1
Chemistry for Engineers (CHEM 114)
PRELIM: Weeks 1 to 3
MODULE 1
Heat and Energy: Thermochemistry
Course Learning
Outcomes
• CLO 1. Discuss
the application of
chemistry in
relation to the
generation of heat
and energy.
•CLO 3. Identify key
chemistry
concepts related
to the specific field
of engineering.
•CLO 4. Learn and
apply concepts in
engineering
materials.
Module Learning
Outcomes
•Recognize a system
as separate from the
universe and its
surroundings
•Explain the energy
flow between a
system and its
surroundings, and
associate it by
solving problems
involving the flow of
energy.
•Apply previous
knowledge on the
flow of energy in
solving calorimetric
problems as an
exchange of heat
energy.
CONTENTS of this MODULE
Topics
Page
Energy Transformation and Conservation
2
Flow of Energy: Heat and Work
3
Heat Quantity Units
5
Exercise Number 1
6
Calorimetry: Sensible Heat and Specific Heat
7
Sensible Heat
7
Specific heat
8
Method of Mixtures
12
Exercise Number 2
14
Exercise Number 3
15
Calorimetry: Phase Changes
17
Latent Heat
17
Exercise Number 4
21
Topic Learning
Outcomes:
•TLO 1. Describe
the energy flow
between a system
and its
surroundings.
•TLO 2. Solve
problems
involving the flow
of energy.
•TLO 3. Define
calorimetry and
solve problems on
exchange of heat.
Time Frame (as applicable)
1 hour
15 minutes
30 minutes
45 minutes
Please put your signature over printed name in the Student’s Honesty Clause at the bottom of each exercise.
Keywords: energy, system, heat, work, calorimetry, sensible heat, specific heat, latent heat
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
2
Energy Transformation and Conservation of Energy
Refer to the figure below:
Universe
Surroundings
System
boundary
Figure 1.1 A system and the universe
In studying energy, we must separate a specific area to be studied. A system is that part
of the universe that is being considered for studying energy flows. Outside the system, is its
surroundings. Beyond the surroundings, is the universe. The system and the surroundings is
separated by a boundary.
The reason why we should consider a system, is because we have to maintain obedience
to the Law of Conservation of Energy: Energy can neither be created nor destroyed, only
transformed, i.e., the energy of the universe is constant.
This is expressed by the equation:
∆𝑬 𝐮𝐧𝐢𝐯𝐞𝐫𝐬𝐞 = ∆𝑬𝒔𝒖𝒓𝒓𝒐𝒖𝒏𝒅𝒊𝒏𝒈𝒔 + ∆𝑬 𝒔𝒚𝒔𝒕𝒆𝒎 = 𝟎
The equation above expresses the First Law of Thermodynamics.
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
3
Some forms of Energy
Energy is the ability to do work. It takes many forms among which are:
o
o
o
o
Potential Energy is energy at rest.
Kinetic Energy is energy in motion;
Electrical energy is the energy from the movement of charges;
Heat energy or thermal energy is the energy that is transferred through
differences in temperature (sensible heat) or is released or absorbed due to a
change in phase (latent heat).
(Your lecturer will provide the equations and units for each of these forms of energy).
Flow of Energy: Heat and Work
Based from Figure 1.1, energy flowing into a system should be equal to energy that flows out of
a system, to obey the law of conservation of energy.
Surroundings
Energy
or Heat
in
System
Energy out
or Work
Done
boundary
In equation form, this applies that energy is expressed as:
∆𝑬 = 𝒒 + 𝒘
Where:
ΔE
q
w
= Energy transferred in a system
Units:
= heat added to (or removed from) a system Joule, J or kiloJoule, KJ
= work done by (or done to) a system
BTU (British Thermal Units
Calorie (cal)
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
4
Take note of the sign conventions:
q is POSITIVE (+) when ADDED TO a system; NEGATIVE (-) when
REMOVED FROM or RELEASED BY a system.
w is POSITIVE (+) when DONE TO a system; NEGATIVE (-) when
DONE BY a system.
