The rate of radiation of a black body at 0^{c | vedclass

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(A) According to Stefan-Boltzmann Law,the rate of radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^

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$16 \ E$

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(A) According to Stefan-Boltzmann Law,the rate of radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.Given:Initial temperature $T_1 = 0^{\circ}C = 0 + 273 = 273 \ K$.Final temperature $T_2 = 273^{\circ}C = 273 + 273 = 546 \ K$.Initial rate of radiation = $E$.Using the ratio formula: $\frac{E_1}{E_2} = \left( \frac{T_1}{T_2} \right)^4$.Substituting the values: $\frac{E}{E_2} = \left( \frac{273}{546} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$.Therefore,$E_2 = 16 \ E$.

The rate of radiation of a black body at $0^{\circ}C$ is $E \ J/s$. The rate of radiation of this black body at $273^{\circ}C$ will be:

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