Two cylindrical rods A and B made of different materials are joined end to end in a straight line. The ratios of lengths, radii and thermal conductivities of these rods are (L_A)/(L_B)=1/2, (r_A)/(r_B)=2 and (K_A)/(K_B)=1/2. The free ends of rods A and B are maintained at 400 K and 200 K respective…

Question

Two cylindrical rods A and B made of different materials are joined end to end in a straight line. The ratios of lengths, radii and thermal conductivities of these rods are (L_A)/(L_B)=1/2, (r_A)/(r_B)=2 and (K_A)/(K_B)=1/2. The free ends of rods A and B are maintained at 400 K and 200 K respectively. The temperature of the rods' interface is K when equilibrium is established.

Accepted Answer

360. Idea: in the steady state the same heat current crosses the joint, so compare the two conductances (Kπ r²)/L rather than averaging the end temperatures. (C_A)/(C_B)=(K_A)/(K_B)((r_A)/(r_B))²(L_B)/(L_A)=1/2×4×2=4. So rod A conducts four times as readily as rod B: 4(400-T)=T-200. 1600-4T=T-200, giving 5T=1800. T=360 K, close to the hot end as it must be, since the resistance is concentrated in rod B.

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