Heat Quantity Units
Calorie, cal - refers to the amount of heat required to raise the temperature of
1 gram of water by 1 C0.
British Thermal Unit, BTU - is defined as the amount of heat required to raise the
temperature of 1 pound of water by 1 F0
Conversion Units for Heat:
1 calorie, cal
1 kilocalorie (kcal)
= 4.186 Joules, J
= 1000 cal
= 4,186 J
= 4.186 kiloJoules, KJ
1 British Thermal Unit, BTU
= 778 ft.lb
= 252 cal
= 1055 J
*The calorie is not a fundamental SI unit. The International Committee on Weights and
Measures recommends using Joule as the basic unit of energy in all forms.
Examples:
1. Convert 2000 calories to Joules
2000 calories
4.186 J
1 cal
=
8, 372 J
2. Convert 300 BTU to Joules
300 BTU
1055 J
1 BTU
= 316, 500 J
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
5
Examples: (to be solved by your lecturer)
1. If 500 J of heat is added to a gas system and it did 215 J of work, how much is its change in
energy? Convert your answer to BTU, then to calories.
2. If a machine does 3.8 x 103 KJ of work after an input of 4.6 x 104 KJ of heat from fuel, how
much is its change in internal energy? Convert your answer to BTU, then to calories.
3. Calculate the heat in a system if it does 27 J of work, and its energy decreases by 40 J.
Answer Exercise No.1 to be uploaded by your faculty-in-charge later.
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
6
PART 2
CALORIMETRY: Heat Transfer, Quantity of Heat (Sensible Heat),
Specific Heat
Thermal Equilibrium - The state where two bodies have the same temperature.
An example of this state is when a hot metal shot is placed in a glass containing water at
room temperature. The heat of the metal shot will be transferred to the water until such time
that there will be a constant temperature reading. The temperature achieved is the
equilibrium temperature, Te, also called mixing temperature, Tm.
Temperature - A physical property that depends on the physical state of a material
and is a quantitative description of its hotness or coldness.
Heat - refers to energy in transit (energy transferred) from one body to another
because of a temperature gradient (temperature difference). It is not the amount of
energy contained within a system.
Sir James Prescott Joule (1818 – 1889)
He studied how water is warmed by stirring the water with a paddle wheel (refer to figure
1.2). The paddle wheel adds energy to the water by doing work on it. He found out that the
temperature rise is directly proportional to the amount of work done on the water.
∆T α W
It is for this reason that the energy transferred
as heat is also the amount of energy consumed as work.
Take note that the units of heat, energy and work are
the same. The basic unit of energy, the Joule, is named
after him.
Figure 1.2: Joule’s paddle wheel
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
7
Quantity of Heat as Sensible Heat, Q
Heat transfer or heat flow takes place solely because of a temperature difference. Heat
flows occurs between bodies when their temperatures are different, from a body with the higher
temperature going towards the body with a lower temperature.
The quantity of heat, Q required to increase the temperature of a mass m of a certain
material is proportional to its temperature change, ∆T. It is also dependent on the nature of the
material. Hence;
Q = mc∆T
Where:
Q
m
c
∆T
= quantity of heat transferred or absorbed
= mass of substance
= specific heat
= temperature difference or temperature gradient
Sign Conventions:
+ Q = add heat
- Q = release or remove heat
Temperature Change, ∆T (also called temperature difference, or temperature gradient).
This refers to the difference between two temperatures, usually final temperature, (Tf)
minus the initial temperature (Ti) or higher temperature (TH) minus the lower temperature (TL).
A temperature change can be calculated using the following formulas:
∆T = Tf - Ti
∆T = TH – TL
Note: change in
temperature is
denoted by the
degree symbol
AFTER the unit
symbol.
Problem-Solving Tips:
✓ For a decrease in temperature, ∆T is negative.
✓ For an increase in temperature, ∆T is positive.
✓ 1 C0 change in temperature is equal to 1 K change.
