Jack Skrainka
Mr. Kark – Purple
February 28, 2013
The Specific Heat of a Metal
Goals:
Measure the specific heat of lead.
Identify an unknown metal from its specific heat (optional).
Procedure:
As you perform the experiment, record your data in Table 15.1.
1. Heat 250ml of water in a 400ml beaker, until it is boiling gently.
2. While the water is heating, determine the mass of a clean, dry 50ml beaker to the nearest
.01g and record. Add between 80g and 120g of lead shot to the beaker. Measure the
combined mass of the beaker and lead to the nearest .01g and record measurement.
3. Transfer the lead shot to a large, dry test tube. Use the utility clamp to suspend the test
tube in the boiling water as shown in Figure 15.1. The lead shot should be below the level
of water in the beaker. Leave the test tube in the boiling water bath for at least 10
minutes.
4. While the lead shot is heating, measure 100ml of distilled water in a graduated cylinder.
Pour the water into a plastic foam cup, and place the cup in a 250ml beaker for support,
as shown in Figure 15.2.
5. Measure and record the temperature of the water in the plastic foam cup and of the water
in the boiling bath.
6. Remove the test tube from the boiling water and quickly pour the lead shot into the water
filled plastic foam cup. Place a thermometer and a glass stirring rod into the cup. Use
the stirring rod to gently stir the lead shot. Do not stir the shot with the thermometer.
Note the temperature frequently, and record the maximum temperature reached.
7. Pour the water off, and return the lead shot to your teacher.
8. (Optional) Follow the same procedure to determine the specific heat of an unknown
metal.
Table 15.1 Data Sample Data
Trial 1
Trial 2
Mass of lead shot
75.31g
75.13g
Initial temperature of water in cup
Initial temperature of lead shot
(temperature of boiling water)
Maximum temperature of lead + water
Mass of water
19.8°C
20.0°C
100.0°C
100.0°C
21.8°C
25.6°C
91.37g
90.26g
Skrainka 2
Calculation Section
Calculating the Energy Gained by the Water:
Calculating the temperature gained by water:
Lead
∆Twater = Tf - Ti
=21.8°C – 19.8°C
=2.0°C
Energy gained by the water:
Q = m x c x ∆T
=91.37g x 4.184J/g°C x 2.0°C
=760J
Copper
∆Twater = Tf - Ti
=25.6°C – 20.0°C
=5.6°C
Q = m x C x ∆T
=90.26g x 4.184J/g°C x 5.6°C
= 2.1 x 103J
Calculating the temperature lost by metal:
∆TPb = Tf - Ti
=21.8°C – 100.0°C
= -78.2°C
∆TCu = Tf - Ti
= 25.6°C – 100.0°C
= -74.4°C
Energy Gained by the water = Energy lost by metal:
Calculating the specific heat capacity:
Lead
CPb = Q/ m x ∆T
=-760J/ 75.31g x -78.2°C
= .13 J/g°C
Copper
CCu = Q/ m x ∆T
= -2.1 x 103J/ 75.13g x -74.4°C
= .38 J/g°C
Calculating % Error:
% Error Pb = |Accepted value – Experimental Value| x 100%
Accepted Value
= |.16 J/g°C - .13 J/g°C| x 100%
.16 J/g°C
= 20% error Pb
% Error Cu = |Accepted value – Experimental Value| x 100%
Accepted Value
= |.385 J/g°C - .38 J/g°C| x 100%
.385 J/g°C
= 1% error Cu
Skrainka 3
Specific Heat Capacities
3000
2500
E
n 2000
e
1500
r
g 1000
y
500
Lead
Copper
0
0
2000
4000
6000
8000
M∆T
Average Percent Error:
Pb % Error = |Accepted value – Experimental value| x 100%
Accepted Value
= |.16 J/g°C - .106 J/g°C| x 100%
.16 J/g°C
= 30% error Pb
Cu % Error = |Accepted value – Experimental value| x 100%
Accepted value
= |.385 J/g°C - .3139 J/g°C| x 100%
.385 J/g°C
=18.5% error Cu
Results Table
Temperature Gained by H2O (°C)
Energy Gained by H2O (J)
Temperature lost by metal (°C)
Specific Heat Capacity (J/g°C)
Percent Error (%)
Average Specific Heat Capacity (J/g°C)
Average Percent Error (%)
Pb
2.0
760
-78.2
.13
20
.106
30
Cu
5.6
2.1 x 103
-74.4
.38
1
.3139
18.5
Skrainka 4
Discussion of Results:
The results that should have occurred were a specific heat capacity of .16 J/g°C for Pb and .385
J/g°C for Cu. The results that actually occurred were .13 J/g°C for Pb and .38 J/g°C for Cu.
There was a percent error of 1% for Pb and 20% of Cu. The results that occurred were lower
than expected. This means the energy of H2O was too small, the final temperature of H2O was
too small, or the change in temperature in H2O was too low. Other problems that occurred were
the change in temperature of Cu was too large. Also if the temperature of the final metal was
lower than expected, then the results would be lower than expected. These things occurred
because the metal was delayed from being put in. To fix this the metal should be put in the H2O
quicker. Another problem was that the metal did not heat up enough. This means it took longer
to heat all the way, and to fix this the metal should be left to heat up longer. Lastly, the metal
was not submerged all the way in the boiling H2O, or it lost heat to the surrounding environment.
To fix these it must be submerged more, and the top of the test tube should be lightly covered.
Conclusion:
The purpose was to measure the specific heat capacity of Pb and Cu. The Pb and Cu were heated
to the temperature of boiling water. The temperature of the metals was taken, and the specific
heat capacity was found. The results that should have occurred were a specific heat capacity of
.16 J/g°C for Pb and .385 J/g°C for Cu. The results that actually occurred were .13 J/g°C for Pb
and .38 J/g°C for Cu. There was a percent error of 1% for Pb and 20% of Cu. The results that
occurred were lower than expected. The major source of error was the metal was not moved
from the boiling water quick enough, and it lost too much heat in the process. The results of Pb
were not supported. The results of Cu were supported.