A black body at 227^o C radiates heat at the | vedclass

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

Vedclass · Accepted Answer

$112$

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(B) According to Stefan-Boltzmann law,the rate of heat radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.Given:Initial temperature $T_1 = 227^o C = (227 + 273) K = 500 K$.Initial rate of radiation $E_1 = 7 \; cal/cm^2 s$.Final temperature $T_2 = 727^o C = (727 + 273) K = 1000 K$.Using the ratio formula:$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} ight)^4$Substituting the values:$\frac{E_2}{7} = \left( \frac{1000}{500} ight)^4$$\frac{E_2}{7} = (2)^4$$\frac{E_2}{7} = 16$$E_2 = 7 	imes 16 = 112 \; cal/cm^2 s$.Therefore,the rate of heat radiated at $727^o C$ is $112 \; cal/cm^2 s$.

$A$ black body at $227^o C$ radiates heat at the rate of $7 \; cal/cm^2 s$. At a temperature of $727^o C$,the rate of heat radiated in the same units will be:

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