The energy emitted per second by a black body | vedclass

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(D) According to Stefan-Boltzmann law,the energy emitted per second $(P)$ by a black body is given by $P = A\sigma T^4$,where $A$ is the surface area,

Vedclass · Accepted Answer

$160$

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(D) According to Stefan-Boltzmann law,the energy emitted per second $(P)$ by a black body is given by $P = A\sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature in Kelvin.Since $A$ and $\sigma$ are constants,we have $P \propto T^4$.Given $T_1 = 27^{\circ}C = 27 + 273 = 300\;K$ and $P_1 = 10\;J/s$.Given $T_2 = 327^{\circ}C = 327 + 273 = 600\;K$.Using the ratio: $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} ight)^4$.Substituting the values: $P_2 = P_1 \left( \frac{600}{300} ight)^4 = 10 	imes (2)^4 = 10 	imes 16 = 160\;J/s$.Therefore,the energy emitted per second is $160\;J$.

The energy emitted per second 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 energy emitted per second will be ......... $J$.

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