Aaron Bunch
CHEM 111 Morning Lab
27 October 2014
Experiment 25: Calorimetry
Conclusion: The unknown metal #14 has a specific heat of 0.36 J/g °C; the heat of neutralization of HCl and
NaOH is -53.0 kJ/mol H2O produced; and the heat of neutralization of HNO3 and NaOH is -55.6 kJ/mol H2O
produced.
Abstract
The specific heat of unknown metal #14 was expected to be less than 0.50 J/g °C, and the heats of
neutralization of hydrochloric acid and nitric acid with sodium hydroxide were expected to lie between -57 and
-58 kJ/mol H2O produced. To test the first hypothesis, 58.2431 g of the metal was heated to thermal equilibrium
with a boiling water bath and then combined with room temperature water in a calorimeter. The temperature
change of the metal was -58.39 °C, and 1.225 kJ of energy were transferred from the metal to the water. The
specific heat of the metal was then determined to be 0.36 J/g °C, which is consistent with our expectations. To
test the second hypothesis, each acid was combined with sodium hydroxide in a calorimeter in two separate trials.
The temperature of the solution increased by 6.282 °C and by 6.590 °C for HCl and HNO3, respectively. This
corresponded to 2.6284 kJ and 2.7573 kJ of heat evolved by the reactions. The heats of neutralization were then
determined to be -53.0 kJ/mol H2O produced and -55.6 kJ/mol H2O produced. These results fall just short of the
expected values.
Introduction
In the first part of this experiment, the specific heat of an unknown metal is determined. In accordance
with the experimental procedure outlined in Beran (208-213), the metal is first brought to thermal equilibrium
with a boiling water bath, and then added to water at room temperature in a double coffee cup calorimeter. The
temperature of the water is recorded at regular intervals and plotted against time. A trend line fitted to the data
estimates the maximum temperature of the water and metal in thermal equilibrium. The calorimeter is assumed to
be an isolated system, since the heat lost to the surroundings is negligible. Therefore, the amount of heat
absorbed by the water is equal to the amount of heat lost by the metal:
𝑞!"#$% = −𝑞!"#$%
[1]
The heat absorbed by the water is the product of the mass of the water, the specific heat of water, and the change
in temperature of the water:
𝑞!"#$% = (𝑚!"#$% )(𝑠!"#$% )(∆𝑇!"#$% )
[2]
The specific heat of the metal is equal to the heat lost by the metal divided by the product of the mass of the metal
and the change in temperature of the metal:
𝑠!"#$% = !!"#$%
(!!"#$% )(∆!!"#$% )
[3]
In the second part of this experiment, the heat of neutralization is determined for two strong acids and a sodium
hydroxide solution. In accordance with Beran’s procedure (208-213), hydrochloric acid and nitric acid are
combined in separate trials with sodium hydroxide in a double coffee cup calorimeter. The temperature of the
mixture is recorded at regular intervals and plotted against time. A trend line fitted to the data estimates the
maximum temperature of the mixture. The calorimeter is assumed to be an isolated system, since the heat lost to
the surroundings is negligible. Therefore, the amount of heat evolved by the reaction is equal to the amount of
heat absorbed by the mixture:
𝑞!"#$%&'( = −𝑞!"#$%&'
[4]
Since the mixture is mostly water, the heat absorbed by the mixture is given by equation [2] above. Dividing the
heat evolved by the reaction by the moles of water produced gives the heat of neutralization in terms of Joules per
mole water produced (J/mol H2O).
2 Calorimetry is the measure of heat. Heat is the spontaneous transfer of energy from a system of higher
energy to a system of lower energy through direct contact (Ebbing & Gammon, 231). The heat transfer stops
when thermal equilibrium is reached, which means both systems have the same temperature. A calorimeter is an
instrument used to measure heat transfer during a physical or chemical change (Ebbing & Gammon, 243). Two
Styrofoam coffee cups, one inserted inside the other, and a loose-fitting Styrofoam lid with a hole for a
thermometer, make an inexpensive constant-pressure calorimeter. The heat transfer occurs inside the coffee cups,
and the temperature change is measured with the thermometer inserted through the lid. If the calorimeter is well
insulated, and the heat transfer occurs relatively quickly, then heat loss to the surroundings can be neglected and
the calorimeter can be treated as an isolated system. In an isolated system, the sum of the heat transfers within the
system equals zero.
In order to measure the heat of a chemical or physical change, one must determine the temperature when
thermal equilibrium is first reached. This temperature is never directly measured. In the time it takes to reach
thermal equilibrium, some small amount of heat is lost to the calorimeter (Beran 212). To account for this heat
loss, the temperature at thermal equilibrium must be extrapolated from a trend line. In this procedure, the
temperature of the mixture in the calorimeter is recorded as it cools well after the equilibrium point is first
reached. A linear trend line is fit to the cooling part of the temperature data, and extended backwards to the time
when the heat transfer began. This point gives a theoretical maximum temperature at thermal equilibrium, as if
the heat were transferred instantly without any loss to the calorimeter. Examples of this procedure are given in
Figures 1, 2, and 3 below.
