Chem 122
Name: Grace Cole
Section DM1
Name of Partner: Samuel Colclough
Spring 2015
Instructor: Dr. Eric Cotton
Date of Expt: 4/16/15
Date of Submission: 4/23/15
Experiment #9:
CALORIMETRY
Abstract: In Part I, the specific heat of an unknown metal, lead, was calculated to be 0.12
J/
g∙°C using a constant-pressure calorimeter. In Part II, the molar heat of neutralization for
hydrochloric acid was determined to be 53 kJ/mol HCl for the neutralization reaction of
hydrochloric acid and sodium hydroxide within a constant-pressure calorimeter.
Introduction: For Part I, the objective was to calculate the specific heat of an unknown
metal. Specific heat is “the amount of heat needed to raise the temperature of one gram of [a
pure substance] by one degree (either Celsius or kelvin).”1 It can be calculated through the
following equation (provided by Hamilton et al.1):
𝑞
(1)
𝑠 = 𝑚∆𝑇
where q
= amount of heat transferred (in J)
m
= mass of substance (in g)
s
= specific heat of the substance (in J∙g-1∙°C-1).
It is a rearrangement of the following equation, the standard formula for calculating the heat
transfer in a reaction:
𝑞 = 𝑚 × 𝑠 × ∆𝑇
Once the specific heat of the unknown metal has been calculated, its identity will be
determined using the following table of common metals and their specific heats provided by
Hamilton et al.1
(2)
Metal
lead
tungsten
tin
copper
zinc
aluminum
(Pb)
(W)
(Sn)
(Cu)
(Zn)
(Al)
Specific Heat
(J/g∙°C)
0.13
0.13
0.21
0.39
0.39
0.91
In Part I of this experiment, the metal is considered the system and the calorimeter is
considered the surroundings. Ideally, the “surroundings” account for the entire universe (minus
the system), but in this case, the calorimeter is assumed to insulate the heat exchange completely
and is also assumed to not take in any heat. Thus, in our calculations, only the water contained
in the calorimeter is used to calculate the value of heat transfer. The following equation
(provided by Hamilton et al.1) relates the energy gained or lost by the surroundings (water in the
calorimeter) to the energy lost by the system (unknown metal).
(3)
𝑞𝑠𝑦𝑠𝑡𝑒𝑚 = −𝑞𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠
This equation is very important because it is not possible to determine the heat of the system
directly. However, the heat of the surroundings is able to be calculated, so it is very important to
be able to relate the heat of the surroundings to the heat of the system.
In Part II, the objective is to find the molar heat of neutralization of hydrochloric acid for
the following reaction1:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
(4)
The molar heat of reaction is found by first calculating the heat of reaction, the heat
transferred during a certain chemical reaction. The heat of reaction is an extensive property
because its value is reliant on the amount of limiting reactant used in the reaction. Molar heat of
reaction is, however, an intensive property because it is instead the amount of heat produced per
mole of a certain reactant, thus can be calculated independently of reactant amount. Hamilton et
al.1 provided the following equation to determine the molar heat of neutralization of hydrochloric
acid.
(5)
𝑚𝑜𝑙𝑎𝑟 ℎ𝑒𝑎𝑡 𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑎𝑙𝑖𝑧𝑎𝑡𝑖𝑜𝑛 =
ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑡𝑟𝑖𝑎𝑙
# 𝑚𝑜𝑙 𝐻𝐶𝑙 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑡𝑟𝑖𝑎𝑙
𝑞
𝑛𝑒𝑢𝑡𝑟
= # 𝑚𝑜𝑙
𝐻𝐶𝑙
Experimental Section
Procedure by Hamilton et al.1 was followed.
Data & Observations:
Table 1. Data table for Part I
Unknown Metal Code #:
Appearance of Metal (shape and color):
1
grey, round pellets
Trial
Mass of
metal (g)
Mass
Mass of
Calorimeter
+ Water (g)
Temperature
Mass of Empty
Calorimeter (g)
Initial Temp
of Water (°C)
Initial Temp of
Hot Metal (°C)
Final Temp of
Water & of
Metal
1
10.785 g
66.841 g
9.763 g
22.1 °C
99.6 °C
22.7 °C
2
10.754 g
66.357 g
9.662 g
22.2 °C
99.7 °C
22.8 °C
Table 2. Data table for Part II
Trial
Total Volume
After Mixing
(mL)
Tinitial of HCL
soln. (°C)
Tinitial of
NaOH soln.
