Question: 4) An aluminum rod with a length of 0.58 meters and weight of 0.60 kg is at a temperature of -10 C…

4. An aluminum rod with a length of 0.58 meters and weight of 0.60 kg is at a temperature of -10 C. It is placed end to end with a brass rod of 0.87 meters and a weight of 2.30 kg at a temperature of 30 C to make a rod that is 1.45 meters long. If there is no heat gained or lost, what is the combined length when the temperature equalizes? (8 marks).

Solution

First, identify the given values: \[ \begin{cases} m_1 = 0.60 \text{ kg} & \\ L_1 = 0.58 \text{ m} & \\ T_1 = -10^\circ\text{C} & \\ c_1 = 900 \text{ J/(kg·°C)} & \\ \alpha_1 = 22.8 \times 10^{-6} \text{ /°C} & \\ m_2 = 2.30 \text{ kg} & \\ L_2 = 0.87 \text{ m} & \\ T_2 = 30^\circ\text{C} & \\ c_2 = 380 \text{ J/(kg·°C)} & \\ \alpha_2 = 19 \times 10^{-6} \text{ /°C} & \\ \end{cases} \] Next, apply the principle of conservation of heat energy: \[ m_1 c_1 (T_f - T_1) = -m_2 c_2 (T_f - T_2) \] Expand and solve for \( T_f \): \[ 0.60 \times 900 \times (T_f - (-10)) = -2.30 \times 380 \times (T_f - 30) \] \[ 540 (T_f + 10) = -874 (T_f - 30) \] \[ 540 T_f + 5400 = -874 T_f + 26220 \] \[ 540 T_f + 874 T_f = 26220 - 5400 \] \[ 1414 T_f = 20820 \] \[ T_f = \frac{20820}{1414} \] \[ T_f \approx 14.73^\circ\text{C} \] Now, calculate the final lengths of each rod using thermal expansion: \[ L_1’ = L_1 \left(1 + \alpha_1 (T_f - T_1)\right) \] \[ L_1’ = 0.58 \left(1 + 22.8 \times 10^{-6} \times (14.73 - (-10))\right) \] \[ L_1’ = 0.58 \left(1 + 22.8 \times 10^{-6} \times 24.73\right) \] \[ L_1’ \approx 0.58 \times 1.000564 \] \[ L_1’ \approx 0.580327 \text{ m} \] \[ L_2’ = L_2 \left(1 + \alpha_2 (T_f - T_2)\right) \] \[ L_2’ = 0.87 \left(1 + 19 \times 10^{-6} \times (14.73 - 30)\right) \] \[ L_2’ = 0.87 \left(1 + 19 \times 10^{-6} \times (-15.27)\right) \] \[ L_2’ \approx 0.87 \times 0.999708 \] \[ L_2’ \approx 0.86976 \text{ m} \] Finally, find the combined length: \[ \text{Total Length} = L_1’ + L_2’ \] \[ \text{Total Length} = 0.580327 + 0.86976 \] \[ \text{Total Length} \approx 1.4501 \text{ m} \] The combined length of the rods when the temperature equalizes is approximately 1.4501 meters.