Two solid cylindrical rods are welded end to end

PHYSICS · Multi-part · A Level · Physics · Further mechanics and thermal physics · Further mechanics and thermal physics

Problem

Two solid cylindrical rods are welded end to end, forming a composite rod of total length 0.60 m0.60\,\text{m}. The first segment, made of copper, is 0.20 m0.20\,\text{m} long; the second, made of brass, is 0.40 m0.40\,\text{m} long. Both segments have the same cross-sectional area 2.0×10−4 m22.0\times10^{-4}\,\text{m}^2. The free end of the copper segment is held at 150 ∘C150\,^{\circ}\text{C}, while the free end of the brass segment is pressed against a 0.25 kg0.25\,\text{kg} block of graphite that is initially at 25 ∘C25\,^{\circ}\text{C}. The lateral surfaces of the rods and the block are perfectly insulated. The graphite block remains in good thermal contact with the rod throughout the process. Useful data: • Copper: thermal conductivity kCu=400 W ⁣⋅ ⁣m−1 ⁣⋅ ⁣K−1k_{\text{Cu}} = 400\,\text{W}\!\cdot\!\text{m}^{-1}\!\cdot\!\text{K}^{-1} • Brass: thermal conductivity kBr=120 W ⁣⋅ ⁣m−1 ⁣⋅ ⁣K−1k_{\text{Br}} = 120\,\text{W}\!\cdot\!\text{m}^{-1}\!\cdot\!\text{K}^{-1} • Graphite: specific heat cgr=710 J ⁣⋅ ⁣kg−1 ⁣⋅ ⁣K−1c_{\text{gr}} = 710\,\text{J}\!\cdot\!\text{kg}^{-1}\!\cdot\!\text{K}^{-1} (a) Calculate the steady-state rate at which thermal energy is delivered to the graphite block through the composite rod. (b) Assuming this rate remains constant, determine the time required for the block to reach 65 ∘C65\,^{\circ}\text{C}. (c) If the cross-sectional area of both rod segments were doubled while all other variables remained unchanged, by what percentage would the heating time found in part (b) change? Justify your answer quantitatively.

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