physics - ch6 temperature and heat
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DAS 12603 TECHNICAL SCIENCE I
CHAPTER 6:
TEMPERATURE AND HEATNORAIHAN BINTI SALLEH HUDIN
DEPT OF SCIENCE AND MATHEMATICS
CENTRE FOR DIPLOMA STUDIES, UTHM
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LEARNING OBJECTIVES
The objectives of this chapter are:
i. To impart students with the basic knowledge in
science especially heat and thermal properties of
matter.
ii. To apply the concept of heat and thermal
properties in science and engineering courses.
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LEARNING OUTCOMES
After completing this chapter, students should be able to:
explain the definition of temperature and heat.
describe the thermal equilibrium and the zeroth law ofthermodynamics.
convert the temperature unit scales.
explain the specific heat (sensible heat) and latent heat.
calculate the problem of heat by using the calorimetricprinciple.
differentiate latent heat of fusion and latent heat ofvaporization.
solve problem of heat when a matter change from onephase to another phase.
List down three types of mechanisms of heat transfer.
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3.1 TEMPERATURE
Definition: the measure of the degree of hotness
and coldness of matter whether in solids, liquids or
gases.
When two or more objects are in contact with each
other and their temperature is equal, they are said
to be in thermal equilibrium.
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3.1.1 THERMAL EQUILIBRIUM
When two matters are in thermal contact to each other, the
energy (heat) from matter with higher temperature will flow tothe matter with lower temperature.
As energy continues to flow, temperature of the two matterswill approach each other.
At equal temperature, there are no longer net flow of energy
between them. The two matters are said to be in thermal equilibrium.
hotter cooler hotter cooler
thermal equilibrium
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3.1.2 THE ZEROTH LAW OF
THERMODYNAMICS
States that:
If two objects are in thermal equilibrium with a third
object, then the two objects are in thermal
equilibrium with each other.
AC C
B
A B
A in thermal equilibrium with C C in thermal equilibrium with B
Therefore, A in thermal equilibrium with B
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Thermometer Thermometry propertiesMercury thermometer The change in expansion of mercury in glass
column with temperature
Ideal gas thermometer The change in glass pressure with temperature
at constant volume OR the change in gas
volume with temperature at constant gaspressure
Thermocouple The change in electric motion force (emf)
through thin wires of different metals (welded
together at the ends to form two junctions) with
temperature
Electric resistance
thermometer (thermistor)
Electrical resistant changes with temperature
Thermogram The intensity of the radiation emitted by an
object (infrared radiation) increases
substantially as the temperature is raised
Types of thermometer with their thermometry properties
Thermometry propertyof a material: property of a material that is
temperature-dependent (change linearly with the changing of
temperature)
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3.1.3 TEMPERATURE SCALES
Temperature scales were built to determine thetemperature of an object quantitatively.
To built a temperature scale, several standard fixedpoints were used as calibration points, which are:
Ice point: the point where an ice (solid) and water
(liquid) are in equilibrium phase (atmospheric pressure,p= 1 atm)
Steam point: the point where the water (liquid) andsteam (gases) are in equilibrium phase (atmosphericpressure,p= 1 atm)
There are three temperature scales: Celsius (C)
Fahrenheit (F)
Kelvin (K)
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Temperature
scale
Symbol Ice Point Steam
Point
Difference
Celsius C 0 100 100 C
Fahrenheit F 32 212 180 F
Kelvin K 273.15 373.15 100 K
Ice point and steam point of temperature scales
The Kelvin temperature scale has been chosen as the
international standard temperature scale.
Thus, Kelvin (K) is the SI unit for temperature.
The temperature 0 K represents the absolute zero of
temperature (there are no temperature below 0 K)
Therefore, there are no negative temperature number in
Kelvin scale.
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The three temperature scales can be related by the
following equations:
Celsius to Fahrenheit:
Fahrenheit to Celsius:
Celsius to Kelvin:
1.8 32F CT T
32
1.8FC
TT
273.15K CT T
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Example:
A friend suffering from flu has a slight fever. His body temperature is
37.20C. What is his temperature in
i) Kelvinii) degree Fahrenheit, F.
