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Explorations in Thermodynamics: Calorimetry, Specific Heat & Neutralization
Del Valle, Alyssa L.
Manuel, Robin Piolo M.
Motas, Hannah Faye A.
Pagaduan, Mark Gerald N.
Ricablanca, Princess Jane D.
Santos, Claude Jayson U.
Seleteria, Richel B.
Torres, Khim Harold O.
GENCHEM 2 11 – Kingston
Group 3
April 23, 2018
_______________________________
1 A scientific paper submitted in partial fulfillment of the requirements in General
Chemistry 2 Laboratory under Mr. Jeoffrey Sanga - Delos Reyes, 2nd sem., 2017 - 2018.
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ABSTRACT
This experiment was done to discover more about heat
transfer in regards to the law of conservation of energy. The
students observed that the heat will increase at first, but it will
decrease after some times. This experiment allowed us to
derive three different values related to heat transfer inside a
calorimeter (Styrofoam cup). The first part allowed us to
calculate the mass and temperature of cold water and hot
water using calorimeter constant. We did this by following a
simple procedure of heating up the water and then transferring
it into a Styrofoam cup containing cold water. By observing
and recording the temperature changes in both the water (cold
and hot) we were able to determine the calorimeter. This
process can be categorized as Exothermic reaction, because
this releases energy during the process. The calorimeter
constant of the 50.0mL distilled water was determined based
on the data collected from the experiment. Major findings
showed that the calorimeter constant is 35.36 J/°C. For the
heat of solution, calcium chloride and urea was used for the
experiment. The students then recorded the temperature of
the solution. The students also determined the heat of
neutralization between sodium hydroxide and hydrochloric
acid. After recording, the students then observed that the
temperature of the mixture at first became warm that reaches
into 40˚C but when the time goes at 30-45 seconds the
temperature changes and remained at 43˚C until it reaches
the time of 5 minutes.
Keywords: calorimeter, heat, neutralization, temperature
INTRODUCTION
One of the timeliest scientific observations made by humans must have been the
warmth provided by fire, and since that time the unexpected utility of heat as a tool has
contributed to many of the most profound technological advances in humanity’s history.
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The science of heat flow, or more generally energy flow, is called thermodynamics.
Many of the fundamental foundations of thermodynamics have been deliberated in our
lectures on gas laws and the kinetic molecular theory of gases where we learned that
heat and kinetic energy (the energy of motion) of gaseous atoms/molecules are
equivalent. Now we are ready to turn our attention to subtler aspects of thermodynamics
relating to thermal energy being extracted from or stored within a wider array of chemical
compounds than gases. In this set of experiment, we will focus our attention upon one
particular area of thermodynamics, namely calorimetry, is the science or act of calculating
alterations in state variables of a body for the resolution of deriving, the heat transfer
connected with changes of its state due, for example, to chemical reactions, physical
changes, or phase transitions under quantified restrictions. Calorimetry is performed with
a calorimeter (Buchholz 2001). The Calorimeter is the method of measuring the heat of
chemical reactions or physical changes as well as heat capacity.
The container we use as a calorimeter should thermally insulate the matter we’re
interested in studying, it should prevent matter from entering or exiting once our
measurement has begun, and it should allow for easy measurement of temperature
changes.
A coffee cup calorimeter is a constant pressure calorimeter. As such, the heat that
is measured in such a device is equivalent to the change in enthalpy. A coffee cup
calorimeter is normally used for solution based chemistry and as such commonly
encompasses a reaction with diminutive or no volume change. (see Figure 1).
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Here is a coffee cup and its parts:
Source:https://ch301.cm.utexas.edu/section2.php?target=thermo/thermochemistry/coffeecup-calorim.html
Experimentally, measuring heat flows is somewhat difficult. We use thermometer
to measure changes in temperature that can be operated as indirect measures of heat
flow. In order to precisely relate measured temperature changes to the flow of particular
amounts of thermal energy, a proportionality constant that relates these two quantities
called the heat capacity. The Specific Heat Capacity is the amount of heat per unit mass
required to raise the temperature by one degree Celsius.
