Physics 2 [Summer 2024 - 2025]
Department of Physics
Faculty of Science & Technology (FST)
American International University-Bangladesh
Course Requirements
➢ Students are expected to attend at least 75-80% of the class.
➢ Students are expected to participate actively in the class.
➢ For both terms, there will be two quizzes based on the theoretical knowledge and conceptual
understanding of the topics discussed in class. Students must attend at least one of them.
➢ Submission of a report based on the given course-related assignments on time.
1
Course outline
• Course Requirements
• Students are expected to attend at least 75-80% of the class.
• Students are expected to participate actively in the class.
• For both terms, there will be at least 2 quizzes based on the theoretical knowledge and conceptual
understanding of the topic discussed in the classes.
• Submit a report based on the given course-related assignments.
• Submission of assignments and projects should be in due time.
Grading and marks distribution
Midterm
ATTENDANCE AND PERFORMANCE
ASSESSMENTS (QUIZZES) : BEST ONE OUT OF TWO
ASSIGNMENT
MIDTERM Written Exam (COUNT ALL)
TOTAL
ATTENDANCE AND PERFORMANCE
Final term
ASSESSMENTS (QUIZZES) : BEST ONE OUT OF TWO
POSTER PRESENTATION (POWERPOINT)
FINAL TERM ASSESSMENTS (COUNT ALL)
Grand Final
10
(10%)
20 (20 %)
20 (20%)
50 (50%)
100
MARKS
10 (10%)
20 (20 %)
20 (20%)
50 (50%)
TOTAL
100
MARKS
MIDTERM 40 %+FINAL TERM 60%
Total 100
Lesson Plan (Midterm)
WEEKS
1
CLASSES
/LECTURES
LECTURE 1
LECTURE 2
2
LECTURE 3
LECTURE 4
3
4
CHAPTERS
ASSESSMENT
Introduction of the course
Introduction to Thermodynamics: Heat, temperature, zeroth
law of thermodynamics, absorption of heat: heat capacity,
specific heat capacity, molar specific heat,
Latent heat of fusion and vaporization, related problems.
Review of Lesson 1 and Lesson 2, related problems
Concept of work done in thermodynamics, First law of
thermodynamics and its applications, related problems.
Quiz 1
Avogadro number, Ideal gas concept, Ideal gas equation, work
done by an ideal gas, and related problems. Pressure,
temperature, and rms speed
Chapter 18
Temperature, heat and the
first law of thermodynamics
(Fundamentals of Physics,
th
10 edition)
Group discussion and oral question
answer
Chapter 19
The Kinetic Theory of Gases
(Fundamentals of Physics,
10th edition)
Quiz 1 Oral question answer
LECTURE 5
Translational kinetic energy and related problems. Internal
energy of monoatomic gas, molar specific heat at constant
volume for an ideal gas.
Group discussion and oral question
answer
LECTURE 6
Molar Specific Heat of Ideal Gases, Degrees of Freedom, and
related problems. Adiabatic expansion of an ideal gas and
related problems.
Review of Chapter 19
Quiz 2
Entropy and the Second Law of Thermodynamics, Entropy in
the Real World: Engines
Work done, entropy change and efficiency calculation of a
Carnot engine, related problems, Comparison: ideal engine,
Carnot engine and real engines.
Review
Oral question answer
LECTURE 7
LECTURE 8
5
CONTENTS
LECTURE 9
EXAM WEEK
Chapter 20
Assignment
Group discussion and oral question
answer
Reference Books:
Fundamentals of Physics (Edition: 10th) Written by Halliday, Resnick and Walker
University Physics (13/14th Edition) written by Hugh D. Young and Roger A. Freedman
• OBE Grading
MID
Final
Grading policy
• https://www.aiub.edu/academicregulations
Chapter 18
Temperature, heat and the first law of thermodynamics
(Fundamental of Physics, 10th edition)
Lecture- 1
18.1 Thermodynamics :
One of the principal branches of physics and engineering is
thermodynamics, which is the study and application of the thermal energy
(often called the internal energy) of systems. One of the central concepts of
thermodynamics is temperature.
Temperature :
Temperature is an SI base quantity related to our sense of hot and cold.
It is measured with a thermometer, which contains a working substance with
a measurable property, such as length or pressure, that changes in a regular
way as the substance becomes hotter or colder.
Recall: Temperature Scales
Here
T= temperature in Kelvin scale
𝑇𝐶 = temperature in Celsius scale
𝑇𝐹 = temperature in Fahrenheit scale
Class Work: Temperature difference
▪ The temperature of an object is 25 °C. After absorbing some
amount of heat the temperature of the object becomes 55 °C.
What would be the temperature difference in Kelvin scale, in this
case?
Solve it by yourself!
