A heat conducting rod, 1.60 m Iong and wrapped in insulation, is made of an aluminum section
that is 0.90 m long and a copper section that is 0.70 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 $30 ^ { \circ } \mathrm { C } \text { and } 170 ^ { \circ } \mathrm { C }$
Respectively. The thermal conductivities of aluminum and copper are 205 W/m . K (aluminum) and $385 \mathrm {~W} / \mathrm { m } \cdot \mathrm { K } \text { (copper). }$ At what rate is heat conducted in the rod under state conditions?

Question

A heat conducting rod,  1.60 m Iong and wrapped in insulation, is made of an aluminum section
that is  0.90 m long and a copper section that is  0.70 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 $30 ^ { \circ } \mathrm { C } \text { and } 170 ^ { \circ } \mathrm { C }$
Respectively. The thermal conductivities of aluminum and copper are  205 W/m . K (aluminum) and $385 \mathrm {~W} / \mathrm { m } \cdot \mathrm { K } \text { (copper). }$ At what rate is heat conducted in the rod under  state conditions?

Quizplus · Accepted Answer

The rate of heat conducted in the rod can be found using the formula for heat conduction, $Q/t = kA(T_2 - T_1)/d$, where $Q/t$ is the heat transfer per unit time, $k$ is the thermal conductivity, $A$ is the cross-sectional area, $T_2 - T_1$ is the temperature difference, and $d$ is the length of the material. Since the rod is made of two different materials, we calculate the heat transfer rate for each section separately and then use the fact that the heat transfer rate must be the same through both sections (steady state conditions) to find the overall rate.For aluminum:$$Q/t = (205 \, \text{W/m} \cdot \text{K}) \times (0.00040 \, \text{m}^2) \times (T_{\text{interface}} - 30) / 0.90 \, \text{m}$$For copper:$$Q/t = (385 \, \text{W/m} \cdot \text{K}) \times (0.00040 \, \text{m}^2) \times (170 - T_{\text{interface}}) / 0.70 \, \text{m}$$Since $Q/t$ is the same for both, we can set the equations equal to each other to solve for $T_{\text{interface}}$, but we don't need to find $T_{\text{interface}}$ to answer the question about the rate of heat conducted.Instead, we can rearrange the formula to express $Q/t$ directly in terms of known quantities. The total temperature difference across the rod is $170^\circ C - 30^\circ C = 140^\circ C$. The overall thermal resistance ($R_{\text{total}}$) of the rod is the sum of the thermal resistances of each section, given by $R = d/(kA)$. Thus, $R_{\text{total}} = R_{\text{Al}} + R_{\text{Cu}}$.$$R_{\text{Al}} = \frac{0.90}{205 \times 0.00040}$$$$R_{\text{Cu}} = \frac{0.70}{385 \times 0.00040}$$$$R_{\text{total}} = R_{\text{Al}} + R_{\text{Cu}}$$Then, the total heat transfer rate is given by:$$Q/t = \frac{\Delta T}{R_{\text{total}}}$$Plugging in the numbers and solving gives a heat transfer rate of approximately 9.0 W, which corresponds to choice E. This approach simplifies the problem by focusing on the overall thermal resistance and temperature difference, avoiding the need to find the intermediate temperature at the interface between the two materials.

A Heat Conducting Rod, 1 0.00040 m20.00040 \mathrm {~m} ^ { 2 }0.00040 m2

A Heat Conducting Rod, 1 $0.00040 \mathrm {~m} ^ { 2 }$