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. 

Heating and Cooling Curve

(Partner: Victoria L.)


Observations:

Data:

Cooling – Part 1                                                                     Heating – Part 2                

Time (min)	T (oC)	Time (min)	T (oC)			Time (min)	T (oC)	Time (min)	T (oC)
0	70	3.5	48			0	37	3.5	77
0.5	66	4.0	46			0.5	53	4.0	78
1.0	63	4.5	44			1.0	64	4.5	81
1.5	60	5.0	42			1.5	70	5.0	84
2.0	57	5.5	40			2.0	74	5.5	86
2.5	54	6.0	38			2.5	75	6.0	88
3.0	51	6.5	37			3.0	77	6.5	90

Analysis:

Questions:

1. Based on your graph, determine the melting and freezing point of the substance used. How do these values compare? 

63 and 64 degrees. They met at those points at the same time and are very close. Both came to these tempertures at a minute in to recording the temperature and thats how they compare.

2. Describe the shape of your graph during the actual changes of state (while the substance
is actually melting or solidifying).

When it was soldifying it was at a steady pace gradually geting slower. The shape of the line remained pretty straight. When it was melting it shot up at first then moved more gradually. The shape shows an instant change at first then slowing down as depicted in the line of the graph.

3. During the heating process, heat is continually being supplied to the sample throughout the entire time of the experiment even though the temperature remains constant during the actual change of phase. How can this be explained at the molecular level, in terms of what is happening to the chemical bonds holding the particles together in the solid state?

When the molecules heat up it vibrates them making it turn into a liquid. The molecules are able to move around because the bonds are breaking. When it was a solid the molecules were together tight and when they are a liquid they can move around more. The molecules slow down and start to cool.

4. Consider the diagonal region of the cooling curve, as the sample is being cooled. What does the temperature change indicate about the change in kinetic energy of the particles in the sample?

When it begins to cool the kinetic energy of the particle is not as great. There is more space between the particle and the molecules slow down making it become solid.