QuestionJuly 31, 2025

5. Two rods one made of brass and other made of copper are to be joined end to end. The length of the brass section is 0 .2 meters long. The length of the copper section is 0.7 meters long. Both rod sections have cross section of 0.004 square meters . The free end of the brass segment is in boiling water. The free end of the copper section is in an ice -water mixture. The pressure is one atmosphere for both rods The sides of the rods are insulated perfectly. No heat can leave the sides of the rods over the length of the rods.Please calculate the temperature of the point where the two rods are connected . Thank you.

5. Two rods one made of brass and other made of copper are to be joined end to end. The length of the brass section is 0 .2 meters long. The length of the copper section is 0.7 meters long. Both rod sections have cross section of 0.004 square meters . The free end of the brass segment is in boiling water. The free end of the copper section is in an ice -water mixture. The pressure is one atmosphere for both rods The sides of the rods are insulated perfectly. No heat can leave the sides of the rods over the length of the rods.Please calculate the temperature of the point where the two rods are connected . Thank you.
5. Two rods one made of brass and other made of copper are to be joined end to end. The length of the brass section is 0 .2 meters long. The length of the copper section is 0.7 meters long. Both rod sections have cross section of 0.004 square meters . The free end of the brass segment is in boiling water. The free end of the copper section is in an ice -water mixture. The pressure is one atmosphere for both rods The sides of the rods are insulated perfectly. No heat can leave the sides of the rods over the length of the rods.Please calculate the temperature of the point where the two rods are connected . Thank you.

Solution
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Answer

The temperature at the junction is approximately 11.19°C. Explanation 1. Identify the heat conduction formula Use **Fourier's Law of Heat Conduction**: Q = \frac{kA(T_1 - T_2)t}{L}, where k is thermal conductivity, A is cross-sectional area, T_1 and T_2 are temperatures, t is time, and L is length. 2. Set up the equation for steady state In steady state, heat flow through brass equals heat flow through copper. Thus, \frac{k_{\text{brass}} A (100 - T)}{0.2} = \frac{k_{\text{copper}} A (T - 0)}{0.7}. 3. Solve for temperature at junction Cancel A from both sides and solve for T: \frac{k_{\text{brass}} (100 - T)}{0.2} = \frac{k_{\text{copper}} T}{0.7}. 4. Insert known values Assume k_{\text{brass}} = 109 \, \text{W/mK} and k_{\text{copper}} = 401 \, \text{W/mK}: \frac{109 (100 - T)}{0.2} = \frac{401 T}{0.7}. 5. Simplify and solve Rearrange and solve for T: 7630 - 109T = 573T, 7630 = 682T, T = \frac{7630}{682}.

Explanation

1. Identify the heat conduction formula<br /> Use **Fourier's Law of Heat Conduction**: $Q = \frac{kA(T_1 - T_2)t}{L}$, where $k$ is thermal conductivity, $A$ is cross-sectional area, $T_1$ and $T_2$ are temperatures, $t$ is time, and $L$ is length.<br /><br />2. Set up the equation for steady state<br /> In steady state, heat flow through brass equals heat flow through copper. Thus, $\frac{k_{\text{brass}} A (100 - T)}{0.2} = \frac{k_{\text{copper}} A (T - 0)}{0.7}$.<br /><br />3. Solve for temperature at junction<br /> Cancel $A$ from both sides and solve for $T$: <br />$$\frac{k_{\text{brass}} (100 - T)}{0.2} = \frac{k_{\text{copper}} T}{0.7}.$$<br /><br />4. Insert known values<br /> Assume $k_{\text{brass}} = 109 \, \text{W/mK}$ and $k_{\text{copper}} = 401 \, \text{W/mK}$:<br />$$\frac{109 (100 - T)}{0.2} = \frac{401 T}{0.7}.$$<br /><br />5. Simplify and solve<br /> Rearrange and solve for $T$:<br />$$7630 - 109T = 573T,$$<br />$$7630 = 682T,$$<br />$$T = \frac{7630}{682}.$$
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