A rod of length l with thermally insulated la | vedclass

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Question

(B) The heat current $H$ through a rod is given by $H = -KA \frac{dT}{dx}$,where $A$ is the cross-sectional area and $K$ is the thermal conductivity.G

Vedclass · Accepted Answer

$\frac{C}{l} \ln \left( \frac{T_2}{T_1} \right)$

Vedclass · Answer

(B) The heat current $H$ through a rod is given by $H = -KA \frac{dT}{dx}$,where $A$ is the cross-sectional area and $K$ is the thermal conductivity.Given $K = C/T$,we have $H = -\frac{C}{T} A \frac{dT}{dx}$.Rearranging the terms,we get $H \frac{dx}{A} = -C \frac{dT}{T}$.Integrating both sides from $x=0$ to $x=l$ and $T=T_1$ to $T=T_2$:$\int_0^l \frac{H}{A} dx = -\int_{T_1}^{T_2} C \frac{dT}{T}$.Since the rod is in steady state,the heat current density $J_H = H/A$ is constant.$J_H \int_0^l dx = -C [\ln T]_{T_1}^{T_2}$.$J_H \cdot l = -C (\ln T_2 - \ln T_1) = C \ln \left( \frac{T_1}{T_2} \right)$.Assuming the magnitude of the heat current density (taking $T_1 > T_2$ for flow),we get $J_H = \frac{C}{l} \ln \left( \frac{T_1}{T_2} \right)$.Note: The options provided in the prompt contain a sign discrepancy; based on standard convention,the magnitude is $\frac{C}{l} \ln \left( \frac{T_1}{T_2} \right)$.

$A$ rod of length $l$ with thermally insulated lateral surface is made of a material whose thermal conductivity $K$ varies as $K = C/T$,where $C$ is a constant. The ends are at temperatures $T_1$ and $T_2$. The heat current density is

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