A tungsten body of diameter 2.3 cm is at 200 | vedclass

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(C) The rate of energy radiation is given by the Stefan-Boltzmann law: $P = e \sigma A T^4$.For a black body,emissivity $e = 1$.Let $R_1$ be the radiu

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$0.63$

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(C) The rate of energy radiation is given by the Stefan-Boltzmann law: $P = e \sigma A T^4$.For a black body,emissivity $e = 1$.Let $R_1$ be the radius of the tungsten body and $R$ be the radius of the black body.The power radiated by the tungsten body is $P_T = e_T \sigma (4 \pi R_1^2) T^4$,where $e_T = 0.3$.The power radiated by a black body of the same radius $R_1$ is $P_B = 1 \cdot \sigma (4 \pi R_1^2) T^4$.Given that the tungsten body radiates $30 \%$ of the energy of a black body of the same radius,we have $P_T = 0.3 P_B$.We want to find the radius $R_{new}$ of a black body that radiates at the same rate as the tungsten body at the same temperature: $P_{new} = 1 \cdot \sigma (4 \pi R_{new}^2) T^4 = P_T$.Since $P_T = 0.3 \cdot \sigma (4 \pi R_1^2) T^4$,we equate the two:$\sigma (4 \pi R_{new}^2) T^4 = 0.3 \sigma (4 \pi R_1^2) T^4$.$R_{new}^2 = 0.3 R_1^2$.$R_{new} = \sqrt{0.3} \cdot R_1$.Given diameter $d = 2.3 \ cm$,so $R_1 = 1.15 \ cm$.$R_{new} = \sqrt{0.3} \cdot 1.15 \approx 0.5477 \cdot 1.15 \approx 0.6299 \ cm \approx 0.63 \ cm$.

$A$ tungsten body of diameter $2.3 \ cm$ is at $2000 \ ^oC$. It radiates $30 \%$ of the energy radiated by a black body of the same radius and temperature. Find the radius of a black body which will radiate energy at the same rate at the same temperature (in $cm$).

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