Two thin metallic spherical shells of radii 2 | vedclass

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(C) Given,radii of two thin metallic spherical shells are $r_1 = 20 	ext{ cm} = 0.2 	ext{ m}$ and $r_2 = 30 	ext{ cm} = 0.3 	ext{ m}$.Temperature

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$\frac{5}{3 \pi}$

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(C) Given,radii of two thin metallic spherical shells are $r_1 = 20 	ext{ cm} = 0.2 	ext{ m}$ and $r_2 = 30 	ext{ cm} = 0.3 	ext{ m}$.Temperature of inner shell $T_1 = 300 	ext{ K}$,temperature of outer shell $T_2 = 310 	ext{ K}$,and rate of heat flow $H = 40 	ext{ W}$.The radial rate of flow of heat through the shell in the steady state is given by Fourier's law of heat conduction:$H = \frac{dQ}{dt} = \alpha A \frac{dT}{dr} = \alpha (4 \pi r^2) \frac{dT}{dr}$Rearranging the terms for integration:$\frac{dr}{r^2} = \frac{4 \pi \alpha}{H} dT$Integrating both sides from $r_1$ to $r_2$ and $T_1$ to $T_2$:$\int_{r_1}^{r_2} \frac{dr}{r^2} = \frac{4 \pi \alpha}{H} \int_{T_1}^{T_2} dT$$[-\frac{1}{r}]_{r_1}^{r_2} = \frac{4 \pi \alpha}{H} (T_2 - T_1)$$\frac{1}{r_1} - \frac{1}{r_2} = \frac{4 \pi \alpha (T_2 - T_1)}{H}$$\frac{r_2 - r_1}{r_1 r_2} = \frac{4 \pi \alpha (T_2 - T_1)}{H}$Solving for $\alpha$:$\alpha = \frac{H(r_2 - r_1)}{4 \pi r_1 r_2 (T_2 - T_1)}$Substituting the given values:$\alpha = \frac{40 	imes (0.3 - 0.2)}{4 \pi 	imes 0.3 	imes 0.2 	imes (310 - 300)}$$\alpha = \frac{40 	imes 0.1}{4 \pi 	imes 0.06 	imes 10} = \frac{4}{2.4 \pi} = \frac{40}{24 \pi} = \frac{5}{3 \pi}$Thus,the value of $\alpha$ is $\frac{5}{3 \pi} 	ext{ J s}^{-1} 	ext{ m}^{-1} 	ext{ K}^{-1}$.

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