The connection between heat and temperature change is usually expressed in the
form shown below where c is the specific heat. On the other hand, the Law of
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5
Conservation of Energy states that the whole energy of an insulated system remains
constant, it is said to be preserved over time. This law means that energy can neither be
produced nor destroyed; rather, it can only be converted from one form to another. Most
of the chemists use a technique called calorimetry to regulate the specific heat of a
substance. Calorimetry, or the measurement of heat transfer, is based on the Law of
Conservation of Energy. This law states that energy is not produced nor destroyed; it is
only converted from one form to another. Mathematically, the proportionality can be
expressed as shown in equation 2.1:
qrxn  qsolution  qcalorimeter 
qrxn  (msolutioncwater T )  Ccal T 
ΔΗ solution 
1 kJ
1000 J
molsolute
qrxn 
Where c is the specific heat capacity (in J/(g ˚C)), q is the quantity of thermal energy
released/absorbed (in J), m is the mass of the sample (in g), and ∆T is the change in
temperature (in ˚C), which is taken as Tfinal-Tinitial for the sample. Since specific heat
capacities are positive constants for a given material, heat flow is directly proportional to
changes in temperature. Thus if the change in temperature (∆T) is positive, then q is also
positive indicating the absorption of heat by the sample (an endothermic process).
Alternatively, if ∆T is negative then q is also negative indicating the loss of heat by the
sample (an exothermic process).
The Heat of neutralization is the change in heat that happens when one equivalent of
an acid and one equivalent of a base undergo a neutralization reaction to form water and
salt. It is defined as the energy released with the formation of 1 mole of water.
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qcalorimeter  qHCl  q NaOH  0
qrxn  [Ccal ΔTcal  mHCl cwater ΔTHCl  mNaOH cwater ΔTNaOH ]
The study was conducted at the PHL Building, 4th Floor, Lyceum of the Philippines –
Laguna, April 13, 2018. This study aimed to conduct a successful experiment about
Calorimetry. The specific objectives were:
1. to calculate the heat the neutralization between sodium hydroxide and hydrochloric
acid is endothermic and exothermic.
2. to find out the heat released or changed in a reaction.
3. to learn how to measure Heat flow and,
4. to learn about Exothermic and Endothermic chemical reactions
MATERIALS AND METHODS
In the observation and procedure, calorimeter was used in measuring the heat
effects and calorimetry. The heat neutralization between an acid and a base can be
calibrated through calorimetry where an insulated tumbler serves as the calorimeter. The
tumbler, being a good insulator, is assumed to have negligible contribution to the heat
capacity of the entire calorimeter. Thus, the reaction will only exchange heat with the
water inside the calorimeter.
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Figure 1. Calorimeter
To begin with the determination of calorimeter constant, the calorimeter has been
assembled using the coffee cup. 50.0 mL cold distilled water was placed into the
Styrofoam cup. Using a separate beaker, the distilled water measuring another 50.0 mL
was heated to about 70⁰C. While the water is heating, the temperature of the cold water
in the calorimeter has been monitored for about 2-3 minutes to make sure that it has
become constant and the temperature of the cold water was recorded.
Next, the water in the beaker that has reached 70⁰C was just set aside for about 12 minutes then the temperature of the warm was also recorded. The lid of the calorimeter
was quickly removed and the warm water has been poured into the cold water. The lid
has been replaced immediately and the water was stirred for 30 seconds. The
temperature of the mixed warm and cold water was monitored and recorded the highest
temperature reached by the water.
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Figure 2. Measuring the distilled water with thermometer
The calorimeter constant has been calculated for the calorimeter. Thema ss of the
empty calorimeter together with the warm and cold water were computed as well as the
final and initial temperature of it.
Afterwards, the heat of the solution was computed. The distilled water measuring
50.0 mL was placed in the calorimeter and recorded the temperature. The calcium
chloride weighing 5.00 g was placed into the calorimeter and then the lid was quickly
replaced. The solution was stirred and the temperature has been determined. The
procedure was repeated using the urea instead of calcium chloride. The Change in
Enthalpy of solution in calcium chloride and urea was measure by using the data
gathered and applying its formula.
For the heat of neutralization, the materials that are needed to be used are
Insulated tumbler which is the Styrofoam cup that was given, A 6 M HCl solution and 6 M
NaOH solution, A digital thermometer and a 100 mL graduated cylinder. The estimated
time frame of doing this experiment is around 20 minutes. The first thing that should be
done before anything else is to put all the materials to be used on a safe place.
To start the heat neutralization, the first thing that should be done is to measure
20mL of 6 M NaOH solution using 100 mL graduated cylinder and then after measuring
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the solution, put it into the beaker. After that, measure and record the initial temperature
of the NaOH solution using digital thermometer.
Figure 3. Measuring distilled water with NaOH
Next is Measure 20mL of 6M HCl solution using also the graduated cylinder and
then measure and record also the initial temperature of the 20mL 6M HCl solution using
the digital thermometer just like what we do from the first step.
