Let's solve the problem **step-by-step** using the correct formula and **carrying out all intermediate results to four digits** to minimize round-off error. --- ### **Step 1: Understanding the Problem** Given: - Resistance at \(^\circ C\) (\(R_\)): **210. Ω** - Resistance at melting substance (\(R\)): **257.8 Ω** - Need to find: **Melting point temperature \(T\) of the substance** The resistance of a platinum resistance thermometer varies linearly with temperature: \[ R_T = R_ [1 + \alpha (T - T_)] \] where: - \(R_T\) = resistance at temperature \(T\) - \(R_\) = resistance at reference temperature \(T_\) (usually \(^\circ C\)) - \(\alpha\) = temperature coefficient of resistance for platinum (\( \approx 3.85 \times 10^{-3} \ ^\circ C^{-1} \)) - \(T_ = ^\circ C\) --- ### **Step 2: Find \(\alpha\) using Ice Bath and Room Temperature** Usually, you solve for \(\alpha\) using a second resistance at a known temperature (often room temperature, \(20.^\circ C\)). But the problem only provides: - \(R_\) at \(^\circ C\) - \(R\) at the unknown melting temperature But the hint says **first determine the resistance at room temperature (20.°C)**, which suggests we’re missing the resistance at 20.°C. Since it’s not provided, let’s double-check the standard approach. #### **Assumption: Use Standard \(\alpha\) for Platinum** If not given, **use standard value**: \[ \alpha = .00385 \ ^\circ C^{-1} \] --- ### **Step 3: Solve for the Melting Point Temperature** From the formula: \[ R = R_ [1 + \alpha (T - T_)] \] \[ \frac{R}{R_} = 1 + \alpha (T - T_) \] \[ \alpha (T - T_) = \frac{R}{R_} - 1 \] \[ T = T_ + \frac{1}{\alpha}\left( \frac{R}{R_} - 1 \right) \] Plug in the values: - \(R = 257.8 \, \Omega\) - \(R_ = 210. \, \Omega\) - \(T_ = ^\circ C\) - \(\alpha = .00385 \ ^\circ C^{-1}\) \[ T = + \frac{1}{.00385} \left( \frac{257.8}{210.} - 1 \right) \] Calculate inside the parentheses first: \[ \frac{257.8}{210.} = 1.228571 \] \[ 1.228571 - 1 = .228571 \] Now, \[ T = \frac{1}{.00385} \times .228571 \] \[ \frac{1}{.00385} = 259.7402597 \] \[ T = 259.7402597 \times .228571 = 59.3926^\circ C \] --- ### **Step 4: Final Answer (Rounded to Proper Sig Figs)** \[ \boxed{59.4^\circ C} \] --- ### **Summary of Steps** 1. **Formula**: \( R = R_ [1 + \alpha (T - T_)] \) 2. **Plug in values**: \( R_ = 210.\,\Omega, R = 257.8\,\Omega, \alpha = .00385\,^\circ C^{-1}, T_ = ^\circ C \) 3. **Solve for \(T\)**: \( T = \frac{1}{.00385} \left( \frac{257.8}{210.} - 1 \right) \) 4. **Calculate**: \( T \approx 59.4^\circ C \) --- **If the problem required you to use a different \(\alpha\), or if a resistance at room temperature was supposed to be used, please provide that value! Otherwise, the above steps should give you the correct answer.** --- #### **Final Answer** \[ \boxed{59.4^\circ C} \]