Two rods with thermal conductivities K and 3K | vedclass

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Question

(C) In a steady state,the rate of heat flow through both rods must be equal.Let $A$ be the cross-sectional area,$L_1 = 1 \ cm$,$L_2 = 2 \ cm$,$K_1 = K

Vedclass · Accepted Answer

$60^{\circ}C$

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(C) In a steady state,the rate of heat flow through both rods must be equal.Let $A$ be the cross-sectional area,$L_1 = 1 \ cm$,$L_2 = 2 \ cm$,$K_1 = K$,$K_2 = 3K$,$T_1 = 0^{\circ}C$,$T_2 = 100^{\circ}C$,and $\phi$ be the interface temperature.The rate of heat flow is given by $H = \frac{KA(T_{high} - T_{low})}{L}$.For the first rod: $H_1 = \frac{K \cdot A \cdot (\phi - 0)}{1}$.For the second rod: $H_2 = \frac{3K \cdot A \cdot (100 - \phi)}{2}$.Since $H_1 = H_2$,we have:$K \cdot A \cdot \phi = \frac{3K \cdot A \cdot (100 - \phi)}{2}$$\phi = \frac{3(100 - \phi)}{2}$$2\phi = 300 - 3\phi$$5\phi = 300$$\phi = 60^{\circ}C$.

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