Now, after measuring the initial temperature of the two solutions, the next
procedure to be done is to pour the HCl solution into the tumbler to mix it with the NaOH
solution. Cover the tumbler immediately and swirl it gently to mix the contents of the two
solution. Then, after mixing the two solutions, ready a timer or a watch to measure its
temperature every 15 seconds for 5 minutes. To do this, slightly lift the cover to insert the
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thermometer. Record the gathered data to the table that is given. Lastly, calculate the
heat of neutralization from the data gathered.
RESULTS AND DISCUSSION
Results showed that styrofoam cups and styrofoam cover were used in the
experiment, because it is a good insulator that prevents heat exchange with the
environment. The students made sure that the cover is in place during the experiment so
that the sytrofoam cover can provide adequate insulation for heat transfer measurements
which enables the accurate determination of the heat involved in chemical processes and
to determine the heat of neutralization and the heat of solution.
Part A: Determination of Calorimeter Constant
Given:
∆𝑇𝑐𝑎𝑙 = 17℃
𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 − 𝑚𝑒𝑚𝑝𝑡𝑦 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟
𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 = 74.92𝑔 − 25.32𝑔
𝑚𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 = 49.6𝑔
∆𝑇𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 = 17℃
𝑚ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟+ℎ𝑜𝑡𝑤𝑎𝑡𝑒𝑟 − 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟
𝑚ℎ𝑜𝑡𝑤𝑎𝑡𝑒𝑟 = 121.92𝑔 − 74.92𝑔
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𝑚ℎ𝑜𝑡𝑤𝑎𝑡𝑒𝑟 = 47𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
∆𝑇ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟 = −21℃
𝑔℃
Solution:
Ccal ΔTcal  mcold cwater ΔTcold water  mhotcwater ΔThot water  0
4.18 𝐽
4.18𝐽
𝐶𝑐𝑎𝑙 (17℃) + (49.6𝑔) (
) (17℃) + (47𝑔) (
) (−21℃) = 0
𝑔℃
𝑔℃
𝐶𝑐𝑎𝑙 (17℃) + (3524.576𝐽) + (−4125.66𝐽) = 0
𝐶𝑐𝑎𝑙 (17℃) − 601.084𝐽 = 0
𝐶𝑐𝑎𝑙 (17℃)
17℃
=
601.084𝐽
17℃
𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
The Calorimeter constant is equal to 35.36 J/°C
Part B: Heat of Solution
Given:
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 + 𝑚𝐶𝑎𝐶𝑙2
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 49.6𝑔 + 5.00𝑔
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 54.6𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
𝑔℃
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = − 82.8 𝑘𝐽/𝑚𝑜𝑙
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∆𝑇 = 𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙
∆𝑇 = 26℃ − 36℃
∆𝑇 = −10℃
𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
Solution:
∆𝐻𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑎𝐶𝑙2
𝑞𝑟𝑥𝑛 = −[𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇) + 𝐶𝑐𝑎𝑙 ∆𝑇]
4.18 𝐽
𝐽
𝑞𝑟𝑥𝑛 = −[(54.6𝑔) (
) (−10℃) + (35.36 )(−10°𝐶)
𝑔℃
°𝐶
𝑞𝑟𝑥𝑛 = −[(−2282.28 𝐽) + (−353.6 𝐽)]
𝑞𝑟𝑥𝑛 = −[−2635.88 𝐽]
𝑞𝑟𝑥𝑛 = 2635.88 𝐽
2635.88𝐽 ×
∆𝐻𝐶𝑎𝐶𝑙2 =
1000𝐽
0.0451𝑚𝑜𝑙
∆𝐻𝐶𝑎𝐶𝑙2 = 58.45
𝑘𝐽
𝑚𝑜𝑙
𝑘𝐽
%𝐸𝑟𝑟𝑜𝑟 = |
1 𝑘𝐽
𝑘𝐽
−82.8 𝑚𝑜𝑙 − 58.45 𝑚𝑜𝑙
𝑘𝐽
−82.8 𝑚𝑜𝑙
%𝐸𝑟𝑟𝑜𝑟 = 1.71%
| × 100%
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The dissolution of calcium chloride is an endothermic process. With the use of
calorimetry formula, we come up with the answer with positive sign. As we all know that if
the answer is positive or if the q, which is the heat, is absorbed then it is an endothermic
process so probably the process that has been found here is an endothermic process.
