A black body of surface area 10 cm^2 is heat | vedclass

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(D) According to Stefan-Boltzmann Law,the rate of loss of heat $P$ is given by $P = A \varepsilon \sigma (T^4 - T_0^4)$.Given:Surface area $A = 10 \ c

Vedclass · Accepted Answer

$0.99$

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(D) According to Stefan-Boltzmann Law,the rate of loss of heat $P$ is given by $P = A \varepsilon \sigma (T^4 - T_0^4)$.Given:Surface area $A = 10 \ cm^2 = 10 	imes 10^{-4} \ m^2 = 10^{-3} \ m^2$.Emissivity $\varepsilon = 1$ (for a black body).Stefan's constant $\sigma = 5.67 	imes 10^{-8} \ W \ m^{-2} \ K^{-4}$.Temperature of the body $T = 127^{\circ}C = 127 + 273 = 400 \ K$.Temperature of the room $T_0 = 27^{\circ}C = 27 + 273 = 300 \ K$.Substituting the values:$P = 10^{-3} 	imes 1 	imes 5.67 	imes 10^{-8} 	imes (400^4 - 300^4)$$P = 5.67 	imes 10^{-11} 	imes (256 	imes 10^8 - 81 	imes 10^8)$$P = 5.67 	imes 10^{-11} 	imes 175 	imes 10^8$$P = 5.67 	imes 1.75 = 9.9225 	imes 10^{-1} \approx 0.99 \ W$.

$A$ black body of surface area $10 \ cm^2$ is heated to $127^{\circ}C$ and is suspended in a room at temperature $27^{\circ}C$. The initial rate of loss of heat from the body at the room temperature will be ...... $W$.

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