Solution:
i)
Celsius to Kelvin conversion
i) Celsius to Fahrenheit conversion
273.15
37.2 273.15
310.35
K CT T
K
1.8 32
1.8(37.20) 32
98.96
F CT T
F
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3.1.4 DETERMINING TEMPERATURE BY
USING THERMOMETRY PROPERTIES
Thermometry property of a material can be exploited to
determine its temperature.
Firstly, the thermometer scale must be calibrated with
the melting point of 0C and boiling point of 100C of
water at atmospheric pressure of 1 atm.
The value of parameterX0andX100are assigned to the
melting and boiling point respectively.
Plot a graph of a straight line between these two points.
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Any unknown temperature Twith their correspondingparameterXTthat lies along the straight line can bedetermined by the graph or by using the equation:
0
100 0
100TX X
T C
X X
X0 XT X100
100C
TC
0C
Temperature, T
Thermometry
property
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Example:
A resistance of thermometer gives a reading of 30.00at
melting point, 41.58at boiling point and 34.59when
immersed into a beaker filled with oil. Calculate the oils
temperature.
Solution:
0
100
30.00
41.58
34.59T
X
X
X
0
100 0
100
34.59 30.00100
41.58 30.0039.64
TX XT CX X
C
C
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3.2 HEAT AND INTERNAL ENERGY
Heat: an energy that flows from a higher-temperature
object to a lower-temperature object because of the
difference in temperature.
SI unit for heat: Joule (J)
Internal energy (thermal energy): the sum of the
molecular kinetic energy, molecular potential energy and
other kinds of molecular energy in a substance.
Molecular kinetic energy: originates from the random motion (or
vibration) of the molecules.
Molecular potential energy: due to forces that acts between the
atom of a molecule and between molecules.
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3.3 HEAT & TEMPERATURE CHANGE.
3.3.1 HEAT CAPACITY
Heat capacity, C: the amount of heat required to raise the
temperature of a substance by 1 unit (JK-1)
The equation of heat capacity which relates heat, Qand
the temperature change,Tis given by:
Specific heat, c: heat capacity per unit mass (Jkg-1K-1)
QC
T
Qc
m T
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Substance Specific Heat, c
(Jkg-1 C-1)
Aluminium 910
Brass 380Copper 390
Lead 130
Mercury 140
Iron 470
Steel 450
Silver 270
Glass 700
Ice 2100
Water 4200
Steam 2000
Glycerine 2500
Alcohol 2430
Specific heat, cof some common substances
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Example:
How much heat is required to raise the temperature of
200g of mercury from 20C to 100C? Given cmercury
= 140
Jkg-1 C-1.
Solution:
(0.2)(140)(100 20)
2240
final initial
Q mc T
mc T T
J
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Example:
How much heat is required to raise the temperature of
200g of mercury from 250K to 330K? Given cmercury
= 140
Jkg-1 C-1.
Solution:
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Example:
A 1.5 kg material requires 20.25 kJ heat to raise its
temperature from 30C to 80C. State the name of the
material.
Solution:
11270
)3080)()(5.1(20250
CJkgc
CCkgJ
TmcQ
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Substance Specific Heat, c
(Jkg-1 C-1)
Aluminium 910
Brass 380Copper 390
Lead 130
Mercury 140
Iron 470
Steel 450
Silver 270
Glass 700
Ice 2100
Water 4200
Steam 2000
Glycerine 2500
Alcohol 2430
Specific heat, cof some common substances
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Example:
A 1.5 kg material requires 20.25 kJ heat to raise its
temperature from 30C to 80C. State the name of the
material.
Solution:
Therefore the material is silver.
11270
)3080)()(5.1(20250
CJkgC
CCkgJ
TmcQ
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Principle of Conservation of Energy: the heat lost by the
hotter object must equal to the heat gained by the coolerobject.
Heat lost = Heat gained
This principle is based on two assumptions:
No thermal energy lost from the system
No external energy comes into the system
.
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Example:
An aluminium cylinder is heated to 98C and then dropped
into a glass filled with 120g of water with initial temperature
20C. The equilibrium temperature of the mixture is 31C.Calculate the mass of the aluminium cylinder? Neglect the
glass heat capacity.
Solution:
heat lost by aluminium = heat gained by water
(910)(31 98) (0.12)(4200)(31 20)
60970 5544
0.091
Al Al Al w w w
Al
Al
Al
m c T m c T
m
m J
m kg
What if the glass is not negligible?
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To be continued.
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3.4 HEAT AND PHASE CHANGE
There forms of material: solid, liquid or gases.
Solid: atoms or molecules are strongly binding each other
in a rigid crystalline structure by their mutual attraction.
Liquid: molecules have more energy, thus more free to
move.
Gases: molecules have even more energy and are free of
one another.
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Heating a material: temperature of the material will
gradually increaseuntil it reaches a certain stage at which
the phase of the material changes:
Melting point: phase change from solid to liquid
Boiling point: phase change from liquid to gases
Cooling a material: temperature of the material will
gradually decrease until it reaches a certain stage at
which the phase of the material changes: Condensing point: phase change from gases to liquid
Freezing point: phase change from liquid to solid
At melting/freezing point and boiling/condensing point, the
temperature of the phase does not change even if theheat is supplied continuously.
This is because the energy absorbed by the material is
used to separate the molecules further apart than its
previous phase.
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Latent heat: the amount of heat, Qrequired to change
the phase of a unit mass of a substance
Latent heat of fusion, Lf: heat per unit mass for the solid-liquid phase transition (in either direction)
Latent heat of vaporization, Lv: heat per unit mass for the
liquid-gases phase transition (in either direction)
SI unit for latent heat: Jkg-1
QL
m
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Substance Melting
Point
(C)
Latent heat of
fusion, Lf
(kJkg-1)
Boiling
Point
(C)
Latent Heat of
Vaporization, Lv
(kJkg-1)
Aluminium 660 377 2467 11390
Copper 1083 134 2595 5075
Lead 327 25 1620 871
Mercury -39 11.7 357 293
Silver 961 88 2193 2340
Water 0 333 100 2264
Oxygen -219 13.8 -183 214
Nitrogen -210 25.5 -196 200
Table of melting and boiling points of few common substance
with their latent heat of fusion and vaporization
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Temperature, T
Quantity of
heat, Q
100 C
0 C
-20 C
Ice
only
Ice+
water
Water
only
Water
+
steam
Steam
only
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Example:
Calculate the energy required to change 100g of ice at -15C tosteam at 110C.
Explanation:
Calculate the energy according to this order:
To increase temperature from -15C to 0C.
To melt the ice at 0C.
To increase temperature from 0C to 100C.
To vaporize the water into steam at 100C.
To raise the temperature from 100C to 110C.
Hint
To calculate energy required to increase temperature, usespecific heat equation, Q= mcT.
To calculate energy require to change phase, use latent heatequation, Q
L m
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3.5 HEAT TRANSFER
Three principle methods by which heat transfer can
occur:
Conduction
Convection
Radiation
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3.5.1 CONDUCTION
Definition: Conduction is the process of heat flow
that occurs in all forms of material whether in solids,
liquids or gases.
Dominantly occurs in solids.
When cylindrical bar is heated at one end, heat will
flow from the hot end to the cold end by means of
the atomic/molecular collision.
Atoms/molecules that are vibrating at the higher
temperature end collide with atoms/molecules at the
lower temperature end resulting in a net transfer of
heat.
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In metal, conduction process is aided by the motion of
free electrons within the substance. Thus, metals are
good heat conductor. Liquid conduct less heat than solids because the forces
between the atoms are weaker.
Gases are even less efficient as heat conductor because
the atoms of gases are further apart. Hence, in order from the most conductive to the least
conductive:
Solids liquids gases
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A cylindrical bar with length dand uniform cross-sectionAismaintained at temperature Thotand Tcoldat its both end.
Evidences from experiments show that the quantity of heat, Q,transferred per unit time, t, is: Directly proportional to the temperature difference between the two ends;
(T=Thot- Tcold)
Directly proportional to the cross-section area,A.
Inversely proportional to the length, d.