Given:
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 + 𝑚𝑢𝑟𝑒𝑎
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 49.6𝑔 + 5.01𝑔
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 54.61𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
𝑔℃
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 14.0
𝑘𝐽
𝑚𝑜𝑙
∆𝑇 = 𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙
∆𝑇 = 26℃ − 24℃
∆𝑇 = 2℃
𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
Solution:
∆𝐻𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑈𝑟𝑒𝑎 (𝐶𝐻4 𝑁2 𝑂)
𝑞𝑟𝑥𝑛 = −[𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇) + 𝐶𝑐𝑎𝑙 ∆𝑇]
𝑞𝑟𝑥𝑛 = −[(54.61𝑔) (
4.18 𝐽
𝐽
) (2℃) + (35.36 )(2℃)
𝑔℃
℃
𝑞𝑟𝑥𝑛 = −[(456.5396 𝐽) + (70.72 𝐽)]
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𝑞𝑟𝑥𝑛 = −[527.2596 𝐽]
𝑞𝑟𝑥𝑛 = −527.2596 𝐽
1 𝑘𝐽
∆𝐻𝑢𝑟𝑒𝑎 =
(−527.2596 𝐽) ×
1000 𝐽
0.0834 𝑚𝑜𝑙
∆𝐻𝑢𝑟𝑒𝑎 = −6.32
𝑘𝐽
𝑚𝑜𝑙
𝑘𝐽
𝑘𝐽
14.0 𝑚𝑜𝑙 − (−6.32 𝑚𝑜𝑙)
%𝐸𝑟𝑟𝑜𝑟 = |
𝑘𝐽
14.0 𝑚𝑜𝑙
| × 100%
%𝐸𝑟𝑟𝑜𝑟 = 1.45%
The dissolution of urea is an exothermic process because as we conduct an
experiment, by the use of calorimetry formula that we also used in the dissolution of
calcium chloride, we found out that it releases heat. The answer we got has a negative
sign. We already know that if the answer is negative or the q, which is the heat, is
released or lose then it is an exothermic process so probably the process that has been
found here is an exothermic process.
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15
Part C: Heat of Neutralization
Temperature versus Time
43,5
43
42,5
42
41,5
41
Ряд1
40,5
40
39,5
39
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
270
285
300
38,5
Table 1. Graph of Temperature Versus Time
As shown in the graph above, the highest temperature attained by the mixture
solution in the graph above is 43˚C and the lowest is 40˚C. As observed in the
experiment, the temperature of the mixture reaches 40˚C at first but when the time goes
at 30-45 seconds the temperature increases and stays at 43˚C until it reaches the given
time of 5 minutes. The highest temperature attained by the mixture solution is 42°C. As
observed, the temperature of the mixture solution remained at 42°C from the start until
the end of the five minutes of temperature reading. The loss of energy to the
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experiment’s surrounding will be less in the styrofoam cup because it is a good insulator
that prevents. Error might occur when some heat escape from the cup. Some of the heat
were absorbed by the water & the cup, and also escape when we did not close the cup
correctly.
Given:
𝑚𝐻𝐶𝑙 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+(𝑁𝑎𝑂𝐻+𝐻𝐶𝑙) − 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑁𝑎𝑂𝐻
𝑚𝐻𝐶𝑙 = 127.67𝑔 − 76.81𝑔
𝑚𝐻𝐶𝑙 = 50.86 𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
𝑔℃
𝑚𝑁𝑎𝑂𝐻 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑁𝑎𝑂𝐻 − 𝑚𝑒𝑚𝑝𝑡𝑦 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟
𝑚𝑁𝑎𝑂𝐻 = 76.81𝑔 − 25.32𝑔
𝑚𝑁𝑎𝑂𝐻 = 51.49𝑔
∆𝑇𝐻𝐶𝑙 = 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑇𝑓𝑖𝑛𝑎𝑙
∆𝑇𝐻𝐶𝑙 = 25℃ − 43℃
∆𝑇𝐻𝐶𝑙 = −18℃
∆𝑇𝑁𝑎𝑂𝐻 = 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑇𝑓𝑖𝑛𝑎𝑙
∆𝑇𝑁𝑎𝑂𝐻 = 25℃ − 43℃
∆𝑇𝑁𝑎𝑂𝐻 = −18℃
𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
Theoretical Value:
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∆𝐻𝑟𝑥𝑛 = −55.8
𝑘𝐽
𝑚𝑜𝑙
Solution:
𝑞𝑟𝑥𝑛 = −[𝐶𝑐𝑎𝑙 ∆𝑇𝑐𝑎𝑙 + 𝑚𝐻𝐶𝑙 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇𝐻𝐶𝑙 + 𝑚𝑁𝑎𝑂𝐻 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇𝑁𝑎𝑂𝐻
𝑞𝑟𝑥𝑛 = − [(35.36
𝐽
4.18 𝐽
4.18 𝐽
) (17℃) + (50.86𝑔) (
) (−18℃) + (51.49𝑔) (
) (−18℃)]
℃
𝑔℃
𝑔℃
𝑞𝑟𝑥𝑛 = −[(601.12 𝐽) + (−3826.7064 𝐽) + (−3874.1076 𝐽)]
𝑞𝑟𝑥𝑛 = −[−7099.694 𝐽]
𝑞𝑟𝑥𝑛 = 7099.694 𝐽
𝑚𝑜𝑙𝑁𝑎𝑂𝐻 𝑢𝑠𝑒𝑑 = 𝑚𝑁𝑎𝑂𝐻 ×
1𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝑁𝑎𝑂𝐻 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝑁𝑎𝑂𝐻 𝑢𝑠𝑒𝑑 = 51.49𝑔 ×
1 𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝑁𝑎𝑂𝐻 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝑁𝑎𝑂𝐻 𝑢𝑠𝑒𝑑 = 0.15447 𝑚𝑜𝑙
𝑚𝑜𝑙𝐻𝐶𝑙 𝑢𝑠𝑒𝑑 = 𝑚𝐻𝐶𝑙 ×
1𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝐻𝐶𝑙 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝐻𝐶𝑙 𝑢𝑠𝑒𝑑 = 50.86𝑔 ×
1 𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝐻𝐶𝑙 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝐻𝐶𝑙 𝑢𝑠𝑒𝑑 = 0.15258 𝑚𝑜𝑙
Limiting Reactant = 0.15258 mol of HCl
∆𝐻𝑟𝑥𝑛 =
𝑞𝑟𝑥𝑛
𝑚𝑜𝑙𝑒𝑠𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
1 𝑘𝐽
∆𝐻𝑟𝑥𝑛 =
7099.694 𝐽 × 1000 𝐽
0.15258 𝑚𝑜𝑙
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∆𝐻𝑟𝑥𝑛 = 46.53 𝑘𝐽
% Error 
H theoretical  H exp erimental
H theoretical
𝑘𝐽
%𝐸𝑟𝑟𝑜𝑟 = |
−55.8 𝑚𝑜𝑙 −
−55.8
100%
46.53𝑘𝐽
𝑚𝑜𝑙
𝑘𝐽
| × 100%
𝑚𝑜𝑙
%𝐸𝑟𝑟𝑜𝑟 = 1.83%
The process or reaction is exothermic.
The process for part C is exothermic because the heat inside the calorimeter was
released as it is measured by the thermometer. As we all know that if the heat was
released then it is an exothermic process. In our experiment, the heat was released as
the temperature goes up in the thermometer. As the heat given off in an exothermic
process will be the same amount of heat gained by the calorimeter. Same as in our
experiment, as time goes on, the temperature stays the same.
All measurements have some degree of uncertainty that may come from a variety of
sources. When we make a measurement, we generally assume that some exact or true
value exists based on how we define what is being measured. While we may never know
this true value exactly, we attempt to find this ideal quantity to the best of our ability with
the time and resources available. Therefore, all experimental results are wrong. As we do
the experiment, during the calibration of calorimeter, the calorimeter must be closed
immediately upon mixing the solution to the water inside it. This is part can be considered
as a source of error especially if the calorimeter is not immediately closed. When this
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19
happen, some heat may have lost to the air, making the heat gained by the calorimeter
and the water lower than the ideal. This may lead to a wrong record of temperature. Also,
the synchronization between starting the time to record data and closing the calorimeter
can also cause error. Ideally, temperature recording must start immediately after mixing
the sample in the calorimeter, then closing it. But in reality, this procedure is very difficult
to do perfectly.
The process or reaction is endothermic.
As seen in the computation above, the computed value of the enthalpy of the of the
reaction is 50.10 kJ. This positive value of the enthalpy of the solution means that the
solution absorbed heat.
In experiments, errors were unavoidable. The possible sources of error were human
error,
improper
calibrations,
measurement
estimation,
andmeasurement
device
limitations. These sources may lead to an inaccurate value from what is expected in the
report.
SUMMARY AND CONCLUSION
The conclusion of this experiment is; materials really affect the experiment. On our
experiment, we did some mistakes like; didn’t close the cup properly and left a small hole
(on the top of the cup where we put the thermometer) and let some heat released. The
use of Styrofoam also affects the experiment. If we use other material like beaker, which
takes less energy to lose heat (lets the heat escape easier). The heat on the glass beaker
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will flow from the water, through the glass, and to our hand. Using paper cup is also not
recommended, because it conducts heat. It will affect the calculation and will give us
different value. To have the best accurate result, we have to make sure that all the
materials are in the best condition. Make sure we have sealed the hole accurately so
there’s no way heat will escape. Just in case, cover the Styrofoam cup with another cup
for safe in case it’s leaking or spilled
If you’re going to have an experiment like this make sure that the cover is unlit so
the heat cannot get out and to measure the exact temperature in order to have a
successful experiment.
