + 273 = 400\,K$$E_1 = 1.0 	imes 10^6\,J/s\cdot m^2$$E_2 = 16.0 	imes 10^6\,J/s\cdot m^2$Using the ratio formula:$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}ight)^4$Substituting the values:$\frac{16.0 	imes 10^6}{1.0 	imes 10^6} = \left(\frac{T_2}{400}ight)^4$$16 = \left(\frac{T_2}{400}ight)^4$$2^4 = \left(\frac{T_2}{400}ight)^4$Taking the fourth root on both sides:$2 = \frac{T_2}{400}$$T_2 = 800\,K$Converting back to Celsius:$T_2 = 800 - 273 = 527^{\circ}C$.

Energy is being emitted from the surface of a black body at $127\,^{\circ}C$ temperature at the rate of $1.0 \times 10^6\,J/s\cdot m^2$. The temperature of the black body at which the rate of energy emission is $16.0 \times 10^6\,J/s\cdot m^2$ will be ......... $^{\circ}C$.

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