A copper rod of length 75 cm and an iron rod | vedclass

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(C) Let the temperature of the copper-iron junction be $	heta$.Since the rods are joined in series and the surfaces are thermally insulated, the rate

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$93$

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(C) Let the temperature of the copper-iron junction be $	heta$.Since the rods are joined in series and the surfaces are thermally insulated, the rate of heat flow through both rods must be the same in the steady state.Let $K_1 = 386.4 \ W/m-K$ (copper), $l_1 = 0.75 \ m$, $T_1 = 100^{\circ} C$.Let $K_2 = 48.46 \ W/m-K$ (iron), $l_2 = 1.25 \ m$, $T_2 = 0^{\circ} C$.The cross-sectional area $A$ is the same for both.The rate of heat flow is given by $H = \frac{KA(T_{high} - T_{low})}{l}$.Equating the heat flow rates: $\frac{K_1 A (T_1 - 	heta)}{l_1} = \frac{K_2 A (	heta - T_2)}{l_2}$.$\frac{386.4 (100 - 	heta)}{0.75} = \frac{48.46 (	heta - 0)}{1.25}$.$515.2 (100 - 	heta) = 38.768 	heta$.$51520 - 515.2 	heta = 38.768 	heta$.$553.968 	heta = 51520$.$	heta = \frac{51520}{553.968} \approx 93^{\circ} C$.

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