For future researchers, mistakes may occur when heat escapes from the cup.
Some of the heat was absorbed by the water and the cup, and some heat escaped when
we didn't closed the cup in the proper way; We accidentally left the hole that we made to
put the thermometer loose and opened, as a result, the heat escaped. This experimental
mistake is commonly caused by incorrect measurements of the chemicals which results
to wrong outcome. The difference between the measurements of the chemicals or
mixtures will result to the wrong measurements between heat capacities after all the
calculations were made. Make sure there will no heat will escape from the cup or else, it
will come out wrong and failed
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1
LITERATURE CITED
Brooks-Cole, (2003) Adapted from Hall, J.F. Experimental Chemistry, 6th Ed.;
Houghton-Mifflin: New York, 2003
McGraw-Hill:Boston, (2007) Standard and Microscale Experiments in General
Chemistry, 5th Ed Adapted from Bishop, Bishop, and Whitten Standard and
Microscale Experiments in General Chemistry, 5th Ed. p 187.
Silberberg, M.S. n.d. Principles of General Chemistry
APPENDICES
Part A: Determination of the Calorimeter Constant
Empty calorimeter
Calorimeter + cold water
Calorimeter + cold water + hot water
Mass (in g)
25.32g
74.92g
121.92g
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Total mass of water in the calorimeter
2
46.6g
Initial
26 °C
64 °C
Temperature (cold water)
Temperature (warm water)
Final
43°C
Part B: Heat of Solution
Mass (in grams)
Calcium chloride
25.32g
74.92g
49.6g
5.00g
Empty calorimeter
Calorimeter + cold water
Cold water
Sample solid used
urea
25.32g
74.92g
49.6g
5.01g
Temperature °C
Initial (water only)
26℃
26℃
Final (solution)
35℃
24℃
T
9℃
-2℃
Part C: Data and Observations
Time After Mixing
(sec)
Temperature After
Mixing (℃)
15
40℃
30
43℃
45
43℃
60
43℃
75
43℃
90
43℃
105
43℃
120
43℃
135
43℃
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150
2
43℃
Time After Mixing
(sec)
Temperature After
Mixing (℃)
165
43℃
180
43℃
195
43℃
210
43℃
225
43℃
240
43℃
255
43℃
270
43℃
285
43℃
300
43℃
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Computations
Part A: Determination of Calorimeter Constant
Given:
∆𝑇𝑐𝑎𝑙 = 17℃
𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 − 𝑚𝑒𝑚𝑝𝑡𝑦 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟
𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 = 74.92𝑔 − 25.32𝑔
𝑚𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 = 49.6𝑔
∆𝑇𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟 = 17℃
𝑚ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟+ℎ𝑜𝑡𝑤𝑎𝑡𝑒𝑟 − 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑐𝑜𝑙𝑑 𝑤𝑎𝑡𝑒𝑟
