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Question

(A) In the steady state,the heat current flowing into the junction must equal the heat current flowing out of the junction.Let $T$ be the temperature

Vedclass · Accepted Answer

$\frac{200}{3}^{\circ} C$

Vedclass · Answer

(A) In the steady state,the heat current flowing into the junction must equal the heat current flowing out of the junction.Let $T$ be the temperature of the junction.The heat current $H$ is given by $H = \frac{KA(T_1 - T_2)}{l}$.Since the dimensions (area $A$ and length $l$) are the same for all rods,the heat current is proportional to the thermal conductivity $K$.According to the figure,the heat flows from the $100^{\circ} C$ source to the junction,and then from the junction to the $50^{\circ} C$ and $0^{\circ} C$ sinks.Applying the principle of conservation of energy (Kirchhoff's law for heat current):$H_{in} = H_{out1} + H_{out2}$$\frac{3KA(100 - T)}{l} = \frac{2KA(T - 50)}{l} + \frac{KA(T - 0)}{l}$Canceling $\frac{KA}{l}$ from both sides:$3(100 - T) = 2(T - 50) + T$$300 - 3T = 2T - 100 + T$$300 - 3T = 3T - 100$$6T = 400$$T = \frac{400}{6} = \frac{200}{3}^{\circ} C$.

Three rods of the same dimensions have thermal conductivities $3 K, 2 K$,and $K$. They are arranged as shown in the figure below. The temperature of the junction in the steady state is:

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