Two thin wires PQ and RS are joined end-to-end at Q and R by soldering. At 10 deg C each wire is exactly 1 m long. The end P is held fixed at 10 deg C while the end S is kept at 400 deg C, and the assembly is insulated so heat flows only along the wires. The thermal conductivity of PQ is t

Question

Two thin wires PQ and RS are joined end-to-end at Q and R by soldering. At 10 deg C each wire is exactly 1 m long. The end P is held fixed at 10 deg C while the end S is kept at 400 deg C, and the assembly is insulated so heat flows only along the wires. The thermal conductivity of PQ is twice that of RS, and the linear expansion coefficient of PQ is 1.2*10⁻⁵ K⁻¹. Determine the increase in length of wire PQ.

Accepted Answer

0.78 mm. Equal heat current through series wires of equal length gives K_PQ*(T-10) = K_RS*(400-T). With K_PQ = 2*K_RS, 2(T-10) = (400-T) so 3T = 420, T = 140 deg C at the junction. The temperature along PQ varies linearly from 10 to 140, so its mean temperature is (10+140)/2 = 75 deg C, a mean rise of 65 K above the original 10 deg C. Thermal expansion: dL = L*alpha*dT = 1*(1.2*10⁻⁵)*65 = 7.8*10⁻⁴ m = 0.78 mm.

Solution

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