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Question

(A) The rate of heat loss per unit area due to radiation,also known as the emissive power $e$,is given by the Stefan-Boltzmann law:$e = \varepsilon \s

Vedclass · Accepted Answer

$595 \, J/m^2 \cdot s$

Vedclass · Answer

(A) The rate of heat loss per unit area due to radiation,also known as the emissive power $e$,is given by the Stefan-Boltzmann law:$e = \varepsilon \sigma (T^4 - T_0^4)$Given:Emissivity $\varepsilon = 0.6$Stefan-Boltzmann constant $\sigma = \frac{17}{3} 	imes 10^{-8} \, W/m^2 K^4$Surface temperature $T = 127^\circ C = 127 + 273 = 400 \, K$Surrounding temperature $T_0 = 27^\circ C = 27 + 273 = 300 \, K$Substituting the values:$e = 0.6 	imes \left( \frac{17}{3} 	imes 10^{-8} ight) 	imes [ (400)^4 - (300)^4 ]$$e = 0.2 	imes 17 	imes 10^{-8} 	imes [ 256 	imes 10^8 - 81 	imes 10^8 ]$$e = 3.4 	imes 10^{-8} 	imes (175 	imes 10^8)$$e = 3.4 	imes 175 = 595 \, J/m^2 \cdot s$

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