Chapter 4 discusses the first law of thermodynamics, focusing on energy conservation and transforming energy forms through thermal and mechanical interactions. It also covers specific heat, calorimetry, latent heat, and methods of heat transfer, explaining the principles guiding these phenomena. Additionally, the chapter introduces concepts of entropy as related to thermodynamic processes and energy availability.

1. Chapter 4: Heat
Phenomenon
By Group 3

2. 4.1 The first Law of Thermodynamics
Location of a point ⊥ energy of the system ⊥ macroscopic
variables

3. Two General Ways of Changing Energy
 Thermal Interaction ( through conduction, convection, and radiation )
Heat absorbed by the system → ∆Q > 0
Heat given off by the system → ∆Q < 0
 Mechanical Interaction
∆W > 0 → work done [by] the system
∆W < 0 → work done [on] the system

5. That is, ∆Q and ∆W depend not only
on he initial and final states but also
on the path taken going from i to f

6. But, if we look instead at the combined quantity ∆Q-∆W along each path, then
we find that
That is, ∆Q-∆W depends only on the initial and final states but not on the path
taken going from i to f .

7. State Variables is a thermodynamic
quantities that depend only on the initial
and final states but not on the path
taken in the pV diagram.
In our present discussion, the relevant
state variable is the internal energy.
E= Q-W
State Variables

8. Corresponding change in internal energy is
∆E = ∆Q - ∆W
Where
∆Q > 0 for heat absorbed
∆Q < 0 for heat given off
∆W > 0 for work done [by] the system
∆W < 0 for work done [on] the system

9. Going back to the perfect gas whose pV diagram is given by Fig.4.1, let us suppose
that
Then
So that

10. It follows that along the
path i1f2i (the curve i1f
then along 2 back to i), we
find that the total change
in internal energy is
Which is simply the law of conservation of energy for a perfect gas.
The First Law of Thermodynamics is just the Law of Conservation of
Energy whose formulation can be stated as follows:
Energy cannot be created nor destroyed but it can be transformed
from one form to another.

11. 4.2 Mechanical and Thermal Interactions
Let us consider a gas confined in
a vertical cylindrical container closed
at one end by a piston light enough
so that it can move up or down a
distance ∆x in spite of its weight. In
addition, let us assume that the gas is
in thermal contact with the laboratory
bench at the bottom of the container
through which it can exchange heat
energy with the environment.
For the mechanical interaction of
the system with he environment
through the movable piston, we find
that

12. In an isochoric process, ∆V=0 and there is no work done (∆W=0).

13. 4.3 Specific Heat, Thermal
Capacity and Heat Capacity

14. Specific Heat (c)
The specific heat is the amount of heat per unit mass required
to raise the temperature by one degree Celsius.

15. UNIT
S
MKS CGS FPS
c kcal/kg * °C cal /g * °C Btu/lb * °F

16. Specific Heat of some materials
NOTE: 1 cal = 4.186J

17. Thermal Capacity (C)
 Thermal capacity is the amount of heat needed to raise the temperature of the
total mass of a substance.
That is:
C=mc

18. UNITS MKS CGS FPS
C kcal / °C cal / °C BTU / °F
NOTE: 1 kcal = 3.96567 BTU

19. Heat Quantity
 Heat Quantity is the total amount of heat needed to change the temperature of
the total mass of a substance when the change in temperature is not equal to 1
°C.
That is:
Q = mcΔT

20. UNITS Joules Calories BTU

21. Example Problem
 It takes 487.5 J to heat 25 grams of copper from 25 °C to 75 °C. What is the specific
heat in Joules/g·°C and cal/g•°C?
Solution
Given: Q = 487.5 J
m = 25 grams
ΔT = 75 °C – 25 °C =50 °C
Find: Specific Heat (c)
Use the formula: Q = mcΔT or c = Q/mΔT
Plug the numbers given to the equation.
c = 487.5J / 25grams•50 °C
c = 0.39 J / g•°C
CONVERT to cal / g•°C
c = 0.39 J / g•°C
= ( 0.39J/g•°C)(1cal/4.186J)
= 0.0931cal/g•°C

