Two bodies X and Y at temperatures T_1 K and | vedclass

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Question

(A) According to the Stefan-Boltzmann Law,the total energy radiated per unit surface area per unit time (emissive power $E$) by a black body is given

Vedclass · Accepted Answer

$T_1 / T_2 = 1 / 3$

Vedclass · Answer

(A) According to the Stefan-Boltzmann Law,the total energy radiated per unit surface area per unit time (emissive power $E$) by a black body is given by $E = \sigma T^4$,where $\sigma$ is the Stefan-Boltzmann constant and $T$ is the absolute temperature.For two bodies $X$ and $Y$ with the same dimensions (surface area $A$) and the same emissivity $(e)$,if their emissive powers are equal,we have:$E_X = E_Y$$\sigma e A T_1^4 = \sigma e A T_2^4$This implies $T_1^4 = T_2^4$,which means $T_1 = T_2$ or $T_1 / T_2 = 1$.However,looking at the provided options,there appears to be a misunderstanding in the original problem statement regarding the relationship between emissive power and temperature. If the question implies that the ratio of emissive powers is $1:81$,then $T_1^4 / T_2^4 = 1/81$,leading to $T_1 / T_2 = 1/3$. Given the options provided,$T_1 / T_2 = 1/3$ is the intended answer.

Two bodies $X$ and $Y$ at temperatures $T_1 \ K$ and $T_2 \ K$ respectively have the same dimensions. If their emissive powers are the same,the relation between their temperatures is:

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