Calorimetry
Logger Pro Version
Thomas M. Moffett Jr., Plattsburgh State University, 2008
Adapted from “Heat It Up” by Roger Sandwick, Plattsburgh State University
This lab will deal with two types of heat -- specific heat and enthalpy of fusion. That’s it, that’s
all the introduction you get! Sorry.
Specific Heat:
Specific heat is the specific heat that a compound has during a specific situation and a specific
temperature. Specifically, it is the amount of heat energy one gram of the material needs to gain
in order to be raised one degree Celsius. Its units are:
_J__
g oC
Here some values for Specific Heats of Certain Substances:
Substance
Iron
Copper
Aluminum
Silver
Lead
Bloburite
Water
Specific Heat (J/g oC)
0.452
0.385
0.920
0.234
0.159
17.08
4.184
(For those who haven’t read the news lately, bloburite is a goo-like substance currently being
developed by the Navy to suck the heat from every woman, man, and child on the face of the
Earth. That’s why you should immediately join the “Stop The Bloburite” movement and send a
check for $1000 to Mr. Thomas Moffett, Chemistry Dept., Plattsburgh State. With your help we
can make this world a safer place to live.)
Here is the question: Based on its specific heat, what is the metal?
In the first part of the lab you will determine the identity of an unknown metal by calculating its
specific heat. It isn’t possible to directly measure specific heat, instead we measure changes in
temperature. Your calorimeter will be filled with water (C = 4.184 J/g oC), and you will add a hot
piece of metal to the calorimeter. The metal will transfer heat to the water, raising the
temperature of the water, and lowering the temperature of the metal.
The heat lost by the metal is equal to the heat gained by the water:
qmetal = - qwater
(1)
At least that is how the equation would look in the ideal world using a perfect insulator as a
calorimeter. Unfortunately there is no such thing as a perfect insulator, which means that we
must account for the heat absorbed by the calorimeter, thus modifying equation 1.
qmetal = - (qwater + qcalorimeter)
(2)
Where qmetal, qwater, and qcalorimeter are defined as follows:
qmetal = mmetal ∆Tmetal Cmetal
(3)
qwater = mwater ∆Twater Cwater
(4)
qcalorimeter = Ccalorimeter∆Tcalorimeter
(5)
Change in temperature is found by subtracting the initial temperature from the final temperature.
Change in temperature is negative for an object that decreases in temperature and positive for an
object that increases in temperature.
∆T = Tfinal - Tinitial
(6)
The calorimeter constant (Ccalorimeter), which has the same symbol as specific heat (for the
purpose of adding to the confusion), is the product of the calorimeter’s mass and specific heat,
and has units of J/ oC. The calorimeter constant allows us to calculate the amount of heat
absorbed by the calorimeter during the experiment.
Substituting equations 3, 4, and 5 into equation two results in:
mmetal ∆Tmetal Cmetal = - (mwater ∆Twater Cwater + Ccalorimeter∆Tcalorimeter) (7)
Equation 5 can then be rearranged to solve for the heat capacity of the metal:
C metal = −
m water ∆Twater C water + C calorimeter ∆Tcalorimeter
m metal ∆Tmetal
(8)
During the experiment the heat is exchanged instantaneously, however the temperature probes do
not respond instantaneously. To compensate for this, we measure the temperatures over a period
of time, and graphically determine the temperature at the time of mixing.
The graph at the left shows an example of data
recorded during a calorimeter constant trial. The
graph includes three data series. A linear regression
has been applied to all three series. The solutions
were mixed at 125 seconds. Thus by solving all three
equations for the temperature at 125 s, we can then
determine the exact temperature changes at the time
of mixing. The lower left line gives the initial
temperature of the cold water and the calorimeter.
The upper left line gives the initial temperature of the
hot water. The line on the right gives the final
temperature of everything (the water and the
calorimeter are all at the same temperature).
See the handout on using Logger Pro for instructions
on creating the graph.