Acid-base neutralization reactions produce heat in addition to water. A strong acid and strong base
neutralize according to the following net ionic equation (Beran 209):
H+(aq) + OH-(aq) → H2O(l) + heat
[5]
The heat of this reaction can be measured in a calorimeter. Since strong acid-base neutralization reactions happen
quickly, and the calorimeter is well insulated, the amount of heat lost to the environment is negligible. Therefore,
we may assume that the calorimeter is an isolated system. In an isolated system, the amount of heat evolved by
the reaction is equal to the amount of heat absorbed by the mixture of solutions. Because the acid and base
solutions are relatively dilute, we may assume that they have the same density (1 g/mL) and specific heat (4.18 J/g
°C) as water (Beran 209). Then the heat absorbed by the solutions can be calculated by equation [2] above. The
heat of neutralization is usually given in terms of Joules per mole water produced by the reaction (Beran 209).
According to equation [5], one mole of water is produced per mole reaction. Therefore, the heat per mole reaction
is equal to the heat of neutralization per mole water produced.
With only a few exceptions, the specific heats of metals lie below 0.50 J/g °C
(http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html). Therefore, in the first part of this
experiment, we expect the specific heat of the unknown metal to fall in this range. In the second part of this
experiment, we expect the heats of neutralization to be between -57 and -58 kJ/mol. Because strong acids and
strong bases ionize completely in water, the net ionic equation for all strong acid-base neutralizations is the same
(equation [5] above), and evolves the same quantity of heat. Other experiments have found this quantity to lie
between -57 and -58 kJ/mol. (http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions
/Enthalpy/Enthalpy_Change_of_Neutralization).
Results
1. Specific heat of an unknown metal
22.0 g of water at 22.8 °C were put into a double coffee cup calorimeter. Meanwhile, 58.2431 g of the
unknown metal were placed in a dry test tube and heated in a boiling water bath for approximately 15 minutes.
3 After 15 minutes, it was assumed that the metal and water bath had reached thermal equilibrium at 94.5 °C.
These initial conditions are summarized in the first four columns of Table 1.
Mass and Temperature of Water and Metal
Table 1: The mass and temperature of the water and metal before they were combined in the calorimeter, and the maximum temperature they reach when combined as extrapolated from their graph. mass of metal
(g)
58.2431
initial
temperature of
metal (°C)
94.5
mass of water
(g)
22.0
initial
temperature of
water (°C)
22.8
maximum temperature
of water and metal
when combined (°C)
36.11
The heated metal was quickly combined with the water in the calorimeter, and stirred with the bulb of a
thermometer inserted through the lid of the calorimeter. The temperature of the water was recorded at irregular
intervals for five minutes. These temperature measurements are summarized in Table 2. Time = 0 is the point at
which the metal was added to the water in the calorimeter.
Water Temperature over Time
Table 2: Water temperature in a coffee cup calorimeter after adding an
unknown metal heated to 94.5 °C. The metal was added to the water at
time = 0.
time (s)
0
36
50
70
94
115
135
155
170
205
230
240
270
300
water temperature (°C)
22.8
35.0
35.8
35.8
35.6
35.5
35.4
35.4
35.3
35.0
34.9
34.9
34.8
34.6
trend line dataset (°C)
35.8
35.6
35.5
35.4
35.4
35.3
35.0
34.9
34.9
34.8
34.6
In Figure 1 below, water temperature is plotted against the time since adding the metal to the water. The
y-axis, t = 0, is the time at which the metal is added to the water. A linear trend line is fitted to the data for the
time after the water has begun cooling. The y-intercept of the trend line gives the extrapolated maximum
4 temperature of the water and metal in thermal equilibrium. This value is given in the final column of Table 1
above, and can be read from the equation of the trend line in the figure below (36.11 °C).
Figure 1: Heated Unknown Metal Combined with Water. The y-axis, time = 0, is the time at which the
metal is added to the water. The trend line is fitted to the data for the time after the water has begun
cooling. The y-intercept of the trend line gives the extrapolated maximum temperature of the water and
metal in thermal equilibrium.
Heated Unknown Metal Combined with Water
40.0
y = -0.005x + 36.11
R² = 0.98385
water temperature (°C)
35.0
30.0
25.0
20.0
0
100
200
300
400
time (s)
The specific heat of the metal is calculated from the temperature change of the metal, the heat lost by the
metal, and the mass of the metal (from Table 1 above). Since we assume the calorimeter to be an isolated system,
the amount of heat lost by the metal is equal to the amount of heat gained by the water. The heat gained by the
water is calculated from the temperature change of the water, the mass of the water (from Table 2 above), and the
specific heat of water (4.184 J/g °C). These results are summarized in Table 1 above and Table 3 below.