(°C)
Tfinal (max T
after mixing)
(Vol HCl s ol n + Vol Na OH s ol n)
1
100.00 mL
23.7 °C
23.8 °C
30.3 °C
2
100.00 mL
23.5 °C
23.5 °C
29.8 °C
Calculations & Results:
Part I: Specific Heat of an Unknown Metal
A. Mass of water =
1. 66.841 𝑔 − 9.763 𝑔 = 𝟓𝟕. 𝟎𝟕𝟖 𝒈
2. 66.357 𝑔 − 9.662 𝑔 = 𝟓𝟔. 𝟔𝟗𝟓 𝒈
B. ∆T of water =
1. 22.7°𝐶 − 22.1°𝐶 = 𝟎. 𝟔°𝑪
2. 22.8°𝐶 − 22.2°𝐶 = 𝟎. 𝟔°𝑪
C. ∆T of metal =
1. 22.7°𝐶 − 99.6°𝐶 = −𝟕𝟔. 𝟗°𝑪
2. 22.8°𝐶 − 99.7°𝐶 = −𝟕𝟔. 𝟗°𝑪
D. qwater =
1. (57.078 𝑔) (4.184
2. (56.695 𝑔) (4.184
𝐽
𝑔∙°𝐶
𝐽
𝑔∙°𝐶
) (0.6°𝐶 ) = 𝟏𝟒𝟎 𝑱
) (0.6°𝐶 ) = 𝟏𝟒𝟎 𝑱
E. qmetal =
1. −(140 𝐽) = −𝟏𝟒𝟎 𝑱
2. −(140 𝐽) = −𝟏𝟒𝟎 𝑱
F. smetal =
1. 𝑠 =
2. 𝑠 =
−140 𝐽
10.785 𝑔 × −76.9°𝐶
−140 𝐽
10.754 𝑔 × −76.9°𝐶
= 𝟎. 𝟏𝟐
= 𝟎. 𝟏𝟐
𝑱
𝒈∙°𝑪
𝑱
𝒈∙°𝑪
G. Average smetal =
(0.12 + 0.12)⁄
2 = 𝟎. 𝟏𝟐
Part II: Molar Heat of Neutralization
A. Mass of Solution After Mixing =
1. 100.00 𝑚𝐿 ×
2. 100.00 𝑚𝐿 ×
0.997 𝑔
1 𝑚𝐿
0.997 𝑔
1 𝑚𝐿
= 𝟗𝟗. 𝟕 𝒈
= 𝟗𝟗. 𝟕 𝒈
B. ∆T of solution =
1. 30.3°𝐶 − 23.8°𝐶 = 𝟔. 𝟓°𝑪
2. 29.8°𝐶 − 23.5°𝐶 = 𝟔. 𝟑°𝑪
C. qsoln =
1. (99.7 𝑔) (4.18
2. (99.7 𝑔) (4.18
𝐽
𝑔∙°𝐶
𝐽
𝑔∙°𝐶
) (6.5°𝐶 ) = 𝟐𝟕𝟎𝟎 𝑱
) (6.3°𝐶 ) = 𝟐𝟔𝟎𝟎 𝑱
D. qneutr =
1. qsoln  qneutr: −(2700 𝐽) = −𝟐𝟕𝟎𝟎 𝑱
JkJ: −2700 𝐽 ×
1 𝑘𝐽
103 𝐽
= −𝟐. 𝟕𝒌𝑱
2. qsoln  qneutr: −(2600 𝐽) = −𝟐𝟔𝟎𝟎 𝑱
JkJ: −2600 𝐽 ×
1 𝑘𝐽
103 𝐽
= −𝟐. 𝟔𝒌𝑱
E. Number of moles of HCl =
1. 50.0 𝑚𝐿 ×
2. 50.0 𝑚𝐿 ×
10−3 𝐿
1 𝑚𝐿
10−3 𝐿
1 𝑚𝐿
×
×
1.00 𝑚𝑜𝑙
1𝐿
1.00 𝑚𝑜𝑙
1𝐿
= 𝟎. 𝟎𝟓𝟎𝟎 𝒎𝒐𝒍𝒆𝒔 𝑯𝑪𝒍
= 𝟎. 𝟎𝟓𝟎𝟎 𝒎𝒐𝒍𝒆𝒔 𝑯𝑪𝒍
F. Molar Heat of Neutralization =
−2.7 𝑘𝐽⁄
𝒌𝑱⁄
1.
0.0500 𝑚𝑜𝑙 𝐻𝐶𝑙 = −𝟓𝟒
𝒎𝒐𝒍 𝑯𝑪𝒍
−2.6 𝑘𝐽⁄
𝒌𝑱⁄
2.
