A rod of length L with sides fully insulated | vedclass

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Question

(A) The heat current $H$ through a rod of cross-sectional area $A$ is given by $H = -KA \frac{dT}{dx}$.Since the rod is insulated,$H$ is constant alon

Vedclass · Accepted Answer

$T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}}$

Vedclass · Answer

(A) The heat current $H$ through a rod of cross-sectional area $A$ is given by $H = -KA \frac{dT}{dx}$.Since the rod is insulated,$H$ is constant along the length.Substituting $K = \frac{\alpha}{T}$,we get $H = -\frac{\alpha A}{T} \frac{dT}{dx}$.Rearranging the terms: $\frac{H}{\alpha A} dx = -\frac{dT}{T}$.Integrating from $x=0$ to $x$ and $T=T_1$ to $T$:$\int_0^x \frac{H}{\alpha A} dx = -\int_{T_1}^T \frac{dT}{T}$.$\frac{H}{\alpha A} x = -(\ln T - \ln T_1) = \ln \frac{T_1}{T}$.At $x=L$,$T=T_2$,so $\frac{H}{\alpha A} L = \ln \frac{T_1}{T_2}$.Thus,$\frac{H}{\alpha A} = \frac{1}{L} \ln \frac{T_1}{T_2}$.Substituting this back: $\frac{x}{L} \ln \frac{T_1}{T_2} = \ln \frac{T_1}{T}$.$\ln \left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \ln \frac{T_1}{T}$.Taking the exponential of both sides: $\left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \frac{T_1}{T}$.Therefore,$T = T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}}$.

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