A metal ball of surface area 200 ; cm^2 and t | vedclass

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(C) The rate of heat loss $P$ by radiation is given by Stefan-Boltzmann Law: $P = \sigma e A (T^4 - T_0^4)$.Given:Surface area $A = 200 \; cm^2 = 200

Vedclass · Accepted Answer

$182$

Vedclass · Answer

(C) The rate of heat loss $P$ by radiation is given by Stefan-Boltzmann Law: $P = \sigma e A (T^4 - T_0^4)$.Given:Surface area $A = 200 \; cm^2 = 200 	imes 10^{-4} \; m^2 = 0.02 \; m^2$.Emissivity $e = 0.4$.Temperature of the ball $T = 527^{\circ}C = 527 + 273 = 800 \; K$.Temperature of the surroundings $T_0 = 27^{\circ}C = 27 + 273 = 300 \; K$.Stefan's constant $\sigma = 5.67 	imes 10^{-8} \; J/(m^2 \cdot s \cdot K^4)$.Substituting the values:$P = 5.67 	imes 10^{-8} 	imes 0.4 	imes 0.02 	imes (800^4 - 300^4)$.$P = 5.67 	imes 10^{-8} 	imes 0.008 	imes (4096 	imes 10^8 - 81 	imes 10^8)$.$P = 5.67 	imes 0.008 	imes (4015)$.$P = 0.04536 	imes 4015 \approx 182.12 \; J/s$.Thus,the rate of loss of heat is approximately $182 \; J/s$.

$A$ metal ball of surface area $200 \; cm^2$ and temperature $527^{\circ}C$ is surrounded by a vessel at $27^{\circ}C$. If the emissivity of the metal is $0.4$,then the rate of loss of heat from the ball is approximately .......... $J/s$. $(\sigma = 5.67 \times 10^{-8} \; J/(m^2 \cdot s \cdot K^4))$

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