✓ 1 C0 change in temperature is equal to 1.8 F0 change.
✓
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
8
Specific Heat, c
Specific heat is the amount of heat required to raise a unit mass of a substance by
one degree of temperature. It is dependent on the nature of material.
The specific heat, c, of water is: 4190 J/kg K = 4.19 KJ/kg K = 1 cal/g C0
= 1 BTU/lb F0
TABLE 1.1
Approximate Specific and Molar Heat Capacities at Constant Pressure
Substance
Specific Heat
Molecular
Molar Heat
Capacity, c
Mass, M
Capacity, C
(J/ kg K)
(kg/mol)
(J/mol K)
Aluminum
910
0.0270
24.6
Beryllium
1970
0.00901
17.7
Copper
390
0.0635
24.8
Ethanol
2428
0.0460
112.0
Ethylene glycol
2386
0.0620
148.0
Ice
2000
0.0180
36.5
Iron
470
0.0559
26.3
Lead
130
0.207
26.9
Marble (CaCO3)
879
0.100
87.9
Mercury
138
0.201
27.7
Salt (NaCl)
879
0.0585
51.4
Silver
234
0.108
25.3
Water (liquid)
4190
0.0180
75.4
Source: University Physics 9th edition (p. 473) 1996 by H.D. Young and R.A. Freeman.
Addison-Wesley Publishing Company, Inc. Massachusetts, USA.
Problems:
Example 1.
An ice cube tray of negligible mass contains 0.450 kg of water at 220C. How
much heat must be removed to cool the water to 00C without freezing?
Solution:
Q = mc ΔT
= (0.450 kg)(4190 J/ kg C0)( 00C - 220C)
= - 41,481 J
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
9
Example 2.
How much heat is required to raise the temperature of 30 kg of silver
from 250C to 900C?
From Table 1.1 : Csilver = 234 J/kg C0
Solution:
Q = mcΔT
= (30 kg)( 234 J/kg C0)( 900C - 250C)
= 456,300 J
Example 3.
How much heat must be removed to cool 8.5 kg of mercury from 1000C
to 450C?
From Table 1.1: c mercury = 138 J/ kg C0
Solution:
Q = mc ΔT
= (8.5 kg)(138 J/ kg C0)(450C - 1000C)
= - 64,515 J
Conceptual question: A cold block of metal feels colder than a cold block of wood at the
same temperature. Likewise, a hot block of metal feels hotter than a hot block of wood at the
same temperature. Why?
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
10
Method of Mixtures
In heat transfer, the hotter object loses heat while the cooler object absorbs the heat
released to reach thermal equilibrium (thermodynamic equilibrium). The final temperature of the
resulting mixture of these two objects is the equilibrium temperature, Tm (or Te). If the masses
of the objects, the initial and equilibrium temperatures are measured, the specific heat of the
materials can be calculated using the Method of Mixtures.
Q lost =Q gained
mcΔTH = mcΔTL
Tip: Te or Tm should
be between TH and
TL
Figure 1.3
Principle of Heat Transfer to
achieve thermal equilibrium
(thermodynamic equilibrium)
Problems:
Example 1.
Steam Burns vs. Water Burns.
What is the amount of heat input to your skin when it receives the heat released by an engine
boiler? Which of them will cause a more severe burn?
a)
20 g of steam initially at 1000C when it is cooled to 340C.
(c of steam= 2,260,440 J/kg C0)
Q = mc ΔT
= (0.02 kg)( 2,260,440 J/ kg C0)(340C - 1000C)
= - 2, 983,780.8 J
The answer is negative because released by the engine boiler.
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
11
b)
20 g of water initially at 1000C when it is cooled to 340C.
Q = mc ΔT
= (0.02 kg)( 4190 J/ kg C0)(340C - 1000C)
= - 5,530.8 J
A more severe burn will be caused by steam.
Example 2.
An engineer is working on a new engine design. One of the moving parts contains
1.40 kg of aluminium and 0.50 kg of iron, and is designed to operate at 1500C. How much heat is
required to raise its temperature from 200C to 1500C?
