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(B) Let the temperature of the junction be $T$. Since the rods are connected at a junction,the net heat flow into the junction must be zero (steady st

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$25$

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(B) Let the temperature of the junction be $T$. Since the rods are connected at a junction,the net heat flow into the junction must be zero (steady state condition).The rate of heat flow through a rod is given by $H = \frac{KA(T_1 - T_2)}{L}$.Let $A$ be the cross-sectional area and $L$ be the length of each rod. The thermal resistance of each rod is $R = \frac{L}{KA}$.For the three rods,the heat currents are:$H_1 = \frac{K A (110 - T)}{L}$$H_2 = \frac{2K A (20 - T)}{L}$$H_3 = \frac{3K A (T - 0)}{L}$According to the principle of conservation of energy at the junction,$H_1 + H_2 + H_3 = 0$.$\frac{KA}{L} (110 - T) + \frac{2KA}{L} (20 - T) + \frac{3KA}{L} (T - 0) = 0$Dividing by $\frac{KA}{L}$,we get:$(110 - T) + 2(20 - T) + 3T = 0$$110 - T + 40 - 2T + 3T = 0$$150 = 0$Wait,re-evaluating the heat flow direction: The heat flows from higher temperature to lower temperature. Let $T$ be the junction temperature.$H_1 = \frac{KA(110 - T)}{L}$$H_2 = \frac{2KA(20 - T)}{L}$$H_3 = \frac{3KA(T - 0)}{L}$Sum of heat currents leaving the junction is zero: $H_1 + H_2 + H_3 = 0$ is incorrect if we assume all are leaving. Let's assume all heat flows towards the junction or away. Actually,the sum of heat currents flowing into the junction is zero: $\frac{KA(110-T)}{L} + \frac{2KA(20-T)}{L} = \frac{3KA(T-0)}{L}$.$110 - T + 2(20 - T) = 3T$$110 - T + 40 - 2T = 3T$$150 - 3T = 3T$$6T = 150$$T = 25\ ^oC$.

Three rods of equal length and cross-sectional area with thermal conductivities $K, 2K$,and $3K$ are joined as shown in the figure. The temperatures of their outer ends are $110\ ^oC, 20\ ^oC$,and $0\ ^oC$ respectively. Find the temperature of the junction in $^oC$.

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