An aluminium rod of length 1 ,m and a steel r | vedclass

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(A) Let the junction temperature be $T \,K$. In the steady state, the rate of heat flow is the same through both rods.Using the formula for heat condu

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$322$

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(A) Let the junction temperature be $T \,K$. In the steady state, the rate of heat flow is the same through both rods.Using the formula for heat conduction, $\frac{Q}{t} = \frac{kA(T_2 - T_1)}{l}$, we have:$\frac{Q}{t} = \frac{k_{steel} A (500 - T)}{2} = \frac{k_{Al} A (T - 300)}{1}$Since the cross-sectional areas $A$ are the same, we can cancel them out:$\frac{50(500 - T)}{2} = \frac{200(T - 300)}{1}$$25(500 - T) = 200(T - 300)$$12500 - 25T = 200T - 60000$$225T = 72500$$T = \frac{72500}{225} \approx 322.2 \,K$Thus, the junction temperature is approximately $322 \,K$.

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