The energy emitted per second by a black body | vedclass

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(D) According to Stefan-Boltzmann Law,the power $P$ (energy emitted per second) radiated by a black body is proportional to the fourth power of its ab

Vedclass · Accepted Answer

$160$

Vedclass · Answer

(D) According to Stefan-Boltzmann Law,the power $P$ (energy emitted per second) radiated by a black body is proportional to the fourth power of its absolute temperature $T$ (in Kelvin).$P \propto T^{4}$Given,$P_{1} = 10 \ J/s$ at $T_{1} = 27^{\circ}C = (27 + 273) \ K = 300 \ K$.When the temperature is increased to $T_{2} = 327^{\circ}C = (327 + 273) \ K = 600 \ K$.Using the ratio method:$\frac{P_{2}}{P_{1}} = \left(\frac{T_{2}}{T_{1}}ight)^{4}$$\frac{P_{2}}{10} = \left(\frac{600}{300}ight)^{4}$$\frac{P_{2}}{10} = (2)^{4} = 16$$P_{2} = 16 	imes 10 = 160 \ J/s$.Therefore,the energy emitted per second will be $160 \ J$.

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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