A copper rod of length 18 cm and a steel rod | vedclass

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(C) In a steady state,the rate of heat flow through both rods must be equal.Let $k_1$ and $k_2$ be the thermal conductivities of copper and steel,resp

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$33$

Vedclass · Answer

(C) In a steady state,the rate of heat flow through both rods must be equal.Let $k_1$ and $k_2$ be the thermal conductivities of copper and steel,respectively. Given $k_1 = 9k_2$.Let $L_1 = 18 \ cm$ and $L_2 = 6 \ cm$ be the lengths of the copper and steel rods.Let $T_x$ be the temperature at the junction.The rate of heat flow is given by $H = \frac{kA(T_H - T_L)}{L}$.Since the rods are in series,$H_{copper} = H_{steel}$.$\frac{k_1 A (100 - T_x)}{L_1} = \frac{k_2 A (T_x - 0)}{L_2}$.Substituting the values: $\frac{9k_2 (100 - T_x)}{18} = \frac{k_2 (T_x)}{6}$.$\frac{100 - T_x}{2} = T_x$.$100 - T_x = 2T_x$.$3T_x = 100$.$T_x = \frac{100}{3} \approx 33.33 \ ^\circ C$.

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