A heat conducting rod, 0.90 m long, is made of an aluminum section that is 0.10 m long, and a copper section that is 0.80 m long. Both sections have cross-sectional areas of 0.00040 m². The aluminum end is maintained at a temperature of 40°C and the copper end is at 150°C. The thermal conductivity of aluminum is 205 W/m∙K and of copper is 385 W/m∙K. Steady state has been reached, and no heat is lost through the well-insulated sides of the rod. The temperature of the aluminum-copper junction in the rod is closest to

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The temperature gradient in the aluminum section is:$$\frac{dT}{dx}=\frac{T_2-T_1}{L_1}=\frac{150\degree C-40\degree C}{0.10	ext{ m}}=1100\degree C/m$$The temperature gradient in the copper section is:$$\frac{dT}{dx}=\frac{T_2-T_1}{L_2}=\frac{150\degree C-40\degree C}{0.80	ext{ m}}=137.5\degree C/m$$The heat flow rate through the rod is:$$Q=kA\frac{dT}{dx}$$where k is the thermal conductivity, A is the cross-sectional area, and dT/dx is the temperature gradient.The heat flow rate through the aluminum section is:$$Q_1=k_1A\frac{dT}{dx}=205	ext{ W/m}\cdot	ext{K}	imes0.00040	ext{ m}^2	imes1100\degree C/m=90.2	ext{ W}$$The heat flow rate through the copper section is:$$Q_2=k_2A\frac{dT}{dx}=385	ext{ W/m}\cdot	ext{K}	imes0.00040	ext{ m}^2	imes137.5\degree C/m=21.6	ext{ W}$$The heat flow rate through the rod is the same in both sections, so we can equate the two expressions for Q:$$Q_1=Q_2$$$$90.2	ext{ W}=21.6	ext{ W}$$This gives us:$$\frac{dT}{dx}=\frac{Q}{kA}=\frac{21.6	ext{ W}}{385	ext{ W/m}\cdot	ext{K}	imes0.00040	ext{ m}^2}=141.3\degree C/m$$The temperature at the aluminum-copper junction is:$$T_j=T_1+\frac{dT}{dx}L_1=40\degree C+141.3\degree C/m	imes0.10	ext{ m}=54.1\degree C$$Therefore, the temperature of the aluminum-copper junction in the rod is closest to 54°C.

A Heat Conducting Rod, 0