5 Specific Heat of the Metal
Table 3: The specific heat of the metal is calculated from the temperature change of the metal, the
heat lost by the metal, and the mass of the metal (from Table 1). The amount of heat lost by the metal
is equal to the amount of heat gained by the water. The heat gained by the water is calculated from the
temperature change of the water, the mass of the water (from Table 1), and the specific heat of water
(4.184 J/g °C).
temperature
change of
water (°C)
heat gained by
water (kJ)
heat lost by
metal (kJ)
temperature
change of metal
(°C)
specific heat of metal
(J/g °C)
13.31
1.225
-1.225
-58.39
0.3602
2. Heat of neutralization of two strong acids with NaOH
In trial #1, 50.0 g of 1.1 M HCl solution at 22.0 °C was added to 50.0 g of standardized 0.9920 M NaOH
solution at 22.0 °C in a double coffee cup calorimeter. The density of both solutions is assumed to be that of
water (1 g/mL). The mixture was stirred with the bulb of a thermometer inserted through the lid of the
calorimeter, and the temperature of the mixture was recorded at irregular intervals for five minutes. These
measurements are given in Table 4.
Temperature of Solution over Time
Table 4: Temperature of solution in a coffee cup calorimeter after
combining equal volumes of HCl and NaOH solutions. The solutions
were mixed at time = 0.
time (s)
temperature (°C)
trend line dataset (°C)
0
22.0
23
27.0
34
27.8
41
28.1
51
28.1
61
28.1
80
28.2
104
28.2
28.2
120
28.1
28.1
140
28.1
28.1
160
28.0
28.0
200
28.0
28.0
220
28.0
28.0
240
28.0
28.0
260
27.9
27.9
280
27.9
27.9
300
27.9
27.9
6 In Figure 2, the temperature of the mixture is plotted against the time since combining the solutions in the
calorimeter. The y-axis, time = 0, is the time when the solutions are mixed. A linear trend line is fitted to the data
for the time after the mixture has begun cooling. The y-intercept of the trend line gives the extrapolated
maximum temperature of the mixture.
Figure 2: HCl Combined with NaOH. The y-axis, time = 0, is the time when the solutions are mixed.
The trend line is fitted to the data for the time after the mixture has begun cooling. The y-intercept of the
trend line gives the extrapolated maximum temperature of the mixture.
HCl Combined with NaOH
temperature of solution (°C)
30.0
y = -0.0013x + 28.282
R² = 0.867
28.0
26.0
24.0
22.0
20.0
0
100
200
300
400
time (s)
In trial #2, 50.0 g of 1.1 M HNO3 at 21.5 °C was added to 50.0 g of standardized 0.9920 M NaOH
solution at 22.0 °C in a double coffee cup calorimeter. The density of both solutions is assumed to be that of
water (1 g/mL). The mixture was stirred with the bulb of a thermometer inserted through the lid of the
calorimeter, and the temperature of the mixture was recorded at irregular intervals for five minutes. These
measurements are given in Table 5 on the next page.
7 Temperature of Solution over Time
Table 5: Temperature of solution in a coffee cup calorimeter after
combining equal volumes of HNO3 and NaOH solutions. The
solutions were mixed at time = 0.
time (s)
0
25
60
80
100
120
140
160
180
200
220
240
260
280
300
temperature (°C)
21.5
27.0
27.8
27.8
27.8
27.8
27.8
27.8
27.7
27.7
27.6
27.6
27.6
27.5
27.5
trend line dataset (°C)
27.8
27.7
27.7
27.6
27.6
27.6
27.5
27.5
In Figure 3 on the next page, the temperature of the mixture is plotted against the time since combining
the solutions in the calorimeter. The y-axis, time = 0, is the time when the solutions were mixed. A linear trend
line is fitted to the data for the time after the mixture has begun cooling. The y-intercept of the trend line gives
the extrapolated maximum temperature of the mixture.
8 Figure 3: HNO3 Combined with NaOH. The y-axis, time = 0, is the time when the solutions are mixed.
The trend line is fitted to the data for the time after the mixture has begun cooling. The y-intercept of the
trend line gives the extrapolated maximum temperature of the mixture.
temperature (°C)
HNO3 Combined with NaOH
29.0
28.0
27.0
26.0
25.0
24.0
23.0
22.0
21.0
20.0
y = -0.002x + 28.09
R² = 0.91746
0
100
200
time (s)
300
400
The results of both trials are summarized in Table 6. The heat of neutralization (kJ/mol H2O) of the
strong acid with NaOH in each trial is the heat evolved by the mixture (kJ) per mole of H2O formed. By the
stoichiometry of the reaction, we know that one mole of H2O is produced for every mole of OH- reacted. The
heat evolved by the mixture is calculated from the total mass of the mixture, the change in temperature of the
mixture, and the specific heat of water (4.184 J/g °C).
