Question

5. In Procedure II of Part 2 of the lab you will add the correct ratio of water to ice so that the final temperature of the water in the calorimeter is 0°C. Suppose you do this and get the following data: Mass of ice 23.5 g Mass of water originally in the calorimeter 73.79 g Initial temperature of water 25°C Final temperature of water and melted ice 0°C Determine ?Hfusion of water given these data. ----->5.92 kJ/mol

6. Given that ?Hfusion of water = 6.0 kJ/mol, determine the percent error with these data

__________%

7. In Procedure I of Part 2 of the lab, you add ice to more water than the minimum required. Questions 7-10 address this portion of the lab. Which of the following best describes what is in the calorimeter when the system reaches equilibrium?

a. There is only water in the calorimeter and the water is greater than 0°C.

b. There is a mixture of water and ice in the calorimeter and the temperature of the system is less than 0°C.

c. There is a mixture of water and ice in the calorimeter and the temperature of the system is 0°C.

d. There is a mixture of water and ice in the calorimeter and the temperature of the system is greater than 0°C.

e. There is only water in the calorimeter and the water is 0°C.

f. There is only water in the calorimeter and the water is less than 0°C.

8. So, you take ice out of the freezer and add it to room temperature water in a calorimeter (more than enough to just melt the ice). We can determine the ?Hfusion of water by assuming we can account for all energy transfer and that “heat lost” is equal to “heat gained”. We can assume that “heat lost” is due to which of the following? Check all that apply. Select all that are True.

a. Raising the temperature of ice to 0°C.

b. Cooling the calorimeter.

c. Melting the ice.

d. Raising the temperature of the melted ice from 0°C to the final temperature.

e. Cooling the water originally in the calorimeter.

9. We can assume that “heat gained” is due to which of the following? Check all that apply. Select all that are True.

a. Melting the ice.

b. Raising the temperature of the melted ice from 0°C to the final temperature.

c. Cooling the calorimeter.

d. Raising the temperature of ice to 0°C.

e. Cooling the water originally in the calorimeter.

10. We will use 29.1 g of ice and add it to 128.04 g of water at 25°C. Let’s assume that the ice starts at 0°C (we will allow the ice to sit out for a bit of time and “dry it” so we will make this assumption). For this PreLab assignment, let’s also assume we have a perfect calorimeter (you will judge this assumption in Part 1 of the lab). So, we have “heat lost” = “heat gained” as Cooling the 128.04 g of water originally in the calorimeter from 25°C to the final temperature = melting the 29.1 g of ice + raising the temperature of 29.1 g of the melted ice from 0°C to the final temperature. Given that ?Hfusion of water = 6.0 kJ/mol, determine the theoretical final temperature of the water left in the calorimeter when the system reaches equilibrium._____ °C

Answer 1

5. From the given data

heat lost by water = mCpdT

                             = 73.79 x 4.18 x 25 = 7.711 kJ

heat gained by ice = 7.711 kJ

heat of fusion of ice = 7.711 x 18/23.5 = 5.91 kJ/mol

6. True value for heat of fusion = 6.0 kJ/mol

% error = (6.0 - 5.91) x 100/6.0 = 1.5%

7. If we added more ice to water than required,

b. There is a mixture of water and ice in the calorimeter and the temperature of the system is less than 0°C.

8. We can asume the heat lost is due to,

a. Raising the temperature of ice to 0°C. (step 1)

b. Cooling the calorimeter. (step 2)

c. Melting the ice. (step 3)

e. Cooling the water originally in the calorimeter. (step 4)

9. Heat gained by ice in the process is due to the following statements,

c. Cooling the calorimeter.

e. Cooling the water originally in the calorimeter.

10.

heat lost by water = heat gained by ice

feeding the given values for water and ice

128.04 x 4.18 x (25 - T) = 2.91 x 6.0 x 1000/18 + 2.91 x 4.18 x (T - 0)

13380.2 - 535.21T = 970 + 12.2T

T = 12410.2/547.41 = 22.7 oC

So final temperature of water in calorimeter is 22.7 oC