𝑚ℎ𝑜𝑡𝑤𝑎𝑡𝑒𝑟 = 121.92𝑔 − 74.92𝑔
𝑚ℎ𝑜𝑡𝑤𝑎𝑡𝑒𝑟 = 47𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
∆𝑇ℎ𝑜𝑡 𝑤𝑎𝑡𝑒𝑟 = −21℃
𝑔℃
Solution:
Ccal ΔTcal  mcold cwater ΔTcold water  mhotcwater ΔThot water  0
4.18 𝐽
4.18𝐽
𝐶𝑐𝑎𝑙 (17℃) + (49.6𝑔) (
) (17℃) + (47𝑔) (
) (−21℃) = 0
𝑔℃
𝑔℃
𝐶𝑐𝑎𝑙 (17℃) + (3524.576𝐽) + (−4125.66𝐽) = 0
𝐶𝑐𝑎𝑙 (17℃) − 601.084𝐽 = 0
𝐶𝑐𝑎𝑙 (17℃) 601.084𝐽
=
17℃
17℃
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𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
The Calorimeter constant is equal to 35.36 J/°C
Part B: Heat of Solution
Given:
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 + 𝑚𝐶𝑎𝐶𝑙2
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 49.6𝑔 + 5.00𝑔
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 54.6𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
𝑔℃
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = − 82.8 𝑘𝐽/𝑚𝑜𝑙
∆𝑇 = 𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙
∆𝑇 = 26℃ − 36℃
∆𝑇 = −10℃
𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
Solution:
∆𝐻𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝐶𝑎𝐶𝑙2
𝑞𝑟𝑥𝑛 = −[𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇) + 𝐶𝑐𝑎𝑙 ∆𝑇]
4.18 𝐽
𝐽
𝑞𝑟𝑥𝑛 = −[(54.6𝑔) (
) (−10℃) + (35.36 )(−10°𝐶)
𝑔℃
°𝐶
𝑞𝑟𝑥𝑛 = −[(−2282.28 𝐽) + (−353.6 𝐽)]
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𝑞𝑟𝑥𝑛 = −[−2635.88 𝐽]
𝑞𝑟𝑥𝑛 = 2635.88 𝐽
2635.88𝐽 ×
∆𝐻𝐶𝑎𝐶𝑙2 =
1 𝑘𝐽
1000𝐽
0.0451𝑚𝑜𝑙
∆𝐻𝐶𝑎𝐶𝑙2 = 58.45
𝑘𝐽
𝑚𝑜𝑙
𝑘𝐽
%𝐸𝑟𝑟𝑜𝑟 = |
𝑘𝐽
−82.8 𝑚𝑜𝑙 − 58.45 𝑚𝑜𝑙
𝑘𝐽
−82.8 𝑚𝑜𝑙
| × 100%
%𝐸𝑟𝑟𝑜𝑟 = 1.71%
Given:
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 𝑚𝑐𝑜𝑙𝑑𝑤𝑎𝑡𝑒𝑟 + 𝑚𝑢𝑟𝑒𝑎
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 49.6𝑔 + 5.01𝑔
𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 = 54.61𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
𝑔℃
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑉𝑎𝑙𝑢𝑒 = 14.0
∆𝑇 = 𝑇𝑓𝑖𝑛𝑎𝑙 − 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙
∆𝑇 = 26℃ − 24℃
𝑘𝐽
𝑚𝑜𝑙
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∆𝑇 = 2℃
𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
Solution:
∆𝐻𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑈𝑟𝑒𝑎 (𝐶𝐻4 𝑁2 𝑂)
𝑞𝑟𝑥𝑛 = −[𝑚𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇) + 𝐶𝑐𝑎𝑙 ∆𝑇]
𝑞𝑟𝑥𝑛 = −[(54.61𝑔) (
4.18 𝐽
𝐽
) (2℃) + (35.36 )(2℃)
𝑔℃
℃
𝑞𝑟𝑥𝑛 = −[(456.5396 𝐽) + (70.72 𝐽)]
𝑞𝑟𝑥𝑛 = −[527.2596 𝐽]
𝑞𝑟𝑥𝑛 = −527.2596 𝐽
1 𝑘𝐽
∆𝐻𝑢𝑟𝑒𝑎 =
(−527.2596 𝐽) ×
1000 𝐽
0.0834 𝑚𝑜𝑙
∆𝐻𝑢𝑟𝑒𝑎 = −6.32
𝑘𝐽
𝑚𝑜𝑙
𝑘𝐽
𝑘𝐽
%𝐸𝑟𝑟𝑜𝑟 = |
14.0 𝑚𝑜𝑙 − (−6.32 𝑚𝑜𝑙)
𝑘𝐽
14.0 𝑚𝑜𝑙
%𝐸𝑟𝑟𝑜𝑟 = 1.45%
| × 100
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Part C: Heat of Neutralization
Given:
𝑚𝐻𝐶𝑙 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+(𝑁𝑎𝑂𝐻+𝐻𝐶𝑙) − 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑁𝑎𝑂𝐻