Heat of Neutralization of a Strong Acid and a Strong Base
Table 6: The heat of neutralization (kJ/mol H2O) of a strong acid with NaOH is the heat evolved by the mixture
(kJ) per mole of H2O formed. One mole of H2O is produced for every mole of OH- reacted. The heat evolved by
the mixture is calculated from the total mass of the mixture, the change in temperature of the mixture, and the
specific heat of water (4.184 J/g °C).
acid
total
mass of
mixture
(g)
initial
temperature
of mixture
(°C)
maximum
temperature
of mixture
(°C)
change in
temperature
(°C)
heat
evolved
by
reaction
(kJ)
moles
of OHreacted
(mol)
moles
of H2O
formed
(mol)
heat of
neutralization
(kJ/mol H2O)
HCl
100.0
22.0
28.282
6.282
-2.6284
0.0496
0.0496
-53.0
HNO3
100.0
21.5
28.090
6.590
-2.7573
0.0496
0.0496
-55.6
9 Discussion
1. Specific heat of an unknown metal
When the heated metal was combined with the room temperature water in the calorimeter, heat
spontaneously transferred from the hotter metal to the cooler water until the metal and the water reached thermal
equilibrium. The recorded value of this temperature is 35.8 °C, but the extrapolated value is 36.11 °C. The
extrapolated value corrects for the heat lost to the calorimeter before the equilibrium temperature reached its
maximum.
Since the calorimeter is assumed to be an isolated system, the amount of heat lost by the metal is equal to
the amount of heat gained by the water. The heat gained by the water was determined to be 1.225 kJ, so the heat
lost by the metal was -1.225 kJ. The specific heat of the metal was then determined to be 0.3602 J/g °C using
equation [3]. As expected, this value (below 0.5 J/g °C) is typical of the specific heat of most metals.
2. Heat of neutralization of two strong acids with NaOH.
When the acids were combined with the NaOH solution in the calorimeter, the H+ ions and the OH- ions
combined to form liquid water. The reaction also released an amount of heat. This heat was absorbed by the
mixtures (mostly water) in the calorimeter, and raised their temperature. The peak temperature was measured to
be 28.2 °C for HCl and 27.8 °C for HNO3, but the extrapolated values are 28.282 °C for HCl and 28.090 °C for
HNO3. The extrapolated values correct for the heat lost to the calorimeter before the maximum temperature was
reached.
Since the calorimeter is assumed to be an isolated system, the amounts of heat evolved by the reactions
are equal to the amounts of heat absorbed by the mixtures. The heats evolved by the reactions were determined to
be -2.6284 kJ for HCl and -2.7573 kJ for HNO3. Dividing by 0.0496 moles of H2O produced gives -53.0 kJ/mol
H2O and -55.6 kJ/mol H2O, respectively.
These values for the heats of neutralization are 1.4 – 4.0 kJ/mol lower than the expected value. This error
can be accounted for by some combination of 1) underestimating the temperature change of the mixture and 2)
overestimating the moles of H2O produced by the reaction. The temperature change of the mixture would be
underestimated if its rate of cooling were underestimated. In that case, the trend line would be flatter, and it
would extrapolate a lower maximum temperature. It is possible that this occurred in the present experiment. The
HNO3 and NaOH mixture cooled only 0.3 °C over three minutes, which seemed suspiciously slow. It is also
possible that the amount of H2O produced was overestimated. This could happen if some of the solution were
inadvertently spilled.
Conclusion
The unknown metal #14 has a specific heat of 0.36 J/g °C; the heat of neutralization of HCl and NaOH is
-53.0 kJ/mol H2O produced; and the heat of neutralization of HNO3 and NaOH is -55.6 kJ/mol H2O produced.
References
Beran, J. A. Laboratory Manual for Principles of General Chemistry, 9E. John Wiley & Sons, Inc: USA, 2011.
Reprinted in Citrus Lab Manual: CHEM 111/112. John Wiley & Sons, Inc: USA, 2014.
Ebbing, Darrell D. and Steven D. Gammon. General Chemistry, 10E. Brooks/Cole, Cengage Learning: USA,
2013.
10 The Engineering ToolBox. Web Resource. <www.EngineeringToolBox.com> Accessed on October 26, 2014.
UC Davis ChemWiki. Web Resource. <http://chemwiki.ucdavis.edu/ > Accessed on October 26, 2014.
11