Q total = Q aluminium + Q iron
= mcΔT aluminium + mcΔT iron
From Table 1.1 : caluminum = 910 J/kg C0
ciron = 470 J/kg C0
Q total = (1.40 kg)(910 J/kg C0)( 1500C - 200C) +
(0.50 kg)( 470 J/kg C0)( 1500C - 200C)
Q total = 196,170 J
Example 3.
A technician measures the specific heat capacity of an unidentified liquid by immersing
an electrical resistor in it. Electrical energy is converted to heat, and is transferred to the liquid
for 120 seconds at a constant rate of 65 W. The mass of the liquid is 0.780 kg, and its temperature
rises from 18.550C to 21.320C. Find the average specific heat of the liquid at this temperature
range.
Solution:
Step 1: Power, P = 65 W = 65 J/s
time, t = 120 seconds
Recall in Physics 1: P = work, W = energy, E
t
t
E = Pt
= (65 J/s)( 120 seconds)
= 7,800 J = Q (amount of heat)
Step 2: Q = mcΔT
c = __Q__
mΔT
= ________7,800 J ________
(0.780 kg)( 21.320C - 18.550C)
= 3,610.11 J/kg C0 or K
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
12
Example 4.
A 0.0356 kg aluminum calorimeter contains 0.200 kg of water at a temperature of
24 C. Lead shots with a mass of 0.241 kg at 1290C were dropped into the water. Find the
equilibrium temperature of the mixture.
0
caluminum = 910 J/kg C0
cwater = 4190 J/kg C0
clead = 130 J/kg C0
Q lost =Q gained
mcΔTH = mcΔTL
mc ΔT lead = mc ΔTaluminum + mc ΔTwater
(0. 241 kg)( 130 J/kg C0)( 1290C – Te) = (0.0356 kg)( 910 J/kg C0)(Te – 240C) +
(0. 200kg)( 4190 J/kg C0)(Te – 240C)
Te = 27.650C
Answer Exercises Nos. 2 & 3 to be uploaded by your faculty-in-charge later.
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
13
PART 3: CALORIMETRY: PHASE CHANGES (Latent Heat)
Refer to the figure below:
Freezing
Evaporation
Condensation
Melting
Sublimation
Deposition
Figure 1.4: Phase Changes (Changes of State)
For any given pressure, a phase change takes place at a definite temperature,
usually accompanied by absorption or emission of heat, and a change in volume
and density.
PHASE CHANGES
Latent Heat of Fusion, Lf
The heat required per unit mass to melt (or freeze) a given material.
Q = mLf
To melt 1 kg of ice at 00 C to 1 kg of liquid water at 00C and normal atmospheric pressure,
requires an absorption of a latent heat of fusion of 3.34 x 105 Joules, or 334,000 J. The process
is reversible, so to freeze water, heat is removed, but the magnitude is the same.
 For any given material at any given pressure, the freezing temperature is the same as
the melting temperature.
 The latent heat of fusion, Lf of water at normal atmospheric pressure is:
3.34 x 105 J/kg = 334,000 J/kg = 80 cal/g = 143 BTU/lb
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
14
Latent Heat of Vaporization, Lv
The heat required per unit mass to evaporate (or condense) a given material.
Q = mLv
 For any given material at any given pressure, the boiling temperature is the same as the
condensation temperature.
 The latent heat of vaporization, Lv of water at normal atmospheric pressure is:
2.256 x 106 J/kg = 2,256,000 J/kg = 540 cal/g = 970 BTU/lb
To vaporize 1 kg of water at 1000 C to 1 kg of water vapour or steam at 1000C and normal
atmospheric pressure, requires an absorption of a latent heat of vaporization of 2.256 x 106
Joules. The process is reversible, so to condense the steam or water vapor, heat is removed,
but the magnitude is the same.
