LAB PARTNER:
Chris Luong
Quiz Section:
Total Points = 60 pts
Notebook: No points are specifically assigned to your notebook for this lab, however, your TA may deduct up to 5 points
for notebook pages that are disorganized, difficult to read, or do not contain complete data and observations.
PURPOSE AND METHOD
DATA AND CALCULATIONS
A: HEAT CAPACITY OF THE CALORIMETER
Run 1
Run 2
Run 3
Voltage, V (J/C)
2.3
2.3
2.3
Current, A (C/s)
2.85
3.04
2.97
Time
(s)
118.75
136.32
146.69
24.7
24.7
24.2
28.9
28.8
28.2
*Eletrical power input into calorimeter (q), J
778.4
953.1
1002.0
185.33
232.3
250.51
5 pts
241.4
Standard Dev
9.1
Help
In Excel type "=average(range of values)"
Instead of entering a range, just click at one end of the values and drag mouse to the other end
For standard deviation, in Excel type "=stdev(range of values)".
NAME and ID Number:
Alice Meng-Ju Hsieh 0734371
Initial temperature,
o
C
Final temperature,
o
C
*Calorimeter Constant, C
cal,
J/
o
C
Average, C
cal
Our goals are to:
i) Determine the heat capacity of the calorimeter
ii) Measure the heat of fusion of ice
iii) Measure the heat of neautralization
iv) Determine the enthalpy of hydration magnesium sulfate.
Explain how each of these is accomplished in this experiment .
(2 pts each = 8 pts):
i). To determine the heat capacity of the calorimeter, several values need to be measured - voltage, current, time, initial and final temperature.
The initial and final temperature allows us to find the change in temperature. To find the heat capacity of the calorimeter, we need to calculate the amount of
heat added first. The equation is given by: qcal = current x voltage x time. Current is measured in amps, voltage is measured in volts while time is measured
in seconds. In this lab, we assume that the amount of heat added into the system is equal to the amount of heat absorbed by the calorimeter because there
is no heat loss. Then, we can calculate the heat capacity of the calorimeter by the euqation "Ccal = qcal / delta T".
ii). To measure the heat of fusion of ice, we need to know the amount of heat that took place. We need to find the total heat lost by the
calorimeter. Also, we know that ice melting is an endothermic reaction, thus heat is absorbed by the system. There is an equation given in the pre-lab
introduction to find ∆Hfusion.
q = Ccal∆Tcal = (mass of ice)(∆Hfusion) + (mass of ice)(specific heat of water)(∆Twater that was ice)
Keep in mind that specific heat of water is 4.184J/g˚C.
iii). In order to measure the heat of neutralization, we need to know the number of moles of the substance and the heat released by the system.