Given:Mass of Gold (g) -51.1557 gVolume of H2O (mL) - 100mLMass of H2O (g) - 99.8gInitial Temperature of H2O (degrees Celsius) - 25 degrees CelsiusInitial Temperature of Gold (degrees Celsius) - 200 degrees CelsiusMax temp of H2O + metal (degrees Celsius) - 27.23 degreesCalculate:The change in temperature: The heat (q) in J gained by the water using the following equation:The changes in temperature of the metal:Remembering that the heat gained by the water is equal to the heat lost by the metal, calculate the specific heat of metal in J/g degrees Celsius.The percent error in the specific heat value that you determined experimentally. The accepted value for the metal is Au= 0.126 J/g degrees Celsius. **Calculation of Temperature Change and Specific Heat**1. **Calculate the Change in Temperature of the Water (\( \Delta T_{\text{water}} \)):** \[ \Delta T_{\text{water}} = T_p - T_i \]2. **Calculate the Heat (\( q \)), in Joules, Gained by the Water:** Use the equation: \[ q_{\text{water}} = m_{\text{water}} \times \Delta T_{\text{water}} \times C_{\text{water}} \]3. **Calculate the Change in Temperature of the Metal (\( \Delta T_{\text{metal}} \)):** \[ \Delta T_{\text{metal}} = T_p - T_i \]4. **Calculate the Specific Heat of the Metal:** Since the heat gained by water is equal to the heat lost by the metal: \[ q_{\text{water}} = -q_{\text{metal}} = m_{\text{metal}} \times \Delta T_{\text{metal}} \times C_{\text{metal}} \] Therefore, the specific heat, \( C_{\text{metal}} \), is: \[ C_{\text{metal}} = q_{\text{metal}} / (m_{\text{metal}} \times \Delta T_{\text{metal}}) \]5. **Calculate the Percent Error in the Specific Heat Value:** Compare the experimentally determined specific heat with the accepted value: \[ \text{Accepted value for the metal is} \, Au = 0.126 \, \text{J/g°C} \]This serves as a step-by-step guide for determining the changes in temperature and calculating specific heat, essential in understanding thermal transfer processes.

Given: Mass of gold = 51.1557 g mass of water = 99.8 g Initial temperature of gold = 200.0 oC Initial temperature of water = 25.0 oC Final temperature of gold and water = 27.23 oC specific heat of water = 4.184 J/g.oC accepted value specific heat of gold = 0.126 J/g.oCCalculation for change in temperature of water: ∆Twater=Ti-Tf∆Twater=27.23 °C-25.0 °C∆Twater=2.23 °CCalculation for heat gain by water: Q=mC∆TQwater=99.8 g×4.184 J/g.°C×(2.23 °C)Qwater=931.166 JCalculation for change in temperature of metal: ∆Tmetal=Ti-Tf∆Tmetal=200.0 °C-27.23 °C∆Tmetal=172.77 °CCalculation for specific heat of metal: Heat loss by metal=Heat gain by waterQmetal=Qwatermmetal×Cmetal×∆Tmetal=931.166 J51.1557 g×Cmetal×172.77 °C=931.166 JCmetal=931.166 J51.1557 g×172.77 °CCmetal=0.105 J/g.°CCalculation for percent error in the specific heat value that determined experimentally: Percent error=accepted value-experimenta valueaccepted value×100% error=0.126 J/g.°C-0.105 J/g.°C0.126 J/g.°C×100% error=16.67 %