REPORT SHEET EXPERIMENT Heat of Neutralization 28 A. Heat Capacity of Calorimeter 1. Temp. of calorimeter and water before mixing 2. Temp. of warm water 3. Masimum temp. determined from your curve Heat lost by warm water temp decrease 50.0gx4.154J/KgI 5. Heat gained by cooler water temp, increase 500gx41841/K-g)- 6. Heat gained by the calorimeter ((4)(5)]- -1047, 7. Heat capacity of calorimeter: heat gained by the calorimeter temperature incnose -372 K 25 25 γι C B. Heat of Neutralization of HCI-NaOH 1. Temp. of calorinscier and NaOH Temp. of HC1 2. AF determined from your curve after adding HCT to the NaOH 3. Heat gained by solution (temperature increase x 100gx4.184 J/Kg) 11757091 4. Heat gained by calorimeter (temperature increase x heat capacity of calorimeter) 1045.32, 5. Total joules released by reaction [(3)+(4)]- 26 34 116,525
The Correct Answer and Explanation is :
To correctly answer and explain the data provided in the report sheet, we will analyze each step and compute the values where possible. Here’s the revised and clarified report with proper explanation:
A. Heat Capacity of Calorimeter
- Temperature of calorimeter and water before mixing: Not provided in the sheet.
- Temperature of warm water: Not provided in the sheet.
- Maximum temperature determined from the curve: Not provided directly; let’s assume values to illustrate the process.
- Heat lost by warm water (calculated using (Q = mc\Delta T)):
[
Q_{\text{lost}} = 50.0 \, \text{g} \times 4.184 \, \text{J/g°C} \times \Delta T_{\text{decrease}}
]
The result depends on the temperature change, which isn’t provided. - Heat gained by cooler water:
[
Q_{\text{gained}} = 500 \, \text{g} \times 4.184 \, \text{J/g°C} \times \Delta T_{\text{increase}}
] - Heat gained by the calorimeter:
[
Q_{\text{calorimeter}} = Q_{\text{lost}} – Q_{\text{gained}} = -1047 \, \text{J (provided)}
] - Heat capacity of the calorimeter:
[
C_{\text{calorimeter}} = \frac{Q_{\text{calorimeter}}}{\Delta T_{\text{increase}}} = -372 \, \text{J/°C (provided)}
]
B. Heat of Neutralization of HCl-NaOH
- Temperature of calorimeter and NaOH: Not provided.
- Temperature of HCl: Not provided.
- (\Delta T) from the curve after adding HCl to NaOH: Not directly provided.
- Heat gained by solution:
[
Q_{\text{solution}} = \Delta T \times 100 \, \text{g} \times 4.184 \, \text{J/g°C} = 11,757.09 \, \text{J (provided)}
] - Heat gained by calorimeter:
[
Q_{\text{calorimeter}} = \Delta T \times C_{\text{calorimeter}} = 1045.32 \, \text{J (provided)}
] - Total joules released by reaction:
[
Q_{\text{total}} = Q_{\text{solution}} + Q_{\text{calorimeter}} = 11,757.09 + 1045.32 = 12,802.41 \, \text{J}
]
Explanation (300 words)
The experiment aims to determine the heat capacity of the calorimeter and the heat of neutralization for the reaction between HCl and NaOH.
In Part A, the calorimeter’s heat capacity is calculated using energy conservation principles. Heat lost by warm water is equal to the heat gained by cooler water and the calorimeter. Since specific heat and mass of water are known, the calorimeter’s heat capacity is computed as (-372 \, \text{J/°C}). This step ensures the calorimeter’s contribution to heat absorption is accounted for in subsequent experiments.
In Part B, the neutralization of HCl and NaOH is exothermic, releasing heat absorbed by the solution and the calorimeter. The heat gained by the solution is calculated using the mass of the solution, its specific heat capacity ((4.184 \, \text{J/g°C})), and temperature change ((\Delta T)). The heat gained by the calorimeter is similarly calculated using its heat capacity from Part A. The total heat released by the reaction is the sum of these two values.
These calculations confirm the importance of accurate temperature measurements and calorimeter calibration for precise heat determination.