0.0500 𝑚𝑜𝑙 𝐻𝐶𝑙 = −𝟓𝟐
𝒎𝒐𝒍 𝑯𝑪𝒍
H. Average Molar Heat of Neutralization =
(−54 𝑘𝐽⁄𝑚𝑜𝑙 𝐻𝐶𝑙 + −52 𝑘𝐽⁄𝑚𝑜𝑙 𝐻𝐶𝑙 )⁄
𝒌𝑱
2 = −𝟓𝟑 ⁄𝒎𝒐𝒍 𝑯𝑪𝒍
Please see attached for Summary of Calculated Results page.
Discussion of Results:
The table below contains the results of the above calculations and the data through which the
average specific heat of unknown metal and the average molar heat of neutralization were found.
Table 3. Results for Part I and II
Part I
Part II
Trial
Specific Heat of Unknown Metal
Molar Heat of Neutralization
#1
0.12 J/g∙°C
54 kJ/mol HCl
#2
0.12 J/g∙°C
52 kJ/mol HCl
Average
0.12 J/g∙°C
53 kJ/mol HCl
In Part I of this experiment, the heat exchange between the unknown metal sample and a
measured amount of water in an insulated coffee-cup calorimeter was observed and measured in
order to find its specific heat. The amount of heat transferred to the water was calculated from
its mass, specific heat, and temperature change. The heat transferred to water was then used to
find the heat transferred from the metal through the 𝑞𝑠𝑦𝑠𝑡𝑒𝑚 = −𝑞𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 relationship.
From the metal’s mass, temperature change, and heat exchange, the specific heat was calculated
to be 0.12 J/g∙°C for both trials. Thus, the average of both trials was 0.12 J/g∙°C. Of the metals
listed on the table provided by Hamilton et al.1, lead and tungsten were the two closest matches
to the experimental specific heat. Lead is a metallic, grey metal, whereas tungsten is a lustrous,
silvery-white metal. The unknown metal was described as grey, round pellets, thus the identity
of the unknown metal must be lead.
In Part II of the experiment, molar heat of neutralization of hydrochloric acid was
calculated by first determining the heat of the neutralization reaction between measured amounts
of hydrochloric acid and sodium hydroxide. Using the mass of solution after mixing (calculated
using the final volume and density of water), its specific heat, and temperature change, the heat
released by the reaction was found. After converting that heat into kilojoules and calculating the
moles of hydrochloric acid used in the reaction, the molar heat of neutralization in kilojoules per
mole of hydrochloric acid was obtained.
The results from both parts of the experiment were close to the accepted values of lead’s
specific heat and hydrochloric acid’s molar heat of neutralization, respectively. There was,
however, a limitation issue in Part I because the calculated ∆T was very small. The small ∆T
allowed for only one significant figure in the calculations, even though the measuring device
should have allowed for two. A solution to this issue would be to measure a greater ∆T by using
either a larger amount of lead or a smaller amount of water in the experiment. The ∆T will then
carry two significant figures and no longer hinder the calculations.
An area where an error could have been introduced during the experiment was in
transferring the metal from Part I from the boiling water to the calorimeter. It was exposed to the
air for about three seconds before being safely contained within the Styrofoam. It is likely that
some heat escaped to the air, not to the water to be measured. The final temperature would have
been too low and forced the calculations off, making the experimental specific heat to be too
low.
A second area where error could have been introduced was in Part II, when measuring the
initial temperature of the solutions. If not given enough time to cool down between trials, the
temperature probes may not have given accurate readings of the actual initial temperatures of the
reactants. The low ∆T will translate to a low heat of reaction. This would skew the calculations
based on equation (5) to yield a low experimental molar heat of neutralization of HCl because
the low value will be in the numerator.
Conclusion: In Part I, an unknown metal (lead) was identified by its specific heat, 0.12
J
/g∙°C, which was determined through calorimetry and appearance. In Part II, the acid-base
reaction between hydrochloric acid and sodium hydroxide was used to calculate the molar heat
of neutralization for hydrochloric acid, 53 kJ/mol HCl.
References:
1. Hamilton, P. Yau, C. and Zaman, K., CHEM 122 Experiments in General Chemistry I
Laboratory, Academx Publishing services: Bel Air, MD, 2013; pp. 105-116
Answers to Post-Lab questions:
1. Examine the initial and final temperatures in Part I. Explain how the temperatures tell
you what type of reaction was involved (endothermic or exothermic). Are the signs of
your qwater and qmetal consistent with this? Explain.
We measured the final temperature within the calorimeter, and it was higher than the
initial temperature, indicating an exothermic reaction. However, the “reaction” here is a
physical exchange of heat, not a chemical reaction, therefore there was no endothermic or
exothermic reaction involved.
