Question

A student determine the calorimeter constant of the calorimenter, using the procedure described in this module. The student added 50.00 mL of heated, distilled water in a Styrofoam cup. The initial temperature of the cold water was 21.00 C and of the hot water, 29.15 C. The maximum temperature of the mixture was found to be 24.81C. Assume the density of water is 1.00 g mL^-1 and the specific heat is 4.184 J g^-1 K^-1.

1)   Determine the DT for the hot water and the cold water.

2)   Calculate the heat lost by the hot water

3)   Calculate the heat gained by the cold water

4)   Calculate the calorimeter constant, using the DT of the cold water

The student then determined DHneutzn for the reaction of sodium hydroxide and acetic acid, using the procedure describes in this module. The student added 100.0 mL of 0.8404M acetic acid. Prior to and fallowing the mixing of the acid and base solutions, the fallowing temperature-time data were collected.

1. Plot the temperature-time data and determine the mean temperature of the unmixed reagents

2. Determine DT from the graph

3. Calculate the heat absorbed by the reaction mixture. You may assume the density if the reaction mixture was 1.02 g mL^-1 and the specific heat of the reaction mixture was 3.822 J g^-1 K^-1

4. Calculate the amount of heat absorbed by the calorimeter, thermometer, and stirrer, using the calorimeter constant calculated in Question (2) 4.

5. Calculate the total heat absorbed.

6. Find the total heat released by the reaction of 100.0 mL of 0.8500M and 100.0 mL of 0.8404M acetic acid.

7. Identify the limiting reagent and briefly explain why it is limiting

8. Find DHneutzn for the reaction

(D= delta)

Answer 1

Given:

The specific heat is 4.184 J g^-1 K^-1

The unit of specific heat is J g^-1 K^-1, so convert all temperature in Kelvin

The initial temperature of the cold water = 21.0⁰C =21.0⁰C + 273.15 = 294.15 K

The initial temperature of hot water = 29.15^oC = 29.15^oC + 273.15 = 302.30 K

The maximum temperature of the mixture = 24.81^oC = 24.81⁰C + 273.15 = 297.96 K

The density of water is 1.00 g mL^-1

1)   Determine the DT for the hot water and the cold water.

DT for the hot water = the maximum temperature of the mixture – The initial temperature of hot water

                                      = 297.96 K - 302.30 K = - 4.34 K

DT for the cold water = the maximum temperature of the mixture - The initial temp of the cold water

                                           =297.96 K - 294.15 K = 3.81K

2)   Calculate the heat lost by the hot water

Q = DT * m * c

Q = - 4.34 K x 50g x 4.184 J g^-1 K^-1

Q = - 907.93J

3)   Calculate the heat gained by the cold water

Q = DT * m * c

Q = 3.81K * 50g* 4.184 J g^-1 K^-1

Q = 797.05 J

4)   Calculate the calorimeter constant, using the DT of the cold water

Energy lost by the hot water = 907.93J

Energy gained by the cold water = 797.05 J

The calorimeter got the rest:

907.93J - 797.05 J = 110.88 J

The heat capacity of the calorimeter = 110.88 J/3.81K = 29.1 J/K

For DHneutzn for the reaction of sodium hydroxide and acetic acid, temperature-time data are not given.