𝑚𝐻𝐶𝑙 = 127.67𝑔 − 76.81𝑔
𝑚𝐻𝐶𝑙 = 50.86 𝑔
𝑐𝑤𝑎𝑡𝑒𝑟 =
4.18 𝐽
𝑔℃
𝑚𝑁𝑎𝑂𝐻 = 𝑚𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟+𝑁𝑎𝑂𝐻 − 𝑚𝑒𝑚𝑝𝑡𝑦 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟
𝑚𝑁𝑎𝑂𝐻 = 76.81𝑔 − 25.32𝑔
𝑚𝑁𝑎𝑂𝐻 = 51.49𝑔
∆𝑇𝐻𝐶𝑙 = 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑇𝑓𝑖𝑛𝑎𝑙
∆𝑇𝐻𝐶𝑙 = 25℃ − 43℃
∆𝑇𝐻𝐶𝑙 = −18℃
∆𝑇𝑁𝑎𝑂𝐻 = 𝑇𝑖𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑇𝑓𝑖𝑛𝑎𝑙
∆𝑇𝑁𝑎𝑂𝐻 = 25℃ − 43℃
∆𝑇𝑁𝑎𝑂𝐻 = −18℃
𝐶𝑐𝑎𝑙 = 35.36 𝐽/℃
Theoretical Value:
∆𝐻𝑟𝑥𝑛 = −55.8
𝑘𝐽
𝑚𝑜𝑙
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Solution:
𝑞𝑟𝑥𝑛 = −[𝐶𝑐𝑎𝑙 ∆𝑇𝑐𝑎𝑙 + 𝑚𝐻𝐶𝑙 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇𝐻𝐶𝑙 + 𝑚𝑁𝑎𝑂𝐻 𝑐𝑤𝑎𝑡𝑒𝑟 ∆𝑇𝑁𝑎𝑂𝐻
𝑞𝑟𝑥𝑛 = − [(35.36
𝐽
4.18 𝐽
4.18 𝐽
) (17℃) + (50.86𝑔) (
) (−18℃) + (51.49𝑔) (
) (−18℃)]
℃
𝑔℃
𝑔℃
𝑞𝑟𝑥𝑛 = −[(601.12 𝐽) + (−3826.7064 𝐽) + (−3874.1076 𝐽)]
𝑞𝑟𝑥𝑛 = −[−7099.694 𝐽]
𝑞𝑟𝑥𝑛 = 7099.694 𝐽
𝑚𝑜𝑙𝑁𝑎𝑂𝐻 𝑢𝑠𝑒𝑑 = 𝑚𝑁𝑎𝑂𝐻 ×
1𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝑁𝑎𝑂𝐻 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝑁𝑎𝑂𝐻 𝑢𝑠𝑒𝑑 = 51.49𝑔 ×
1 𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝑁𝑎𝑂𝐻 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝑁𝑎𝑂𝐻 𝑢𝑠𝑒𝑑 = 0.15447 𝑚𝑜𝑙
𝑚𝑜𝑙𝐻𝐶𝑙 𝑢𝑠𝑒𝑑 = 𝑚𝐻𝐶𝑙 ×
1𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝐻𝐶𝑙 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝐻𝐶𝑙 𝑢𝑠𝑒𝑑 = 50.86𝑔 ×
1 𝑚𝐿
1𝐿
3.0 𝑚𝑜𝑙
×
×
1𝑔 𝐻𝐶𝑙 1000𝑚𝐿
1𝐿
𝑚𝑜𝑙𝐻𝐶𝑙 𝑢𝑠𝑒𝑑 = 0.15258 𝑚𝑜𝑙
Limiting Reactant = 0.15258 mol of HCl
∆𝐻𝑟𝑥𝑛 =
𝑞𝑟𝑥𝑛
𝑚𝑜𝑙𝑒𝑠𝑙𝑖𝑚𝑖𝑡𝑖𝑛𝑔𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡
1 𝑘𝐽
∆𝐻𝑟𝑥𝑛 =
7099.694 𝐽 × 1000 𝐽
0.15258 𝑚𝑜𝑙
∆𝐻𝑟𝑥𝑛 = 46.53 𝑘𝐽
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6
Lyceum of the Philippines
University - Laguna
International School
% Error 
H theoretical  H exp erimental
H theoretical
𝑘𝐽
%𝐸𝑟𝑟𝑜𝑟 = |
−55.8 𝑚𝑜𝑙 −
100%
46.53𝑘𝐽
𝑚𝑜𝑙
𝑘𝐽
−55.8 𝑚𝑜𝑙
| × 100%
%𝐸𝑟𝑟𝑜𝑟 = 1.83%
Figure 1. Coffee cup calorimeter without the thermometer
Figure 2. Measuring of 50.0 mL of distilled water using the graduated cylinder
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7
Lyceum of the Philippines
University - Laguna
International School
Figure 3. Measuring distilled water with NaOH
Page
8
Lyceum of the Philippines
University - Laguna
Page
International School
9
CONTRIBUTION TO THE SCIENTIFIC PAPER
Member
Contribution
Del Valle, Alyssa L.
Materials and Methods
Manuel, Robin Piolo M.
Recommendation
Motas, Hannah Faye A.
Results and Discussion
Pagaduan, Mark Gerald N.
Abstarct and Introduction
Ricablanca, Princess Jane D.
Santos, Claude Jayson U.
Results and Discussion;
Abstract
Introduction
Seleteria, Richel B.
Materials and Methods
Torres, Khim Harold
Summary and Conclusion
.
Signature