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°. 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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Quizplus · Accepted Answer

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 heat flow rate through the aluminum section is:$$Q_1=kA_1\frac{dT}{dx}=(205	ext{ W/m$\cdot$K})(0.00040	ext{ m}^2)(1100\degree C/m)=90.2	ext{ W}$$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 copper section is:$$Q_2=kA_2\frac{dT}{dx}=(385	ext{ W/m$\cdot$K})(0.00040	ext{ m}^2)(137.5\degree C/m)=21.9	ext{ W}$$Since the heat flow rate is the same through both sections, we can equate the two expressions for Q:$$90.2	ext{ W}=21.9	ext{ W}$$Solving for T_1, the temperature at the aluminum-copper junction, we get:$$T_1=40\degree C+\frac{21.9	ext{ W}}{90.2	ext{ W}}(0.10	ext{ m})=58\degree C$$Therefore, the temperature of the aluminum-copper junction in the rod is closest to 58°C.

A Heat Conducting Rod, 0