2. In Part I, we see that copper and zinc have the same specific heat (See table in the
Introduction.) If you obtained an experimental value of 0.39 J∙g-1∙°C-1, how might you
determine which metal you have as an unknown? Explain.
We recorded the appearance of the unknown metal, as well as determining its specific
heat. If the metal was a silvery-grey, then we could assume that the metal was zinc. If
the metal was a coppery color, then we could assume that the metal was copper. We will
not base our conclusion on only specific heat.
3. We assumed that no heat is lost to the surroundings beyond the nested coffee cups. In
Part I, obviously there would have been some loss in heat as the hot metal is transferred
to the calorimeter. How does that unavoidable heat loss affect your calculated specific
heat of the metal? Would your calculated specific heat be too high or too low due to this
error? Explain fully.
The calculated specific heat would be too low due to this error. The calculated
change in temperature for the water and the calculated transferred heat would be
too low and the calculated change in temperature for the metal would be too high.
For example, in an unknown metal experiment, the final temperature was 30.32°C
but was incorrectly recorded as 29.64°C.
First, the change in temperature for water will be calculated to be too low.
Correct calculations of ∆Twater: 30.32°𝐶 − 23.80°𝐶 = 𝟔. 𝟓𝟐°𝑪
Incorrect calculations of ∆Twater: 29.64°𝐶 − 23.8°𝐶 = 𝟓. 𝟖𝟒°𝑪
Second, the heat in Joules will be calculated to be too low.
𝐽
Correct calculations of qwater: (100.0 𝑔) (4.184 𝑔∙°𝐶 ) (6.52°𝐶) = 𝟐𝟕𝟑𝟎 𝑱
𝐽
Incorrect calculations of qwater: (100.0 𝑔) (4.184 𝑔∙°𝐶 ) (5.84°𝐶) = 𝟐𝟒𝟒𝟎 𝑱
Third, the heat transfer value is still incorrect.
Correct calculations of qmetal: −(2730 J) = −𝟐𝟕𝟑𝟎 𝐉
Incorrect calculations of qmetal: −(2400 𝐽) = −𝟐𝟒𝟒𝟎 𝑱
Fourth, the change in temperature for the metal will be calculated to be too high.
Correct calculations of ∆Tmetal: 30.32°𝐶 − 100.00°𝐶 = 𝟔𝟗. 𝟔𝟖°𝑪
Incorrect calculations of ∆Tmetal: 29.64°𝐶 − 100.00°𝐶 = 𝟕𝟎. 𝟑𝟔°𝑪
Fifth, the specific heat of the metal is calculated to be too low.
−2730 𝐽
𝑱
Correct calculations of smetal: 𝑠 = 10.000 𝑔 × −69.68°𝐶 = 𝟑. 𝟗𝟐 𝒈∙°𝑪
−2440 𝐽
𝑱
Incorrect calculations of smetal: 𝑠 = 10.000 𝑔 × −70.36𝐶 = 𝟑. 𝟒𝟕 𝒈∙°𝑪
4. In Part II, we assume that the density and specific heat of the solution is the same as that
of water. What justifications do we have to make that assumption? Explain.
In Part II, the reactants, hydrochloric acid and sodium hydroxide, reacted together to form
salt and water. Both reactants were 1 M solutions, thus the amount of salt produced in
neutralization compared to the water from the reagents plus the water produced in the
reaction is negligible. Thus, the final solution is mainly composed of water or “dilute”,
and we can assume that the density and specific heat of the solution are the same as that
of water.
5. Using the molar heat of neutralization obtained in your experiment (assuming it is
correct), calculate how much heat you would expect to be produced if you mixed 50.0 mL
of 0.250 M HCl with 50.0 mL of 0.250 M NaOH. Show your calculations. (Hint: How
many moles of HCl are involved?)
In the proposed experiment, there would be the same number of moles of hydrochloric
acid as of sodium hydroxide. Since, in equation (4), the two reactants have a
stoichiometric relationship of 1:1, there is no limiting reagent to worry about. Thus,
according to the molar heat of neutralization obtained in my experiment, fifty milliliters
of 0.250 M hydrochloric acid will produce 0.663 kilojoules when mixed with sodium
hydroxide.
50.0 𝑚𝐿 𝐻𝐶𝑙 ×
10−3 𝐿 𝐻𝐶𝑙 0.250 𝑚𝑜𝑙 𝐻𝐶𝑙
−53 𝑘𝐽
×
×
= −𝟎. 𝟔𝟔 𝒌𝑱
1 𝑚𝐿
1 𝐿 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐻𝐶𝑙