The above statements can be expressed in form of equation:
k= thermal conductivity coefficients of material (Wm-1K-1)
= temperature gradient.
.
A
Direction of heat flow
d
Thot Tcold
hot cold T TQ kAt d
hot cold T T
d
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Substance Thermal Conductivity, k
(Wm-1K-1)
Silver 418
Copper 385
Aluminium 238
Brass 122
Iron 80
Steel 46
Lead 38Mercury 8
Concrete 1.7
Glass (pyrex) 1.1
Brick 1
Water 0.6
Asbestos 0.17
Dry soil 0.14
Wood 0.13
Air 0.023
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Example:
A block of concrete has a cross section of 5m2and
a thickness of 10cm. One side is at 40C and theother is at 20C.What is the rate of heat transfer?
Given kconcrete= 1.7 Wm-1K-1.
Solution:
40 20(1.7)(5)
0.1
1700
hot cold T TQ
kAt d
W
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3.5.2 CONVECTION
Convection of thermal energy occurs in a fluid
(liquids or gases) when warm fluids flow, carrying
energy with them as to replace cooler fluids.
Cannot happen in solids since atoms/molecules in
solids are tightly bound together in their position.
Convection occurs in two forms:
Natural convection, e.g. atmospheric convection and
daily weather variations.
Forced convection, e.g. by means of pumps, fans, etc.to propel fluids and create artificially induced convection
current.
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Heated water at the bottom of the beaker will expandand become more buoyant due to the increase ofvolume per unit mass.
This will cause the density of the hot water to decrease.Hot water will rise towards the upper side of the beaker.
The cooler and denser water from the top will go downto replace the hot water.
So the pattern of circulation is formed within the beakerof water.
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3.5.3 RADIATION Radiation is the heat transfer by the emission of
electromagnetic (EM) waves which carry energy away from theemitting object.
In contrary to the conduction and convection, radiation requiresno medium to transfer heat.
The radiation can travel through vacuum, e.g. from the sun to
the earth. The rate, P, at which an object emits energy via EM radiation is
given by:
(unit: Watt,W)
A= surface area (m2)
T = absolute temperature (K)
= Stefan-Boltzmann constant = 5.67 x 10-8 Wm-2K-4.
= emissivity of the emitting object which has the value
between 0 and 1.
4P AT
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An surface with the maximum emissivity of 1 is said to be
a blackbody radiator, but such surface is ideal and does
not occur in nature.
Any object with absolute temperature above 0K emits
thermal radiation, at the same time it absorbs thermal
energy from surroundings.
Thus, net energy radiated per second by such object isgiven by:
4 4EP A T T
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Example:
A spherical body with diameter of 6.0 cm is maintained at
200C. Assuming that the spherical body is an idealblackbody radiator, at what rate (in Watt) is energy
radiated from the sphere?
Solution:Temperature, T = 200C = 473.15 K
Surface area,A= 4r2= 4(0.03m)2 = 1.13 x 10-2m2
Emissivity for blackbody, = 1
4
8 2(5.67 10 )(1)(1.13 10 )(473)
32
P AT
W
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EXERCISE1. How much heat must be added to 3kg of water to raise its
temperature from 20C to 80C?
2. A total of 0.8kg of water at 20C is placed in a 1.2kW electrickettle. How long a time is needed to raise the temperature of thewater to 100C?
3. In preparing tea, 600g of water at 90C is poured into a 200g chinapot (cpot= 0.84 kJkg
-1C-1) at 20C.What is the final temperature ofthe water?
4. Five kilograms of water at 40C is poured on a large block of ice at0C. How many kilograms of ice melts?
5. 500 kcal of heat is added to 2kg of water at 80C. How muchsteam is produced?
6. Find the minimum amount of ice at -10C needed to bring thetemperature of 500g of water at 20C down to 0C.
7. A 30g ice cubes at 0C is dropped into 200g of water at 30C.What is the final temperature?
8. How much steam at 150C is needed to melt 1kg of ice at 0C?
9. A copper ball of 2cm radius is heated in a furnace at 400C. If itsemissivity is 0.3, at what rate does it radiate energy?
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