A solid cylinder of radius r_1=2.5 cm, lengt | vedclass

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(A) The rate of heat transfer by radiation is given by the Stefan-Boltzmann law: $P = e \sigma A (T^4 - T_0^4)$. Since the temperature of the body, th

Vedclass · Accepted Answer

$3.35$

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(A) The rate of heat transfer by radiation is given by the Stefan-Boltzmann law: $P = e \sigma A (T^4 - T_0^4)$. Since the temperature of the body, the temperature of the surroundings, and the nature of the surface (emissivity $e$) remain constant, the rate of heat transfer is directly proportional to the surface area $A$ of the cylinder.$\Rightarrow P \propto A$$\Rightarrow \frac{P_2}{P_1} = \frac{A_2}{A_1} \Rightarrow P_2 = P_1 	imes \frac{A_2}{A_1} \quad \dots(i)$Since the volume of the cylinder remains constant during stretching, $V = \pi r_1^2 l_1 = \pi r_2^2 l_2$.$l_2 = l_1 \left( \frac{r_1}{r_2} ight)^2 = 5.0 	imes \left( \frac{2.5}{0.5} ight)^2 = 5.0 	imes 25 = 125 \ cm$.The total surface area of a cylinder is $A = 2\pi r l + 2\pi r^2 = 2\pi r(l + r)$.$A_1 = 2\pi (2.5)(5.0 + 2.5) = 2\pi (2.5)(7.5) = 37.5\pi \ cm^2$.$A_2 = 2\pi (0.5)(125 + 0.5) = 2\pi (0.5)(125.5) = 125.5\pi \ cm^2$.Substituting these into Eq. $(i)$:$P_2 = 1.0 	imes \frac{125.5\pi}{37.5\pi} = \frac{125.5}{37.5} \approx 3.346 \ W \approx 3.35 \ W$.

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