Hess' Law: Additivity of Heats of Reaction

(Partners: Victoria L.)


Observations:

Reaction 1

Reaction 2

Reaction 3

Calculations:

Calories in Food

(Partners: Victoria L., Zac L.)

Observations:

Data:

Mass of Water: 100g                   Initial Temp of water: 18_̊C

Mass of food before: 1.52g          Final Temp of water: 42.5 _̊C

Mass of food after: .22g               Change in water Temp: 24.5_̊C

Mass of food burned: 1.3g           Calories on package: 5.36Cal

Questions/Calculations:
1. Heat absobed by water (J): (use q=mcΔT) = 10250.8J

2. Heat released per gram of your food (J/g). (Remember: qH2O = -qfood) = -7885.2J/g

3. Heat released by your food in calories (cal): 1 calorie = 4.184J = 1623.2cal

4. Heat released by your food in Calories (cal): 1 calorie = 1000 cal = 1.6232cal

5. Percent Error=Actual - Theoretical/Theoretical x100 = 40.2%

6. Do you think this experiment would be appropriate for all types of food? Why or Why not?
          No, because not all foods burn the same way. So when you burn another food the measurement of calories could be off and not be rational.

7. Describe 3 possible sources of error and how they affected the outcome of the experiment.
          1. We did not let the cheese doodle completely cool to room temperature.
          2. We did not collect the complete sample of the cheese doodle because some broke off.
          3. Some of the heat was lost because the cheese doodle was not properly positioned.


8. A student performed this experiment using a salted peanut. Before it was burned, the peanut weighed 0.353g. After burning, the residue weighed 0.016g. The energy released by the combustion increased the temperature of 200.0mL of water in the calorimeter by 7.2̊C.

          A) Calculate the mass of peanut consumed in the combustion. .337g
          B) Calculate the total energy absorbed by the water in this experiment. 9,037.44J
          C) Calculate the energy released per gram of the peanut. 26817.33J

Solubility and Temperature

(Partners: Erin M., Victoria L., Dan A.)

Observations:

Data:

Sample		Mass of empty test tube	Mass of test tube plus KNO3	Mass of test tube plus KNO3 plus water	Saturation Temperature
Series 1	A	8.76 g	9.21 g	10.26 g	50°
	B	8.77 g	9.47 g	10.44 g	42°
	C	8.82 g	10.03 g	11.23 g	32°
Series 2	A	18.95 g	19.31 g	20.32 g	32°
	B	24.04 g	25.01 g	25.97 g	56°
	C	18.78 g	20.29 g	21.16 g	60°

Post-Lab Calculations and Analysis:

1-2) Series 1: A: 0.45g KNO3 / 1.05g H2O (0.43g)

                      B: 0.70g KNO3 / 0.97g H2O (0.72g)

                      C: 1.21g KNO3 / 1.20g H2O (1.01g)

        Series 2: A: 0.36g KNO3 / 1.01g H2O (0.36g)

                      B: 0.97g KNO3 / 0.96g H2O (1.01g)

                      C: 1.51g KNO3 / 0.87g H2O (1.74g)

3) Series 1: A: 42.86g KNO3/100g H2O

                  B: 72.16g KNO3/100g H2O

                  C: 100.83g KNO3/100g H2O

    Series 2: A: 35.64g KNO3/100g H2O

                  B: 101.04g KNO3/100g H2O

                  C: 173.56g KNO3/100g H2O

4)

  

5) A. 0g

    B. 42.86g

    C. n/a

6) A. 32 degrees C

    B. 13 degrees C

7) Saturated- being a mixture that is unable to absorb or dissolve any more of a substance at a given temperature and pressure

Unsaturated- capable of absorbing or dissolving more of something
Supersaturated- containing an amount of something greater than the amount required for saturation by having been cooled from a higher temperature to a temperature below that at which saturation occurs

8) A. supersaturated?

    B. unsaturated?

9) The solubility would be higher if water from the test tube evaporated because there would be less water for the KNO3 to dissolve into. The saturation temperature would then be higher. 

10) The solubility of the solution would be lower because the temperature was too low.