Procedure:
Part A: Calorimeter Constant
1. Make a coffee cup calorimeter by placing a coffee cup into a second coffee cup. The coffee
cup calorimeter should have a lid of some sort to prevent heat from exiting through the
surface. Determine the mass of your coffee cup calorimeter (without the top and the
thermometer).
2. Place approximately 50 mL of room temperature water into the calorimeter, determine the
mass of the calorimeter and water (without the top). Place one of the temperature probes into
the calorimeter. You may need to clamp the temperature probe to prevent it from tipping over
the calorimeter.
3. In a beaker heat approximately 50 mL of water to a temperature that is 30 to 40 oC greater
than room temperature. Transfer the water to a Styrofoam cup. Place the other temp probe
into the cup.
4. Open Logger Pro, set up the experiment to collect data for 400 seconds at a rate of 0.2
samples/ second (5 seconds/ sample). See the handout on using Logger Pro for instructions.
5. To start collecting data click on Collect, at the same time start a stopwatch. The computer
will now begin collecting data from both temperature probes.
6. At exactly 100 seconds, add the hot water to the cold, put the lid on the calorimeter. The
computer will continue to collect data until the 400 s mark is reached, at which point it will
automatically stop.
7. Determine the mass of the calorimeter (without the top) and the water.
8. Create and print the graph (See instructions in the Logger Pro handout). You may also save
your data at this time if desired.
9. Repeat steps 2 through 6 two additional times. Important: Make sure that you have created
and printed the graph before starting the next run. When you click on Collect again, your
first run will be deleted.
Part B: Specific Heat of a Metal
One thing that should already be apparent at this point in your chemical career -- heat lost by
one material will be gained by another!
1. Weigh the unknown metal then place the piece of metal in hot boiling water. Let it sit until it
has reached the temperature of the boiling water.
2. Place approximately 100 mL of room temperature water into the calorimeter, determine the
mass of the calorimeter and water (without the top). Place one of the temperature probes into
the calorimeter.
3
Measure the temperatures of the metal (boiling water) by placing the second temperature
probe into the beaker of boiling water. Clamp the probe to ensure that it is not touching the
glass.
4
Start collecting data.
5
At exactly 200 seconds, use tongs to remove the piece of metal and place it in the coffee cup
calorimeter. Add the lid and monitor the temperature rise until 400s has been reached.
6
Create and print the graph.
7. Repeat the procedure two additional times. Important: Make sure that you have created and
printed the graph before starting the next run. When you click on Collect again, your first run
will be deleted.
Post Lab Data Analysis and Questions
Part A
•
•
•
Prepare a graph for each trial
Determine the calorimeter constant for each trial
Determine appropriate statistics (mean, standard deviation)
Part B
•
•
•
•
Prepare a graph for each trial
Determine the specific heat of the metal for each trial
Determine the appropriate statistics
Determine the %-error
Q1.) A heater in a hot water tank gives off 50,000 KJ per hour. The initial temperature of 450 kg
of water in the tank is 18.0 oC. If the tank is heated for 90 minutes, determine the final
temperature of the water.
Q2.) How would the following errors affect the calculated specific heat of the metal?
a.) The student placed the metal on the benchtop for 20 seconds while transferring the
metal from the hot water bath to the calorimeter.
b.) While doing this experiment at an altitude of 1600 m, the student didn’t measure the
temperature of the hot water bath. He assumed that the initial temperature of the metal
was 100 oC (the boiling point of water at 0 m).
c.) The calorimeter was wet when the empty mass was determined.
Pre-Lab Questions
1.) How much energy is needed to warm a 45.0 g iron sample from 20.0 oC to 37.5 oC.
2.) Why is the calorimeter made out of two Styrofoam cups instead of two metal cups?
3.) Derive an equation to calculate the calorimeter constant, start with the following equation:
(
q hot H2O = - q cold H2O + q calorimeter
)
Appendix A - A Graphical Guide to Using Logger Pro for the Calorimetry Lab
Thomas M. Moffett Jr., SUNY Plattsburgh, 2008.
Everyone gets the pun in the title right?
