Heat loss takes place from a body maintained | vedclass

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(D) Given that the rate of heat loss by convection is equal to the rate of heat loss by radiation.Let $h$ be the convection coefficient, $A$ be the su

Vedclass · Accepted Answer

$0.66$

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(D) Given that the rate of heat loss by convection is equal to the rate of heat loss by radiation.Let $h$ be the convection coefficient, $A$ be the surface area, $e$ be the emissivity, $\sigma$ be the Stefan-Boltzmann constant, $T$ be the body temperature, and $T_0$ be the surrounding temperature.$T = 400 + 273 = 673 \ K$$T_0 = 30 + 273 = 303 \ K$The rate of heat loss by convection is $P_{conv} = hA(T - T_0)$.The rate of heat loss by radiation is $P_{rad} = eA\sigma(T^4 - T_0^4)$.Equating the two: $hA(T - T_0) = eA\sigma(T^4 - T_0^4)$.$20(673 - 303) = e(5.67 	imes 10^{-8})(673^4 - 303^4)$.$20(370) = e(5.67 	imes 10^{-8})(2.049 	imes 10^{11} - 0.0084 	imes 10^{11})$.$7400 = e(5.67 	imes 10^{-8})(2.0406 	imes 10^{11})$.$7400 = e(11570.2)$.$e = 7400 / 11570.2 \approx 0.6397$.Rounding to the nearest option, $e \approx 0.66$.

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