Calorimetry Lab Report

Topics: Energy, Calorie, Heat Pages: 6 (1143 words) Published: October 12, 2014
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: Amount of energy absorbed by water = 8.45 kcals

In order to find the amount of energy per gram:
Energy per gram =
The change in mass for this sample was 8.85g.
Energy per gram =
Energy per gram = 0.95 kcals

So, finally: According to my experiment, in a 10 gram sample of Pringles, how many kcals are there ? 10g x 0.95 kcals = 9.5 kcals

Sample 2:
In order to find the amount of energy absorbed by the water: q = m x c x ΔT
q = 225 x 4.18 x 40.1
q = 37714.05 J

Now, to convert this to calories:
1 calorie = 4.18 Joules
So: 37714.05 J / 4.18 = 9022.50 calories

Finally, to convert calories into kilocalories, we divide by 1000: Amount of energy absorbed by water = 9.02 kcals

In order to find the amount of energy per gram:
Energy per gram =
The change in mass for this sample was 9.05g.
Energy per gram =
Energy per gram = 0.99 kcals

So, finally: According to my experiment, in a 10 gram sample of Pringles, how many kcals are there ? 10g x 0.99 kcals = 9.9 kcals

Sample 3
In order to find the amount of energy absorbed by the water: q = m x c x ΔT
q = 225 x 4.18 x 35°C
q = 32917.50 J

Now, to convert this to calories:
1 calorie = 4.18 Joules
So: 32917.50 J / 4.18 = 7875 calories

Finally, to convert calories into kilocalories, we divide by 1000: Amount of energy absorbed by water = 7.88 kcals

In order to find the amount of energy per gram:
Energy per gram =
The change in mass for this sample was 8.05g.
Energy per gram =
Energy per gram = 0.98 kcals

So, finally: According to my testing, in a 10 gram sample of Pringles, how many kcals are there ? 10g x 0.98 kcals = 9.8 kcals

Sample 4:
In order to find the amount of energy absorbed by the water: q = m x c x ΔT
q = 225 x 4.18 x 32.5°C
q = 30566.25 J

Now, to convert this to calories:
1 calorie = 4.18 Joules
So: 30566.25 J / 4.18 = 7312 calories

Finally, to convert calories into kilocalories, we divide by 1000: Amount of energy absorbed by water =...
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