A sphere of surface area 4 m^2 at temperatur | vedclass

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(C) The net rate of energy exchange of the sphere is given by the Stefan-Boltzmann Law: $P = \varepsilon \sigma A (T^4 - T_s^4)$.Given:Emissivity $\va

Vedclass · Accepted Answer

$2721.6$

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(C) The net rate of energy exchange of the sphere is given by the Stefan-Boltzmann Law: $P = \varepsilon \sigma A (T^4 - T_s^4)$.Given:Emissivity $\varepsilon = 0.5$Stefan-Boltzmann constant $\sigma = 5.67 	imes 10^{-8} \ W \ m^{-2} \ K^{-4}$Surface area $A = 4 \ m^2$Temperature of the sphere $T = 400 \ K$Temperature of the environment $T_s = 200 \ K$Substituting the values into the formula:$P = 0.5 	imes (5.67 	imes 10^{-8}) 	imes 4 	imes [400^4 - 200^4]$$P = 2 	imes 5.67 	imes 10^{-8} 	imes [(4 	imes 10^2)^4 - (2 	imes 10^2)^4]$$P = 11.34 	imes 10^{-8} 	imes [256 	imes 10^8 - 16 	imes 10^8]$$P = 11.34 	imes 10^{-8} 	imes [240 	imes 10^8]$$P = 11.34 	imes 240 = 2721.6 \ W$.

$A$ sphere of surface area $4 \ m^2$ at temperature $400 \ K$ and having emissivity $0.5$ is located in an environment of temperature $200 \ K$. The net rate of energy exchange of the sphere is (Stefan-Boltzmann constant $\sigma = 5.67 \times 10^{-8} \ W \ m^{-2} \ K^{-4}$) (in $W$)

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