Sign Conventions:
Positive (+)
= absorption of heat
Negative (-)
= release of heat
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
15
Quantity of Heat with Phase Changes and Temperature Changes, Q
Phase change caused by the release or absorption of heat involves the masses of the
materials, and the latent heat of fusion, Lf and latent heat of vaporization, Lv. The process also
involves the transfer of heat across a temperature gradient (temperature difference) between
the hotter and colder objects (refer to figure 1.5).
Q gained + Q change of phase = Q lost + Q change of phase
mcΔTL + mLf (or Lv) = mcΔTH+ mLv(or Lf)
Figure 1.5 Concept diagram of Specific Heat and Latent Heat for water
During change of state in a substance, for example, water becomes ice, its internal energy
changes, so the kinetic energy of its particles changes. As it is changing from one state to another,
the change in energy is reflected in the bonds between the particles, and therefore the
temperature of the object doesn’t change. These bonds either break or bind to melt (or
vaporize) or solidify (or condense) the substance. Once the change of state is complete, however,
changes in energy are again observed in the form of changes in temperature.
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
16
Substance
Helium
Hydrogen
Nitrogen
Oxygen
Ethanol
Mercury
Water
Sulfur
Lead
Antimony
Silver
Gold
Copper
TABLE 1.2
Heats of Fusion and Vaporization
Normal Melting Heat of Fusion, Normal Boiling
Point
Lf (in J/ kg)
Point
K
*
13.84
0
C
*
259.31
63.18
209.97
54.36
218.79
159
-114
234
-39
273.15
0
392
119
600.5
327.3
903.65 630.50
1233.95 960.80
1336.15 1063
1356
1083
Heat of
Vaporization,
Lv (in J/ kg)
0
*
58.6 x 103
K
4.216
20.26
C
-268.93
-252.89
20.9 x 103
452 x 103
25.5 x 103
77.34
-195.8
201 x 103
13.8 x 103
90.18
-183.0
213 x 103
x 103
x 103
x 103
x 103
x 103
x 103
x 103
x 103
x 103
351
630
373.15
717.75
2023
1713
2466
2933
1460
78
357
100
444.60
1750
1440
2193
2660
1187
854
272
2256
326
871
561
2336
1578
5069
104.2
11.8
334
38.1
24.5
165
88.3
64.5
134
x 103
x 103
x 103
x 103
x 103
x 103
x 103
x 103
x 103
Source: University Physics 9th edition (p. 475)1996 by H.D. Young and R.A. Freeman.
Addison-Wesley Publishing Company, Inc. Massachusetts, USA.
•
A pressure in excess of 25 atmospheres (25 atm) is required to make helium solidify.
At 1 atmosphere pressure, helium remains a liquid down to absolute zero (-273 K)
Problems:
Example 1. How much heat is required to convert 8 grams of ice at -150C to steam
at 1000C?
Given: cice below 00C = 2000 J/kg K
Q total = Q gained + Q change of phase
= mcΔT
+ mLf
+
mcΔT
Ice at -150C to
ice at 00C
Melting of
ice at 00C
Melted ice heated
to water at 1000C
+ mLv
water at 1000C
to steam at 1000C
Q total = (0. 008 kg)( 2000 J/kg K)(00C - - 150C) + (0. 008 kg )(3.34 x 105 J/kg)
+ (0. 008 kg)( 4190 J/kg K)(1000C – 00C) + (0. 008 kg )( 2.256 x 106 J/kg)
Q total = 24,312 J
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
17
Example 2.
A calorimeter (vessel whose walls is thermally insulated) contains 2.10 kg of water and
0.250 kg of ice, all at a temperature of 00C. The outlet of a tube coming from a boiler in which
water is boiling at atmospheric pressure is inserted into the water. How many grams of steam
must condense inside the vessel to raise the temperature of the system to 340C? Neglect the
heat transferred to the container.
Figure 1.6
Concept of a
calorimeter
containing water
and ice.
TH = 1000C
Tube from
a boiler
m steam = ?
