The surface of a black body is at a temperature 727^ C and its cross-section is 1m^2. Heat radiated from this surface in one minute in joules is (Stefan's constant =5.7 10^-8W/m^2/k^4 )

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Correct answer: A. (a) Given, Temperature of black body surface, T=727^ C=727^ +273^ =1000K . Cross-sectional area of black body, A=1m^2 Stefan's constant, =5.7 10^-8W/m^2/K^4 From the Stefan-Boltzmann law, energy radiated by a body is, E= σAT^4=5.7 10^-8 1 (1000 )^4 ⇒E=57000J . ∴ Heat radiated from black body surface in one minute, E=57000 60 ⇒E=3.42 10^6 or34.2 10^5J .

$The surface of a black body is at a temperature 727^{\circ} C and its cross-section is 1 m^{2} . Heat radiated from this surface in one minute in joules is (Stefan's constant = 5.7 \times 10^{- 8} W / m^{2} / k^{4})$

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