Chem 152 Experiment #4: Thermodynamics I (Calorimetry)
By signing below, you certify that you have not falsified data, that you have not plagiarized any part of this lab report, and that all Note:
calculations and responses other than the reporting of raw data are your own independent work. Failure to sign this declaration will All sections of
result in 5 points being deducted from your report score. this report must
be typed
Total Points = 60 pts (10 notebook, 50 report)
Signature:
Data Entry 2 pts
DATA AND CALCULATIONS
A: Heat Capacity of the Calorimeter
Run 1 Run 2 Run 3
Voltage, V (J/C) 2.2 2.2 2.2
Current, A (C/s) 2.24 2.28 2.25
Time (s) 420 410 420
o
Initial temperature, C 25.7 26.4 26.7
o
Final temperature, C 30.4 31.0 31.3
*Energy into calorimeter from power supply, (q), J 2070 2060 2080
o
*Heat Capacity of Calorimeter, Ccal, J/ C 440 447 452
o
Average, Ccal 446 J/ C
o
Standard Dev 5.81 J/ C
*Type an example of the calculations you are performing for q and Ccal. (3 pts)
q = current * voltage * time = 2.24 C/s * 2.2 J/C * 420 s = 2069.76 J ≈ 2070 J
= q / T = 2069.76
CcalEnthalpy o
J / (30.4ofCWater o o
- 25.7 C) = 440.374 J/ C ≈ 440 J/ C o Help with Formulas in Excel
B: of Fusion
In Excel type "=average(range of values)" but
Run 1 Run 2 instead of entering a range, just click at one end
Ccal(Average), J/ C
o of the values and drag mouse to the other end.
446 446
Mass of calorimeter, stir bar and water, For standard deviation, in Excel type
g 113 113 "=stdev(range of values)".
Mass of calorimeter, stir bar, and water
+ mass of added ice (or melted ice), g
115 115
Mass of ice, g 2.52 2.68
Temperature of calorimeter before ice
addition, oC 26.4 26.3 *Type examples of the calculations you perform for the following:
Temperature of ice before addition to (a) q cal (2 pts)
the calorimeter, oC 0.0 0.0 (b) q for heating the water that was ice (warming melted ice) (2 pts)
Temp. of the calorimeter after addition (c) Hfusion (melting the ice) (2 pts)
and melting of the ice, oC 23.4 23.0
Tcal -3.00 -3.30
-qcal = -CcalT = -446.469 J/oC * -3oC = 1339.407 J ≈ 1340 J
* (a) Heat lost by calorimeter (q cal in J) -1,340 -1,470
water from ice 23.4 23.0 qcal = -1340 J
Specific heat of water, J/g.oC 4.18 4.18 qwater that was ice = 4.184 J/goC * 2.52 g * 23.4oC = 246.722 J ≈ 247 J
Heat gained by ice cube to warm water
* (b) from ice temperature (q in J)
247 258 H
fusion = (1339.407 J - 246.722 J) / 2.52 g = 433.605 J/g ≈ 434 J/g
*(c) Enthalpy of fusion of ice, H fusion in J/g 434 453
Average H 443 J/g
Standard Dev 13.4 J/g
By signing below, you certify that you have not falsified data, that you have not plagiarized any part of this lab report, and that all Note:
calculations and responses other than the reporting of raw data are your own independent work. Failure to sign this declaration will All sections of
result in 5 points being deducted from your report score. this report must
be typed
Total Points = 60 pts (10 notebook, 50 report)
Signature:
Data Entry 2 pts
DATA AND CALCULATIONS
A: Heat Capacity of the Calorimeter
Run 1 Run 2 Run 3
Voltage, V (J/C) 2.2 2.2 2.2
Current, A (C/s) 2.24 2.28 2.25
Time (s) 420 410 420
o
Initial temperature, C 25.7 26.4 26.7
o
Final temperature, C 30.4 31.0 31.3
*Energy into calorimeter from power supply, (q), J 2070 2060 2080
o
*Heat Capacity of Calorimeter, Ccal, J/ C 440 447 452
o
Average, Ccal 446 J/ C
o
Standard Dev 5.81 J/ C
*Type an example of the calculations you are performing for q and Ccal. (3 pts)
q = current * voltage * time = 2.24 C/s * 2.2 J/C * 420 s = 2069.76 J ≈ 2070 J
= q / T = 2069.76
CcalEnthalpy o
J / (30.4ofCWater o o
- 25.7 C) = 440.374 J/ C ≈ 440 J/ C o Help with Formulas in Excel
B: of Fusion
In Excel type "=average(range of values)" but
Run 1 Run 2 instead of entering a range, just click at one end
Ccal(Average), J/ C
o of the values and drag mouse to the other end.
446 446
Mass of calorimeter, stir bar and water, For standard deviation, in Excel type
g 113 113 "=stdev(range of values)".
Mass of calorimeter, stir bar, and water
+ mass of added ice (or melted ice), g
115 115
Mass of ice, g 2.52 2.68
Temperature of calorimeter before ice
addition, oC 26.4 26.3 *Type examples of the calculations you perform for the following:
Temperature of ice before addition to (a) q cal (2 pts)
the calorimeter, oC 0.0 0.0 (b) q for heating the water that was ice (warming melted ice) (2 pts)
Temp. of the calorimeter after addition (c) Hfusion (melting the ice) (2 pts)
and melting of the ice, oC 23.4 23.0
Tcal -3.00 -3.30
-qcal = -CcalT = -446.469 J/oC * -3oC = 1339.407 J ≈ 1340 J
* (a) Heat lost by calorimeter (q cal in J) -1,340 -1,470
water from ice 23.4 23.0 qcal = -1340 J
Specific heat of water, J/g.oC 4.18 4.18 qwater that was ice = 4.184 J/goC * 2.52 g * 23.4oC = 246.722 J ≈ 247 J
Heat gained by ice cube to warm water
* (b) from ice temperature (q in J)
247 258 H
fusion = (1339.407 J - 246.722 J) / 2.52 g = 433.605 J/g ≈ 434 J/g
*(c) Enthalpy of fusion of ice, H fusion in J/g 434 453
Average H 443 J/g
Standard Dev 13.4 J/g