TL = 00C
Solution:
Q gained + Q change of phase
Te = 340C
2.10 kg water
+ 0.250 kg ice
=
Q lost + Q change of phase
Water and ice
mcΔT
water
+
mLf
Melting of
ice at 00C
steam
+
mcΔT
Melted ice to
water at Te = 340C
=
mLv
steam at
1000C
+
mcΔT
steam to
water at
Te = 340C
(2.10 kg)( 4190 J/kg K)(340C - 00C) + (0.250 kg )( 3.34 x 105 J/kg) + (0.250 kg)( 4190 J/kg
K)(340C - 00C)
= (m steam)(2.256 x 106 J/kg)+ (m steam) ( 4190 J/kg K)(1000C – 340C)
By factoring:
(2.10 kg)( 4190 J/kg K)(340C - 00C) + (0.250 kg )( 3.34 x 105 J/kg)
= (m steam) [(2.256 x 106 J/kg)+ ( 4190 J/kg K)(1000C – 340C) ]
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
18
m steam = (2.10 kg)( 4190 J/kg K)(340C - 00C) + (0.250 kg )( 3.34 x 105 J/kg)+ (0.250 kg)(
4190 J/kg K)(340C - 00C)
[(2.256 x 106 J/kg)+ ( 4190 J/kg K)(1000C – 340C) ]
m steam = 0.165163 kg ≈ 165.16 grams of steam
Answer Exercises No. 4 to be uploaded by your faculty-in-charge later.
SUMMARY:
In studying the flow of energy, the system must be identified first as separate from
its surroundings and the universe. Energy is the ability to do work, in the same way that
work is the use of energy. Hence, the change of energy is the sum (or difference) of these
two.
Calorimetry demonstrates the exchange of energy in two ways: as sensible heat and
as latent heat. As sensible heat, the temperature difference (temperature gradient)
between hot and cold substances, and their specific heats are considered. In latent heat,
the amount of heat absorbed (or released) by substances to change their state or phase,
are considered. In most instances, sensible heat and latent heat occur together in
calorimetry.
References:
Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
Potter, Merle and Somerson, Craig (2006). Schaum’s Outline of Thermodynamics for Engineers.
2nd Edition. McGraw-Hill Companies Inc. USA
https://phet.colorado.edu and https://www.merlot.org for laboratory simulations
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
19
Enrichment Activity/Web Based Task:
Laboratory Simulation on “Calorimetry”
In real engineering practice, a bomb calorimeter is used to measure the heat generated
by fuels. It is a thick-walled vessel filled with water, with an inner vessel wherein the fuel sample
is placed with an ignition wire. The heat generated by the ignited fuel sample is measured as the
change in the temperature of the surrounding water. This laboratory simulation demonstrates the
said measurement.
1.
Connect to this link:
http://employees.oneonta.edu/viningwj/sims/calorimetry_s.html
2.
Using this laboratory simulation, provide data for the following table: (remember to “reset”
every time you change the substance used, before clicking “ignite”)
Final Water
Change in
Substance
Mass
Mass of water Initial Water
Temperature
Temperature
Water
in the
in
Temperature
in degree Celsius
in degree Celsius
milligrams calorimeter
0
0
( C)
( C)
in Celsius
in grams
degree (C0)
Benzoic acid
500
1000
20
Trinitrotoluene
500
1000
20
Nitroglycerin
500
1000
20
3.
Based on your gathered data, write a conclusion for this experiment.
Conclusion:_______________________________________________________________
________________________________________________________________________.
Engr.
M. C. Javier-Tala,
MSEE.CLAUSE:
Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.
Student’s
HONESTY
I hereby declare that all solutions and answers in this activity were done by me. I pledge to practice
the highest degree of Academic Honesty at all times, as expected from all students, indicated in the Course
Policies of the syllabus for the course, Chemistry for Engineers (CHEM114).
20
Engr. M. C. Javier-Tala, MSEE. Reference: Javier, M.C. (2016). Module in Natural Science 2E: Applied Physics. 3rd Edition.