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(D) The rate of energy emission (power) from a black body is given by the Stefan-Boltzmann law: $P = A \sigma T^4$,where $A$ is the surface area,$\sig

Vedclass · Accepted Answer

$\frac{81}{64} E$

Vedclass · Answer

(D) The rate of energy emission (power) from a black body is given by the Stefan-Boltzmann law: $P = A \sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.Given initial state: $A_1 = 8 \ cm 	imes 4 \ cm = 32 \ cm^2$,$T_1 = 127 + 273 = 400 \ K$,$P_1 = E$.Given final state: $A_2 = (8/2) \ cm 	imes (4/2) \ cm = 4 \ cm 	imes 2 \ cm = 8 \ cm^2$,$T_2 = 327 + 273 = 600 \ K$.Since $A_2 = A_1 / 4$,the ratio of power is:$\frac{P_2}{P_1} = \frac{A_2}{A_1} 	imes \left( \frac{T_2}{T_1} ight)^4$$\frac{P_2}{E} = \frac{8}{32} 	imes \left( \frac{600}{400} ight)^4$$\frac{P_2}{E} = \frac{1}{4} 	imes \left( \frac{3}{2} ight)^4 = \frac{1}{4} 	imes \frac{81}{16} = \frac{81}{64}$Therefore,the new rate of energy emission is $P_2 = \frac{81}{64} E$.

$A$ rectangular surface of a black body at $127^{\circ}C$ with dimensions $8 \ cm \times 4 \ cm$ emits energy at a rate of $E$. If the length and width are halved and the temperature is increased to $327^{\circ}C$,find the new rate of energy emission.

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