Calorimetry Lab Report
CHEMISTRY IA: Processing
CALORIMETRY
QUANTITATIVE DATA TAKEN (07.05.14 and 21.15.14)
Initial Pringle Mass
H20 Amount
Final Pringle Mass
Δ Pringle Mass
Initial H2O Temp (± 0.5°C)
Max H2O Temp (± 0.5°C)
Δ H2O Temp (± 0.5°C)
SAMPLE 1
10g
225mL
1.85g
10g-1.85g=8.85g
22.3°C
59.8°C
37.5°C
SAMPLE 2
10g
225mL
0.95g
10g-0.95g=9.05g
21°C
61.1°C
40.1°C
SAMPLE 3
10g
225mL
1.95g
10g-1.95g=8.05g
23°C
58°C
35°C
SAMPLE 4
10g
225mL
1.85g
10g-1.85g=8.15g
50.5°C
83°C
32.5°C
SAMPLE 5
10g
225mL
1.95g
10g-1.95g=8.05g
60°C
90°C
30°C
Note: SAMPLEs 1 and 2 data was taken on 07.05.14 (Room Temp= 21.5°C), and SAMPLEs 3, 4, and 5 data was taken on 21.15.14 (Room Temp= 22°C).
QUALITATIVE DATA: same water used for last first 2 samples, and same water used for last 3 samples pringles become black color with white dusting pool of oil/unknown substance fills bottom of petri dish [filmy brown] glass beaker enveloped with soot/smoke residue chips released thick blackish-grey smoke pringles are fluffy and soft [after burning], collapse at touch sizzling oil sound pringle becomes [in chronological order: golden, brown, black]
PROCESSING THE DATA:
In order to find the Δ Mass and Temperature for each trial, this equation was used:
X = Mass or Temperature
ΔX= Final X - Initial X
[We must not, however, forget that there is an ± 0.5°C absolute uncertainty, and this is taken into count.]
Knowing that:
- Specific Heat Capacity of water (H2O) = 4.186 joule/gram °C
- 1 mL of H2O = 1g of H2O
- 225g of water was used
We can use this equation to find the amount of calories used in the food:
q = m x c x ΔT
CALCULATIONS
Sample 1:
In order to find the amount of energy absorbed by the water: q = m x c x ΔT q = 225 x 4.186 x 37.5 q = 35319.38 J
Now, to convert this to calories:
1 calorie = 4.18 Joules
So: 35319.375 J / 4.18 = 8449.60 calories
Finally, to convert calories into kilocalories, we divide by 1000:
CALORIMETRY
QUANTITATIVE DATA TAKEN (07.05.14 and 21.15.14)
Initial Pringle Mass
H20 Amount
Final Pringle Mass
Δ Pringle Mass
Initial H2O Temp (± 0.5°C)
Max H2O Temp (± 0.5°C)
Δ H2O Temp (± 0.5°C)
SAMPLE 1
10g
225mL
1.85g
10g-1.85g=8.85g
22.3°C
59.8°C
37.5°C
SAMPLE 2
10g
225mL
0.95g
10g-0.95g=9.05g
21°C
61.1°C
40.1°C
SAMPLE 3
10g
225mL
1.95g
10g-1.95g=8.05g
23°C
58°C
35°C
SAMPLE 4
10g
225mL
1.85g
10g-1.85g=8.15g
50.5°C
83°C
32.5°C
SAMPLE 5
10g
225mL
1.95g
10g-1.95g=8.05g
60°C
90°C
30°C
Note: SAMPLEs 1 and 2 data was taken on 07.05.14 (Room Temp= 21.5°C), and SAMPLEs 3, 4, and 5 data was taken on 21.15.14 (Room Temp= 22°C).
QUALITATIVE DATA: same water used for last first 2 samples, and same water used for last 3 samples pringles become black color with white dusting pool of oil/unknown substance fills bottom of petri dish [filmy brown] glass beaker enveloped with soot/smoke residue chips released thick blackish-grey smoke pringles are fluffy and soft [after burning], collapse at touch sizzling oil sound pringle becomes [in chronological order: golden, brown, black]
PROCESSING THE DATA:
In order to find the Δ Mass and Temperature for each trial, this equation was used:
X = Mass or Temperature
ΔX= Final X - Initial X
[We must not, however, forget that there is an ± 0.5°C absolute uncertainty, and this is taken into count.]
Knowing that:
- Specific Heat Capacity of water (H2O) = 4.186 joule/gram °C
- 1 mL of H2O = 1g of H2O
- 225g of water was used
We can use this equation to find the amount of calories used in the food:
q = m x c x ΔT
CALCULATIONS
Sample 1:
In order to find the amount of energy absorbed by the water: q = m x c x ΔT q = 225 x 4.186 x 37.5 q = 35319.38 J
Now, to convert this to calories:
1 calorie = 4.18 Joules
So: 35319.375 J / 4.18 = 8449.60 calories
Finally, to convert calories into kilocalories, we divide by 1000: