Chapter-4 (6).pptx

Heat and Thermodynamics
G. Beyene (Dr.) (ASTU) Chapter four Cover Page - 1/30 June-2022
Chapter Four
Dr. Gashaw Beyene (PhD)
June, 2022
ASTU

The Concept of Temperature
G. Beyene (Dr.) (ASTU) Chapter Four The concept of Temperature - 2/30
 Matter is made up of moving particles (molecules, atoms, and ions).
 Heat is the flow of energy from hotter region to the cooler region.
 When the substance heat up, two things could happen;
𝑄 = Σ𝐸𝑘 + Σ𝑈
 The particles of the substance gain kinetic energy and so move more rapidly.
 The bonds between the particles in the substance are broken and the potential energy
of the substances increases. When this happens, the substances changes state.
June-2022

The Concept …(Cont’d)
 Temperature is a measure of hotness and coldness of a substance.
 The temperature of a substance is a measurement of the average kinetic energy of the
particles within the substance.
 As a substance absorbs heat energy, the particles vibrate more (in a soild) or move
faster (in a liquid or gas) as the heat energy is converted into the kinetic energy of the
particles as the temperature rises.
𝑇𝑐 = 𝑇𝑘 − 273.15
𝑇𝐹 = 1.8𝑇𝑐 + 32
June-2022

The Concept …(Cont’d)
 The zeroth law of thermodynamics (the law of equilibrium):
 If objects A and B are separately in thermal equilibrium with a third object C, then
A and B are in thermal equilibrium with each other.
 If object A is in thermal equilibrium with object B, and if object B is in thermal
equilibrium with object C, then objects A and C are also in equilibrium. This is sort
of a “transitive property of heat.”
hot cold
heat
26°C 26°C
No net heat flow
June-2022

Specific Heat and Latent Heat
G. Beyene (Dr.) (ASTU) Chapter Four Specific Heat and Latent Heat - 5/30
 The calorie (𝑐𝑎𝑙) is defined as the energy necessary to raise the temperature of 1 g
of water by 1℃.
1 𝑐𝑎𝑙 = 4.186 𝐽
 The amount of heat energy required to raise temperature depend on;
 Types of substance being heated.
 The amount of substance (mass)
 Temperature
 Each substance has a specific heat capacity (𝑐 (
𝐽
𝑘𝑔.𝐾
)), define as;
 the heat energy required to raise a temperature of 1 𝑘𝑔 of a given substance by 1 𝐾.
𝑐 = 𝑄/𝑚∆𝑇
June-2022

G. Beyene (Dr.) (ASTU)
Chapter Four Specific Heat and Latent Heat - 6/30
𝑄 = 𝑚𝑐∆𝑇
 The quantity of heat energy (𝑄) required to raise a temperature of a substance.
 Heat energy can flow from hot region
(substance) to cold region (substance).
−𝑄ℎ= 𝑄𝑐
 Hot substance lose its (∆𝑇 = −𝑣𝑒) heat energy whereas cold substance gain heat
energy (∆𝑇 = +𝑣𝑒) from hot substance.
𝐻𝑒𝑎𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 𝑏𝑦 ℎ𝑜𝑡𝑡𝑒𝑟 𝑏𝑜𝑑𝑦 = 𝐻𝑒𝑎𝑡 𝑒𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑐𝑜𝑙𝑑𝑒𝑟 𝑏𝑜𝑑𝑦
June-2022

‽ A 0.05kg ingot of metal is heated to 200°C and then dropped into a calorimeter containing
0.4kg of water initially at 20°C. The final equilibrium temperature of the mixed system is
22.4°C. Find the specific heat of the metal.
 Example 1
June-2022

 If solid and liquid get heat energy, their particles gain energy and begin to vibrate faster
and move further apart as its temperature increases. This continues until the solid melts
and liquid boils.
The specific latent heat (𝐿) of a substance is defines as: the quantity of heat energy
required to change 1 𝑘𝑔 of a substance from one state to another, at a constant
temperature.
𝐿 = 𝑄/𝑚 𝑄 = 𝑚𝐿
 There are two changes of state to consider, solid to liquid and liquid to gas.
 We use two different versions of latent heat, the latent heat of fusion (melting) (𝐿𝑓)
and latent heat of vaporization (boiling) (𝐿𝑣).
June-2022

 Specific latent heat of fusion (𝐿𝑓) is the quantity of heat energy required to change
1 k𝑔 of a substance from solid to liquid at a constant temperature.
 Specific latent heat of vaporization (𝐿𝑣) is the quantity of heat energy required to
change 1 k𝑔 of a substance from liquid to solid at a constant temperature.
Now we use specific heat capacities and specific latent heats, to calculate the total energy
required to change the state and raise the temperature.
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒
= 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑡ℎ𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒
+ 𝑒𝑛𝑒𝑟𝑔𝑦 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑐ℎ𝑎𝑛𝑔𝑒 𝑠𝑡𝑎𝑡𝑒
𝑄𝑡𝑜𝑡 = 𝑚𝑐∆𝑇 + 𝑚𝐿
June-2022

‽ Calculate the heat energy required to increase the temperature of 50g of water from
250𝐶 to 125𝑜𝐶.
 Example 2
1. Heat energy required to heat 50g
water from 250
𝐶 to 100𝑜
𝐶
(∆𝑇 = 75𝑜
𝐶 = 75𝑘)
𝑄1 = 𝑚𝑐∆𝑇 = 0.05 × 4200 × 75 = 15750𝐽
2. Heat energy required to boil 50g
water at 1000𝐶.
𝑄2 = 𝑚𝐿𝑣 = 0.05 × 2501000 = 125050𝐽
3. Heat energy required to heat 50g steam
from 1000
𝐶 to 125𝑜
𝐶 (∆𝑇 = 25𝑜
𝐶 = 25𝑘)
𝑄3 = 𝑚𝑐∆𝑇 = 0.05 × 2080 × 25 = 2600𝐽
So, the total amount of heat energy required to
this process;
𝑄 = 𝑄1 + 𝑄2 + 𝑄3 = 143.4 𝑘J
June-2022

Thermal Expansion of Solids and Liquids
Chapter Four Thermal Expansion of Solids and Liquids - 11/30
 As the particles gain more energy, we can see that they move further apart from each
other, which means the substance will expand.
 The expansion of substances on heating is called
thermal expansion. This happens in solids, liquids
and gases.
 In thermal equilibrium the heat loss from hot substance is equal to heat gained by cold
substance, so that there is no net movement of energy between two substances.
 If two bodies are in thermal equilibrium, they will also be at the same temperature.
June-2022

 Ordinarily a substance expands when heated. If an object has an initial length 𝑙𝑐 at some
temperature and undergoes a change in temperature ∆𝑇, its linear dimension changes by
the amount ∆𝑙, which is proportional to the object’s initial length (𝑙𝑐) and the temperature
change:
∆𝑙 = 𝛼𝑙𝑐∆𝑇
∆𝑙 = 𝑙ℎ − 𝑙𝑐
 𝛼 (1
𝐾 𝑜𝑟 𝐾−1
) is also known as the coefficient of linear expansion for the solid .
𝑙ℎ = 𝑙𝑐(1 + 𝛼∆𝑇)
June-2022

 As the plate/surface is heated to cause an increase in temperature,
∆𝑇, it expands in width and height such that the surface area when
heated .
∆𝐴 = 𝐴ℎ − 𝐴𝑐
∆𝐴 = 𝛽𝐴𝑐∆𝑇
𝐴ℎ = 𝐴𝑐(1 + 𝛽∆𝑇)
 𝛽 (1
𝐾 𝑜𝑟 𝐾−1) is also known as the coefficient of
surface expansion for the solid .
June-2022

𝐴ℎ = 𝑙ℎ
2
= (𝑙𝑐 1 + 𝛼∆𝑇 )2
𝐴ℎ = 𝑙ℎ
2
= 𝑙𝑐
2
(1 + 2𝛼∆𝑇 + 𝛼2∆𝑇2) 1 + 2𝛼∆𝑇 ≫ 𝛼2∆𝑇2
𝐴ℎ = 𝐴𝑐(1 + 2𝛼∆𝑇)
 Therefore,
𝐴ℎ = 𝐴𝑐 1 + 2𝛼∆𝑇 = 𝐴𝑐(1 + 𝛽∆𝑇)
𝛽 = 2𝛼
June-2022

 Consider the expansion of cube.
∆𝑉 = 𝑉ℎ − 𝑉
𝑐 ∆𝑉 = 𝛾𝑉
𝑐∆𝑇
𝑉ℎ = 𝑉
𝑐(1 + 𝛾∆𝑇)
 𝛾(1
𝐾 𝑜𝑟 𝐾−1) is also known as the coefficient
of volume expansion for the solid .
June-2022

𝑉ℎ = 𝑙ℎ
3
= (𝑙𝑐 1 + 𝛼∆𝑇 )3
𝑉ℎ = 𝑙ℎ
3
= 𝑙𝑐
3
1 + 3𝛼∆𝑇 + 3𝛼2∆𝑇2 + 𝛼3∆𝑇3
1 + 3𝛼∆𝑇 ≫ 3𝛼2∆𝑇2 + 𝛼3∆𝑇3
𝑉ℎ = 𝑉
𝑐(1 + 3𝛼∆𝑇)
 Therefore,
𝑉ℎ = 𝑉
𝑐 1 + 3𝛼∆𝑇 = 𝐴𝑐(1 + 𝛾∆𝑇)
𝛾 = 3𝛼 =
3
2
𝛽
Chapter Four Thermal Expansion of Solids and Liquids - 16/30 June-2022

‽ A segment of steel railroad track has a length of 30 m when the temperature is 0𝑜𝐶. A) What is its length
when the temperature is 40𝑜𝐶? B) What is the thermal stress set up in the rail if its temperature is raised
to 40𝑜𝐶?
 Example 3
June-2022

‽ A sample of lead has a mass of 20 kg and a density of 11.3 × 103𝑘𝑔/𝑚3 at 0𝑜𝐶. (a) What is the density
of lead at 90𝑜𝐶? (b) What is the mass of the sample of lead at 20𝑜𝐶?
 Example 4
June-2022

 Volume expansion of liquid;
 ∆𝑉 = change in volume.
 𝛾 = volume expansion coefficient
𝛾𝑟𝑒𝑎𝑙 = 𝛾𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 + 𝛾𝑣𝑒𝑠𝑠𝑒𝑙
∆𝑉 = 𝛾𝑉
𝑐∆𝑇
 𝛾𝑟𝑒𝑎𝑙 = the real/actual expansion of liquid.
 𝛾𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 = liquid expansion
 𝛾𝑣𝑒𝑠𝑠𝑒𝑙 =thermal expansion of container

 The volume expansion of gas is affected by the factors;
𝑃𝑉 = 𝑛𝑅𝑇
• 𝑉 = volume of gas in 𝑚3
• 𝑛 = number of moles in 𝑚𝑜𝑙
• 𝑅 = universal gas constant (8.314
𝐽
𝐾.𝑚𝑜𝑙
)
• 𝑇 = absolute temperature in 𝐾
 Pressure (𝑃)
 The amount (number of moles) of gas present (𝑛)
 Temperature (𝑇)
 At fixed/constant pressure and amount of gas, 𝑉 ∝ 𝑇.
𝛾𝑠𝑜𝑙𝑖𝑑 < 𝛾𝑙𝑖𝑞𝑢𝑖𝑑 < 𝛾𝑔𝑎𝑠

 The internal energy includes kinetic and potential energy associated with the random
translational, rotational, and vibrational motion of the particles that make up the
system, and any potential energy bonding the particles together.
 Heat involves a transfer of internal energy from one location to another.
 Internal energy is the energy associated with the atoms and molecules of the
system.
 Heat is the transfer of energy between a system and its environment due to a
temperature difference between them.
 Temperature is a quantitative measurement of object’s relative the degree of hotness
and coldness.
Transfer of Heat Energy
Chapter Four Transfer of Heat Energy - 21/30 June-2022

Chapter Four Transfer of Heat Energy - 22/30
 Energy is always conserved. It can change forms: kinetic, potential, internal etc., but
the total energy is a constant. Another way to say it is that the change in thermal
energy of a system is equal to the sum of the work done on it and the amount of heat
energy transferred to it.
 The First Law of Thermodynamics
 The change in internal energy of a closed system, ∆𝐸, will be equal to the energy added
to the system by heating (𝑄) minus the work done by the system (𝑊) on the
surroundings.
∆𝐸 = Q − W
 The work is done on the system, 𝑾 will be positive.
∆𝐸 = Q + W
June-2022

Chapter Four Transfe of Heat Energy - 23/30
For an isolated system, no work is done and no heat enters or leaves the system, so 𝑾
= 𝑸 = 𝟎, and hence 𝜟𝑬 = 𝟎.
 A system to which you cannot add matter is called closed.
Though you cannot add/remove matter to a closed system, you can still
add/remove heat.
 A system to which neither matter nor heat can be added/removed is called isolated.
June-2022

Thermal Conductivity
Chapter Four Thermal Conductivity - 24/30
 The process of energy transfer by heat can also be called conduction or thermal
conduction.
 Conduction occurs only if there is a difference in temperature between two parts of the
conducting medium.
 The energy 𝑄 transfers in a time interval ∆𝑡 from the hotter
face to the colder one.
This means that the rate of flow of heat (𝑃) energy 𝑄/𝑡
in watts (J/s) along the bar will be given by:
𝑃 =
𝑄
𝑡
= 𝑘𝐴
∆𝑇
∆𝑥
= 𝑘𝐴
𝑇ℎ − 𝑇𝑐
∆𝑥
June-2022

Thermal Conductivity
Chapter Four Thermal Conductivity - 25/30
 The constant 𝑘 (
𝑊
𝑚.𝑘
) depends on which material the bar is made from. It is known as the
material’s thermal conductivity.
 Thermal conductivity a measurement of the ability of a material to conduct heat.
‽ Two slabs of thickness 𝐿1 and 𝐿2 and thermal conductivities 𝑘1 and 𝑘2 are in thermal
contact with each other. Determine the temperature at the interface and the rate of
energy transfer by conduction through an area A of the slabs in the steady-state
condition.
 Example 5
June-2022

Thermal Convection
Chapter Four Thermal Convection - 26/30
Convection : Energy is transferred from one body to a cooler one
via currents in a fluid (a gas or liquid).
 Energy transferred by the movement of a warm substance is said
to have been transferred by convection, which is a form of
matter transfer.
 Convection is the process whereby heat flows by the mass movement of
molecules from one place to another.
June-2022

 Unlike convection and conduction which require the presence of matter as a medium to
carry the heat from the hotter to the colder region a heat transfer through radiation occurs
without any medium at all.
Thermal Radiation
Chapter Four Thermal Radiation - 27/30
 Radiation consists essentially of electromagnetic waves.
 All objects radiate energy continuously in the form of electromagnetic waves produced
by thermal vibrations of the molecules.
 The rate at which the surface of an object radiates energy is proportional to the fourth
power of the absolute temperature of the surface.
𝑃 = 𝜎𝐴𝜀𝑇4
𝑃 is the power in watts of electromagnetic waves
radiated from the surface of the object
June-2022

Thermal Radiation
• 𝜎 = 5.67 × 10−8𝑊𝑚−2𝑘−4
• 𝐴 is the surface area of the object in square meters,
• 𝜀 is the emissivity (between 1 and 0), and
• 𝑇 is the surface temperature in kelvins.
 An ideal absorber is defined as an object that absorbs all the energy incident on it, and
for such an object, 𝜀 = 1 (the object is black body).
☺If an object is at a temperature 𝑇 and its surroundings are at an average temperature
𝑇0, the net rate of energy gained or lost by the object as a result of radiation is
𝑃 = 𝜎𝐴𝜀(𝑇4 − 𝑇0
4
June-2022

‽ A student is trying to decide what to wear. His bedroom is at 20𝑜
𝐶. His skin temperature
is 35𝑜𝐶. The area of his exposed skin is 1.5𝑚2. People all over the
world have skin that is dark in the infrared, with emissivity about 0.9. Find the net energy
transfer from his body by radiation in 10 𝑚𝑖𝑛.
Thermal Radiation
 Example 6
𝑃 = 𝜎𝐴𝜀 𝑇4 − 𝑇0
4
= 5.67 × 10−8 × 1.5 × 0.9( 308.15 4 − (293.15)4)
𝑃 ≅ 5.67 × 10−8 × 1.5 × 0.9(9 × 109 − 7.4 × 109)
𝑃 ≅ 122.47 𝑊 𝑄 = 𝑃 × 𝑡 = 122.47 × 600 ≅ 7.35 × 105𝐽
June-2022

G. Beyene (Dr.) (ASTU) August - 2021
End
Chapter Five End - 30/30

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