Let's carefully solve this problem step by step. --- ### **Given Data:** - Initial temperature, \( T_0 = 30^\circ C \) - Final length increase, \( \Delta L = 2.0\% \) of original length - Coefficient of linear expansion for brass, \( \alpha \) (commonly approximately \( 19 \times 10^{-6} /^\circ C \)) --- ### **Step 1: Understand the relationship between temperature change and length change** The linear expansion formula: \[ \Delta L = L_0 \times \alpha \times \Delta T \] where: - \( \Delta L \) is the change in length - \( L_0 \) is the original length - \( \alpha \) is the coefficient of linear expansion - \( \Delta T = T_{final} - T_0 \) is the temperature change Since the problem asks for the temperature at which the rod is **2% longer** than at 30°C, the length at this temperature is: \[ L_{final} = L_0 \times (1 + 0.02) = 1.02 L_0 \] The change in length: \[ \Delta L = L_{final} - L_0 = 0.02 L_0 \] --- ### **Step 2: Set up the equation** Using the expansion formula: \[ 0.02 L_0 = L_0 \times \alpha \times \Delta T \] Divide both sides by \( L_0 \): \[ 0.02 = \alpha \times \Delta T \] Solve for \( \Delta T \): \[ \Delta T = \frac{0.02}{\alpha} \] --- ### **Step 3: Substitute the value of \( \alpha \)** Using \( \alpha = 19 \times 10^{-6} /^\circ C \): \[ \Delta T = \frac{0.02}{19 \times 10^{-6}} = \frac{0.02}{0.000019} \] Calculate: \[ \Delta T \approx 1052.63^\circ C \] ### **Step 4: Find the final temperature** Since the initial temperature is 30°C: \[ T_{final} = T_0 + \Delta T = 30^\circ C + 1052.63^\circ C \approx \boxed{1082.63^\circ C} \] --- ### **Final Answer:** **You would need to heat the brass rod to approximately \(\boxed{1083^\circ C}\) for it to be 2% longer than it is at 30°C.**