The Zeroth Law of Thermodynamics:
Suppose that, as in following Fig (a) , we put a thermoscope (which we shall call body T) into
intimate contact with another body (body A).The entire system is confined within a thick-walled
insulating box.
“If bodies A and B are each in thermal equilibrium with a third body T, then A and B are in
thermal equilibrium with each other.”
In less formal language, the message of the zeroth law is: “Every body has a property called
temperature. When two bodies are in thermal equilibrium, their temperatures are equal. And vice
versa.”
18.4 ABSORPTION OF HEAT :
Temperature and Heat
A change in temperature is due to a change in
the thermal energy of the system because of a
transfer of energy between the system and the
system’s environment. The transferred energy is
called ‘ HEAT’ and is symbolized Q.
Heat is negative when energy is transferred from
a system’s thermal energy to its environment (we
say that heat is released or lost by the system).
Heat is positive when energy is transferred to a
system’s thermal energy from its environment (we
say that heat is absorbed by the system).
Heat:
Heat is the energy transferred between a system
and its environment because of a temperature
difference that exists between them.
1 cal = 3.968 × 10-3 Btu = 4.1868 J.
The Absorption of Heat by Solids and Liquids
Heat Capacity
The heat capacity, C of an object is the proportionality constant between the heat Q that the object
absorbs or loses and the resulting temperature change ΔT of the object; that is,
Q ∝ ΔT
Q = C ΔT = C ( Tf – Ti )
C = Q/ ΔT
in which Ti and Tf are the initial and final temperatures of the object, respectively.
Unit : cal/ C° or J/K
Specific Heat Capacity
Two objects made of the same material—say, marble—will have heat capacities proportional to their
masses. It is therefore convenient to define a “heat capacity per unit mass” or specific heat c that refers
not to an object but to a unit mass of the material of which the object is made. [c = C/m
C = mc]
Q = C ΔT = mcΔT
c = Q/mΔT
Unit : cal/g-C°
or J/kg-K
Molar Specific Heat
In many instances the most convenient unit for specifying the amount of a substance is the mole
(mol), where 1 mole = NA= 6.02 × 1023 elementary units (atoms or molecule) of any substance.
The molar specific heat of a material is the heat capacity per mole, which means per 6.02 × 1023
elementary units of the material.
Thus 1 mol of aluminum means 6.02x1023 atoms (the atom is the elementary unit), and 1 mol of
aluminum oxide means 6.02 x1023 molecules (the molecule is the elementary unit of the compound).
When quantities are expressed in moles, specific heats must also involve moles (rather than a mass
unit); they are then called molar specific heats.
The molar specific heat, cm of a material is the heat capacity per mole.
Q = ncm ΔT
cm = Q/nΔT
Unit : cal/mol-C°
or J/ mol-K
Problem 23 : A small electric immersion heater is used to heat 100 g of water for
a cup of instant coffee. The heater is labeled “200 watts” (it converts electrical
energy to thermal energy at this rate). Calculate the time required to bring all this
water from 23° C to 100° C, ignoring any heat losses.
Solution:
m = 0.100 kg
P = 200 W
c = 4190 J/kg-K
Ti = 23°C = 23+273 = 296 K
Tf = 100°C = 100 +273 K= 373 K
∆T = Tf –Ti = 373 – 296 K=77 K
Q = cm ∆T
P = W/t
P = Q/t
𝑸
𝒎𝒄∆T
𝟎.𝟏𝟎𝟎( 𝟒𝟏𝟗𝟎)(𝟕𝟕)
Now, t =
=
=
= 160 sec (Ans)
𝑷
𝑷
𝟐𝟎𝟎
Problem 24 : A certain substance has a mass per mole of 50.0 g/mol. When 314 J is
added as heat to a 30.0 g sample, the sample’s temperature rises from 25.0° C to
45.0° C. What are the (a) specific heat and (b) molar specific heat of this substance?
(c) How many moles are in the sample?
Solution :
Given, m = 30 g = 30 × 10 -3 kg ; Tf = 45°C ; Ti = 25°C ; Q = 314 J
∆T = (45 + 273)K – (25+273)K = (45 -25 )K = 20 K
(a) Q = mc∆T
c=
𝑄
𝑚 ∆T
(c) Mass of sample, m = Msam = 30 g = 30 × 10 -3 kg
314
30 × 10 −3 ×20
= 523 J/kg-K
And [ Mass per mole, M = 50 g =50 × 10 -3 kg ]
So, Msam = nM
Now, number of mole, n =
=
𝑀𝑠𝑎𝑚
=
𝑀
30 × 10 −3
= 0.600 mol
50 × 10 −3
(b) Q = ncm∆T
cm =
𝑄
𝑛 ∆T
=
314
0.600 ×20
= 26.2 J/mol-K