The coefficient of thermal conductivity of co | vedclass

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(A) In a steady state,the rate of heat flow through the copper rod must equal the rate of heat flow through the steel rod.Let $	heta$ be the temperat

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$75$

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(A) In a steady state,the rate of heat flow through the copper rod must equal the rate of heat flow through the steel rod.Let $	heta$ be the temperature at the junction.Let $K_{Cu}$ and $K_S$ be the thermal conductivities,$A$ be the cross-sectional area,and $L_{Cu}$ and $L_S$ be the lengths of the copper and steel rods respectively.Given: $K_{Cu} = 9K_S$,$L_{Cu} = 18 	ext{ cm}$,$L_S = 6 	ext{ cm}$,$T_{Cu} = 100^\circ C$,$T_S = 0^\circ C$.The rate of heat flow is given by $H = \frac{KA(T_1 - T_2)}{L}$.Equating the heat flow rates: $\frac{K_{Cu} A (100 - 	heta)}{L_{Cu}} = \frac{K_S A (	heta - 0)}{L_S}$.Substituting the values: $\frac{9K_S (100 - 	heta)}{18} = \frac{K_S (	heta - 0)}{6}$.Simplifying: $\frac{9(100 - 	heta)}{18} = \frac{	heta}{6}$.$\frac{100 - 	heta}{2} = \frac{	heta}{6}$.$3(100 - 	heta) = 	heta$.$300 - 3	heta = 	heta$.$4	heta = 300$.$	heta = 75^\circ C$.

The coefficient of thermal conductivity of copper is nine times that of steel. In the composite cylindrical bar shown in the figure,what will be the temperature at the junction of copper and steel in $^oC$?

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