22. 4.4 Calorimetry And Method of Mixture

23. Calorimetry And Method of Mixture
 The measurement of quantities of heat is called calorimetry. This is done with the
help of an instrument called a calorimeter. A calorimeter is a thermally insulated
container that is shown below.
The inner vessel is supported in the outer vessel by an
insulated cover, on which the thermometer and stirrer
can be inserted. The inner vessel is the one that
contains the substance whose heat quantities are
determined, while the outer vessel minimizes the
transfer of heat from the inside to the surrounding or
vice versa.

24. Method of Mixture
 The instrument works on the principle of the method of mixture, wherein a
substance at a lower temperature is mixed with another substance at a higher
temperature so that the heat gained by the cold substance is equal to the heat
lost by the hot substance.
Heat gained Qg = Heat lost Ql
 Heat gained characterized by an increase in the temperature of the substance
while heat lost is characterized by decrease in temperature.

25.  Let:
 mc = Mass of the calorimeter
 mm = mass of the metal
 mw =mass of the water inside the calorimeter
 cc = specific heat of calorimeter
 cm = specific heat of the metal
 Tw = initial temperature of water and calorimeter
 Tmix = temperature of the mixture
 Tm = temperature of hot metal
Then it follows:
Qg = Ql → (mccc + mw)(Tmix – Tw) = mmcm(Tm – Tmix)

26. 4.5Change of Phase and
Latent Heat
By Andrei Matias and Jet Estilles

27. 4.5.1 Change of Phase
1. Liquefaction
a. Melting – the solid state to liquid state process of changing
the phase of a substance
b. Condensation – the gas state to liquid state of a substance.
2. Solidification
a. Freezing – the reverse process of melting; it is the liquid state to
solid state process.
3. Vaporization
a. Evaporation – the liquid to gas change of phase process of a
substance.
b. Sublimation – the solid to gas change of phase process of a
substance without passing through the liquid form.

28. Latent Heat
The amount of heat required to
effect a complete change of a unit
mass of a substance without a
change in temperature.

29. 2 Most Common Latent Heats
Ex.
Lf of ice = 80 cal/g = 80 k cal/kg = 144BTU/lb
1. Latent Heat of Fusion, Lf – is the amount of heat needed to
change a unit mass of a solid to liquid or vice versa without any
change in temperature (ΔT=0).
𝐿 𝑓 =
𝑄
𝑚
𝑄 = 𝑚𝐿𝑓

30. Latent Heat of Vaporization, Lv – the amount of
heat needed to change a unit mass of liquid to
or vice versa without any change in
temperature(ΔT = 0).
𝐿 𝑣 =
𝑄
𝑚
𝑄 = 𝑚𝐿𝑣
Ex.
Lv of water = 540 cal/g = 540k cal/kg = 972 BTU/lb

31. Table 4.2 Latent Heat of Some Substances
SUBSTANCE BOILING TEMP
(°C)
VAPORIZATION
(cal/g)
MELTING TEMP
(°C)
FUSION
(cal/g)
Nitrogen - 196 48 -210 6.1
Oxygen -183 51 -219 3.3
Ethanol +78 204 -114 25
Water 100 539 0 80
Mercury 357 71 -39 2.7
Lead 1750 205 327 5.9

32. Summary
1. When ΔT ≠ 0 , and there is NO CHANGE in phase,
𝑸 = 𝒎𝒄 𝜟𝑻
2. When ΔT = 0, and there is A CHANGE in phase,
𝑸 = 𝒎𝑳𝒇 or 𝑸 = 𝒎𝑳𝒗
3. When ΔT = 0, and there is NO CHANGE in phase,
𝑸 = 𝟎

33. 4.6 Sample Problems with Solutions
 Calculate the total amount of heat needed to change the
temperature of a 500 g piece of copper from 20°C to 120°C. (the
specific heat of copper is 0.093 cal/g-°C)
Solution:
Q = mc ΔT
= 500 g (0.093 cal g-°C)(120 – 20)°C
= 4.650 cal

34.  An aluminum (Al) container has a mass of 50 g and holds 200 g of water at
20°C. Find the final temperature of the resulting mixture when a mass of 500 g of
Lead (Pb) at 95°C is placed in the water ( cAl=0.22 cal/g-°C; cPb= 0.031 cal/g-°C).
Solution:
Heat lost by Lead = Heat gained by the aluminum and water
QPb = QAl + Qwater
(mcΔT)Pb = (mcΔT)Al + (mcΔT)water
Substituting the given values and solving for the final temperature
Tmix of the mixture, We get
500(0.031)(95 – Tmix) = [50(0.22) + 200(1.0)][Tmix -20]
Tmix = 25.13°C

35.  How much heat is needed to change 10 g of ice at -20°C to steam at 140°C ,
assuming no heat losses due to the surrounding ?
Solution:
Qf = Q1 + Q2 + Q3 + Q4 + Q5
Q1 = mc ΔT the heat needed to raise the temperature of ice from
-20°C to 0°C
= 10g (0.5 cal/g-°C)(0 – 20) °C
= 100cal.
Q2 = mLf heat needed to melt the ice at 0°C to ice water still
at 0°C
= 10 g (80 cal/g)
= 800 cal.
Q3 = mc ΔT heat needed to raise the temperature of water
from 0°C to 1 00°C
= 10 g (1.0 cal/g-°C)(100 – 0) °C
= 1000 cal.

36. Q4 = mLv heat needed to change the water at
100°C to steam at 100°C
= 10 g (540 cal/g) = 5400 cal.
Q5 = mc ΔT heat needed to raise the temperature of
steam at 100°C to 1 40°C
= 10 g (0.48 cal/g°C)(140 – 100) °C
= 192 cal.
Finally we get
QT = 100 + 800 + 1000 + 5400 + 192
= 7492 calories

37.  Four ice cubes of mass 15 g and at 0°C are placed in a 50 g glass
which contains 200 g of water. If the glass a and the water are
initially at 25°C , find the temperature of the mixture and the mass
of the ice remaining if there is any.
Solution:
Mass of ice = 4 (15g) =60 g
Assuming that the final temperature of the mixture is 0°C, we find
the following:
Heat needed to melt the ice = mLf =60 g (80cal/g)= 4800 cal
Heat available to melt the ice = (mc ΔT)glass + = (mc ΔT) water
= 50 g 90.16 cal/g-°C)(25 – 0°C +200g (1.0 cal/g-°C)(25 – 0) °C
= 5200 cal

38. Therefore,
Heat available > Heat needed All the ice will be melted.
Solving for the final temperature for the Tmix of the mixture
Heat loss by glass and water = Heat gained by ice
(mc Δ T)glass + (mc Δ T)water = (mc Δ T)ice + mLf
50(0.16)(25 – Tmix) + 200 (1.0)(25 – Tmix) = 60 (0.5)(Tmix – 0) +60(80)
Tmix = 1.49°C

39.  A 0.300 kg lead ball was taken from a bath of hot oil and immediately placed in a 0.150kg
copper vessel containing 0.200 kg water at a temperature of 20°C. After stirring the
mixture attained the final equilibrium temperature of 25°C. What was the temperature of
hot oil from which the lead ball was taken? The specific heat of lead, copper, and water
are 0.031 cal/g-°C ,0.093 cal/g-°C and 1.0 cal/g-°C respectively.
Solution:
Heat lost by the lead ball = Heat gained by the water and the copper container
QPb = Qwater + Qcu
(mc Δ T )Pb = (mc Δ T )water + (mc Δ T)Cu
Substituting the given values and solving for the final temperature Tmix of the
mixture we get,
300 (0.031)(Tmix – 25) = 200 (1.0)(25 -20) + 150 (0.093)(25 -
20)
Tmix = 140 °C

40. 4.7 Understanding Heat
Transfer, Conduction,
Convection and Radiation

41. Heat Transfer
 Heat always moves from a warmer place to a cooler place.
 Hot objects in a cooler room will cool to room temperature.
 Cold objects in a warmer room will heat up to room temperature.

