A black body radiates heat energy at the rate | vedclass

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Question

(B) According to Stefan-Boltzmann Law,the rate of heat radiation $E$ is proportional to the fourth power of absolute temperature $T$,i.e.,$E = \sigma

Vedclass · Accepted Answer

$527$

Vedclass · Answer

(B) According to Stefan-Boltzmann Law,the rate of heat radiation $E$ is proportional to the fourth power of absolute temperature $T$,i.e.,$E = \sigma T^4$.Given:$E_1 = 2 	imes 10^5 \, J/s \cdot m^2$$T_1 = 127 + 273 = 400 \, K$$E_2 = 32 	imes 10^5 \, J/s \cdot m^2$Using the ratio formula:$\frac{E_2}{E_1} = \left(\frac{T_2}{T_1}ight)^4$$\frac{32 	imes 10^5}{2 	imes 10^5} = \left(\frac{T_2}{400}ight)^4$$16 = \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$.

$A$ black body radiates heat energy at the rate of $2 \times 10^5 \, J/s \cdot m^2$ at a temperature of $127 \, ^\circ C$. The temperature of the black body at which the rate becomes $32 \times 10^5 \, J/s \cdot m^2$ is ....... $^\circ C$.

Similar Questions

$A$ spherical black body of radius $r$ radiates power $P$ and its rate of cooling is $R$. Which of the following relations are correct?
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$(iii) \ R \propto r^2$
$(iv) \ R \propto \frac{1}{r}$

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