The energy emitted by a black body at 27^{cir | vedclass

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(D) According to the Stefan-Boltzmann law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $E \prop

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(D) According to the Stefan-Boltzmann law,the power radiated by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.Given:Initial temperature $T_1 = 27^{\circ}C = 27 + 273 = 300 \ K$.Final temperature $T_2 = 327^{\circ}C = 327 + 273 = 600 \ K$.Initial energy $E_1 = 10 \ J$.Using the ratio formula:$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} ight)^4$$\frac{E_2}{10} = \left( \frac{600}{300} ight)^4$$\frac{E_2}{10} = (2)^4 = 16$$E_2 = 16 	imes 10 = 160 \ J$.

The energy emitted by a black body at $27^{\circ}C$ is $10 \ J$. If the temperature of the black body is increased to $327^{\circ}C$,the rate of energy emission per second will be ...... $J$.

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