A. Setting up the Data Collection
1. Two temperature probes should be connected to the interface box (sensor 1 and
sensor 2). Start Logger Pro.
2. Click on Experiment
Data Collection, the following window will appear:
3. Set the length and sampling rates as directed in the procedure and then click Done.
4. The graph can be set up at this time as well. Double click on the graph, the following
window will appear.
5. You can enter a title in the text box at the top. You’ll also want to make sure that the
appearance settings match those in the above image. Next click on the Axes Options
tab, the window will now appear as:
6. In the X-Axis area change scaling to manual, set the range to match the length of your
experiment. Click on Done, the program is now ready to collect data.
B. Formatting the Graph and Analyzing Data
1. After the experiment has completed, you will need to prepare the graph in order to
extract the needed data. Currently the window should appear as follows.
2. The temperature recorded by the probe connected to Sensor 2 (Temperature 2) is
meaningless after the solutions have been mixed (125 s in this example). These data
points can be removed by first highlighting the data cells recorded after mixing in the
table.
3. Next click on Edit
appear as follows.
Strike Through Data Cells Alt +-, the window should now
4. Now you can add the trend lines to the three different areas of the graph that require
linear fits. On the graph use the mouse to highlight the data that should be included in
the linear regression. Notice that in the example below the highlighted area applies to
both Temp 1 and Temp 2.
5. Click on Analyze Linear fit, the following window will appear. Select the data to
which you want to apply a linear regression to. Note that in the example shown we
want to add a line to both Temperature 1 and Temperature 2. When you do this
procedure for the data recorded after mixing, select only the appropriate data series.
6. After all of the trend lines have been added, it should appear as follows. The graph
can now be printed by selecting File
Print Graph.
Appendix B - Calorimetry Calculations
The following data and graph were collected in order to determine the calorimeter constant.
Mass of calorimeter
3.164 g
Mass of cold water + calorimeter
51.934 g
Mass of mixed water + calorimeter 97.651 g
To determine the temperature at mixing (t = 125 s), solve each line for Temperature (Y) at this
time (X). These temperatures are then used to determine the temperature change of the hot water
and the cold water.
Finding the temperature of the cold water at 125 s.
Y = -0.0001614(oC/s)x +23.62 oC
Y = -0.0001614(oC/s)(125 s) +23.62 oC = 23.60 oC
Doing the same for the hot water and the mixed water gives the following temperatures:
Hot: 42.44 oC
Mixed: 32.55 oC
Temperature change is found by subtracting the initial temperature (hot or cold water)
from the final temperature (mixed water), at the time of mixing.
Tcold = Tmixed - Tcold = 32.55 oC – 23.60 oC = 8.95 oC
Thot = Tmixed – Thot = 32.55 oC – 42.44 oC = -9.89 oC
Since the cold water is in the calorimeter, the temperature change of the calorimeter is the
same as the temperature change of the cold water.
Tcal = Tcold = 8.95 oC
The mass of the cold water is found by subtracting the mass of the calorimeter from the mass of
the calorimeter and cold water.
mcold = 51.934 g – 3.164 g = 48.770 g
The mass of the hot water is determined by subtracting the mass of the cold water and
calorimeter from the mass of the mixed water and calorimeter.
mhot = 97.561 g – 51.934 g = 45.627 g
Knowing that the heat lost by the hot water is equal to the sum of the heat gained by the cold
water and the calorimeter, we can write the following equation.
qhot = -(qcold + qcal)
This equation can be rewritten as:
mhot ThotChot = – mcold TcoldCcold – Ccal Tcal
Rearranging this equation to determine the calorimeter constant (Ccal) gives
C cal =
m hot ∆Thot C hot + m cold ∆Tcold C cold
− ∆Tcal
Substituting in data gives:
C cal =
(45.627 g)(-9.89 o C)(4.184 J g oC ) + (48.770 g)(8.95 oC )(4.184 J g oC )
- 8.95 oC
Ccal = 6.93 J/ oC
The same general procedure is used to determine the specific heat of the unknown metal (Cmetal).