3. An irregular lump of an unknown metal has a measured density of \( 2.97 \mathrm{~g} / \mathrm{mL} \). The metal is heated to a temperature of \( 173^{\circ} \mathrm{C} \) and placed in a graduated cylinder filled with 25.0 mL of water at \( 25.0^{\circ} \mathrm{C} \). After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.0 mL , and the temperature is recorded as \( 40.6^{\circ} \mathrm{C} \). What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings. What is the metal used in the experiment? Specific heat of water \( =4.184 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of granite \( =0.79 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of copper \( =0.38 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of iron \( =0.45 \mathrm{~J} / \mathrm{g} \mathrm{C} \) Specific heat of lead \( =0.13 \mathrm{~J} / \mathrm{g} \mathrm{C} \)

Let \[ m_{\text{metal}} \quad \text{= mass of the metal}, \] \[ c_{\text{metal}} \quad \text{= specific heat of the metal}, \] \[ T_{i,\text{metal}} = 173^\circ\text{C}, \quad T_{f} = 40.6^\circ\text{C}, \] \[ m_{\text{water}} = 25.0\text{ g} \quad (\text{since } 25.0\text{ mL of water} \approx 25.0\text{ g}), \] \[ c_{\text{water}} = 4.184\text{ J/(g}^\circ\text{C)}, \] \[ T_{i,\text{water}} = 25.0^\circ \text{C}. \] 1. **Find the volume and mass of the metal:** The metal is placed in a graduated cylinder initially containing 25.0 mL of water. After the metal is added, the final volume is 34.0 mL. The displaced water equals the metal’s volume: \[ V_{\text{metal}} = 34.0\text{ mL} - 25.0\text{ mL} = 9.0\text{ mL}. \] The density of the metal is given as \( \rho = 2.97 \text{ g/mL} \). Hence, the mass of the metal is \[ m_{\text{metal}} = \rho \times V_{\text{metal}} = 2.97 \text{ g/mL} \times 9.0 \text{ mL} = 26.73\text{ g}. \] 2. **Set Up the Energy Conservation Equation:** Assuming no heat is lost to the surroundings, the heat lost by the metal equals the heat gained by the water. That is, \[ m_{\text{metal}}\, c_{\text{metal}}\, (T_{i,\text{metal}} - T_f) = m_{\text{water}}\, c_{\text{water}}\, (T_f - T_{i,\text{water}}). \] 3. **Solve for \( c_{\text{metal}} \):** Rearranging gives \[ c_{\text{metal}} = \frac{m_{\text{water}}\, c_{\text{water}}\, (T_f - T_{i,\text{water}})}{m_{\text{metal}}\, (T_{i,\text{metal}} - T_f)}. \] Now substitute the given values: \[ c_{\text{metal}} = \frac{25.0\,\text{g} \times 4.184\,\text{J/(g}^\circ\text{C)} \times (40.6^\circ\text{C} - 25.0^\circ\text{C})}{26.73\,\text{g} \times (173^\circ\text{C} - 40.6^\circ\text{C})}. \] Calculate the temperature differences: \[ (T_f - T_{i,\text{water}}) = 40.6^\circ\text{C} - 25.0^\circ\text{C} = 15.6^\circ\text{C}, \] \[ (T_{i,\text{metal}} - T_f) = 173^\circ\text{C} - 40.6^\circ\text{C} = 132.4^\circ\text{C}. \] Then the numerator becomes \[ 25.0 \times 4.184 \times 15.6 \approx 25.0 \times 65.2704 \approx 1631.76\,\text{J}. \] The denominator is \[ 26.73 \times 132.4 \approx 3539.05\,\text{g}^\circ\text{C}. \] Thus, \[ c_{\text{metal}} \approx \frac{1631.76}{3539.05} \approx 0.461\,\text{J/(g}^\circ\text{C)}. \] 4. **Identify the Metal:** The computed specific heat of approximately \( 0.46\,\text{J/(g}^\circ\text{C)} \) is closest to the provided value for iron, which is \( 0.45\,\text{J/(g}^\circ\text{C)} \). Therefore, the unknown metal is identified as iron. \[ \boxed{c_{\text{metal}} \approx 0.46\,\text{J/(g}^\circ\text{C)} \quad \text{and the metal is iron.}} \]