42. Heat Transfer Methods
Heat transfers in three ways:
Conduction
Convection
Radiation

43. Conduction
When you heat a metal strip at one end, the heat
travels to the other end.
As you heat the metal, the particles vibrate, these
vibrations make the adjacent particles vibrate, and so on
and so on, the vibrations are passed along the metal and
so is the heat. We call this? Conduction

44. Convection
What happens to the particles in a liquid or a
gas when you heat them?
The particles spread out and
become less dense.
This effects fluid movement.What is a fluid?A liquid or gas.

45. Fluid movement
Cooler, more d____, fluids
sink through w_____, less
dense fluids.
In effect, warmer liquids and
gases r___ up.
Cooler liquids and gases s___.
ense
armer
ise
ink

46. Cold air sinks
Where is the
freezer
compartment
put in a fridge?
Freezer
compartment
It is put at the
top, because
cool air sinks,
so it cools the
food on the
way down.
It is warmer
at the
bottom, so
this warmer
air rises and
a convection
current is
set up.

47. The third method of heat transfer
How does heat energy get
from the Sun to the Earth?
There are no particles
between the Sun and the
Earth so it CANNOT
travel by conduction or
by convection.
?
RADIATION

48. Radiation
Radiation travels in straight lines
True
Radiation can travel through a vacuum
True
Radiation requires particles to travel
False
Radiation travels at the speed of light
True

49. Emission experiment
Four containers were filled with warm water. Which
container would have the warmest water after ten minutes?
Shiny metal
Dull metal
Dull black
Shiny black
The __________ container would be the warmest after ten
minutes because its shiny surface reflects heat _______ back
into the container so less is lost. The ________ container
would be the coolest because it is the best at _______ heat
radiation.
shiny metal
radiation
dull black
emitting

50. Absorption experiment
Four containers were placed equidistant from a heater. Which
container would have the warmest water after ten minutes?
The __________ container would be the warmest after ten
minutes because its surface absorbs heat _______ the best.
The _________ container would be the coolest because it is
the poorest at __________ heat radiation.
dull black
radiation
shiny metal
absorbing
Shiny metal
Dull metal
Dull black
Shiny black

52. Thermodynamics is the study of
heat and energy. There are three
laws around which
thermodynamics revolves:
The Zeroth Law
The First law of Thermodynamics
The Second law of
Thermodynamics

53. The Second Law of
Thermodynamics
 Can be stated in different ways. But the basic principles which
the law holds can be summarized in two statements.
(1)Natural processes go In a direction that maintains or increases
the total entropy of the universe.
(2) When all systems taking part in a process are included, the
total entropy either remains constant if the process is
reversible of increases if the process is irreversible.

54. Entropy (time’s arrow)
(1) Is a measure of the disorder in a system.
(2) Is a measure of how much energy in unavailable
for conversion into work.
(3) Points out the forward direction of the flow of
events.

55. To determine the change in entropy in
a process that is approximately
reversible (since a reversible one does
not occur in nature), the following
equation in used:
ΔS = Q
T
where: ΔS = is change in entropy; the
unit for entropy is J/K
Q = heat energy
T = temperature (in kelvin)

56. Sample problem:
A 6.25-kg block of ice is placed on a
concrete floor which has a constant
temperature of 25.5ºC. The ice melts
as it absorbs 2.09 x 10^6 J of heat.
What is the change in entropy of the
ice? Of the floor? Of the universe?
(Consider that the heat absorbed by
ice comes only from the floor.)