A heat conducting rod, 1.60 m long and wrapped in insulation, is made of an aluminum section that is 0.90 m long and a copper section that is $0.70 \mathrm {~m}$ long. Both sections have a cross-sectional area of $0.00040 \mathrm {~m} ^ { 2 }$ The aluminum end and the copper end are maintained at temperatures of $30 ^ { \circ } \mathrm { C }$ and $170 ^ { \circ } \mathrm { C }$ respectively. The thermal conductivities of aluminum and copper are 205 W/m ∙ K (aluminum) and 385 W/m ∙ K (copper). At what rate is heat conducted in the rod under steady state conditions?

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Quizplus · Accepted Answer

The rate of heat conduction is given by:$$Q = \frac{kA(T_h - T_c)}{L}$$where:* Q is the rate of heat conduction (in W)* k is the thermal conductivity (in W/m·K)* A is the cross-sectional area (in m²)* L is the length (in m)* $T_h$ is the higher temperature (in K)* $T_c$ is the lower temperature (in K)For the aluminum section:$$Q_a = \frac{k_aA_a(T_{ha} - T_{ca})}{L_a}$$$$Q_a = \frac{(205 \mathrm{W/m·K})(0.00040 \mathrm{m}^2)(303 \mathrm{K} - 303 \mathrm{K})}{0.90 \mathrm{m}}$$$$Q_a = 0 \mathrm{W}$$For the copper section:$$Q_c = \frac{k_cA_c(T_{hc} - T_{cc})}{L_c}$$$$Q_c = \frac{(385 \mathrm{W/m·K})(0.00040 \mathrm{m}^2)(443 \mathrm{K} - 303 \mathrm{K})}{0.70 \mathrm{m}}$$$$Q_c = 9.0 \mathrm{W}$$Therefore, the rate of heat conduction in the rod under steady state conditions is:$$Q = Q_a + Q_c = 0 \mathrm{W} + 9.0 \mathrm{W} = 9.0 \mathrm{W}$$

A Heat Conducting Rod, 1 0.70 m0.70 \mathrm {~m}0.70 m Long Both Sections Have a Cross-Sectional Area Of Wrapped

A Heat Conducting Rod, 1 $0.70 \mathrm {~m}$ Long Both Sections Have a Cross-Sectional Area Of Wrapped