Three rods AB,BC,and BD made of the same mate | vedclass

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(B) Let the temperature of the junction $B$ be $	heta$. Since the rods $BC$ and $BD$ are connected to the same temperature $80^{\circ}C$ at their end

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$60$

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(B) Let the temperature of the junction $B$ be $	heta$. Since the rods $BC$ and $BD$ are connected to the same temperature $80^{\circ}C$ at their ends $C$ and $D$,they act as parallel thermal resistances. Let $R$ be the thermal resistance of each rod. The equivalent resistance of the parallel combination of rods $BC$ and $BD$ is $R_{eq} = \frac{R 	imes R}{R + R} = \frac{R}{2}$. According to the principle of steady-state heat flow,the heat current flowing through rod $AB$ must equal the sum of the heat currents flowing through rods $BC$ and $BD$. Using the formula for heat current $\frac{Q}{t} = \frac{\Delta T}{R}$,we have: $\frac{	heta - 20}{R} = \frac{80 - 	heta}{R/2}$ $	heta - 20 = 2(80 - 	heta)$ $	heta - 20 = 160 - 2	heta$ $3	heta = 180$ $	heta = 60^{\circ}C$.

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