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ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
CHAPTER 3: TEMPERATURE AND HEAT
3.1 Temperature and Thermometers
Temperature is a physical property of all bodies and is a relative measure or
indication of hotness and coldness of objects.
Thermometers are devices used to measure the temperature of a system.
Three common scales of a temperature measurement:
1. Celsius
2. Fahrenheit
3. Kelvin
Table shows a comparison of the Celsius, Fahrenheit and Kelvin temperature
scale
Celsius Fahrenheit Kelvin
scale scale scale
Steam point 100 C 212 F 373.15 K
Ice point 0 C 32 F 273.15 K
On the Celsius temperature scale, there are 100 equal divisions between the
ice point (0C) and the steam point (100C).
On the Fahrenheit temperature scale, there are 180 equal divisions between
the ice point (32F) and the steam point (212F).
The relationship between Celsius, TC and Fahrenheit, TF temperature scale is
9
TF TC 32
5
3.2 The Kelvin Temperature Scale
For scientific work, the Kelvin temperature scale is the scale of choice.
As shown in the table above, the ice point (0C) occurs at 273.15K on the
Kelvin scale.
Thus, the Kelvin temperature, T and the Celsius temperature are related by,
T TC 273 .15
Thus, the Kelvin temperature, T and the Fahrenheit temperature are related
by,
TF 9 T 273.15 32.0
5
The number 273.15 in the equation is an experimental result, obtained in
studies that utilize a gas-based thermometer.
Gas thermometer is a standard instrument because its measurement is
independent from any type of gas.
One type of gas thermometer is the constant-volume gas thermometer.
The physical property used in a constant-volume gas thermometer is the
pressure variation with temperature of a fixed volume of gas.
A constant-volume gas thermometer consists of a gas-filled bulb to which a
pressure gauge is attached, as in figure below.
DEPARTMENT OF ENGINEERING 41
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
Figure shows a constant-volume gas thermometer
The basic idea of this thermometer is that the volume of gas is held constant
by raising or lowering the right column of the U tube manometer in order to
keep the mercury level in the left column at the same reference level.
With the mercury level that can be adjusted, the gas occupies a constant
volume and its pressure is related to
P2= P1 + gh
where is the density of mercury
P1 = 0 Pa (evacuated)
P2 = Pgas
As the temperature changes, the pressure changes. The temperature
increases when a gas confined to a fixed volume is heated and decreases
when the gas is cooled.
If the results are plotted on a pressure versus temperature graph, a straight
line is obtained as in figure below.
Figure shows a plot of absolute pressure versus temperature for a low
density gas at constant volume
If the straight line is extended to lower temperatures where the pressure is
zero, the line crosses the temperature axis at -273.15C. This is the location
of absolute zero.
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ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
EXERCISE 1
(a) “Room temperature” is often taken to be 68°F. What is this on the Celsius
scale?
(b) The temperature of the filament in a lightbulb is about 1800°C. What is this on
the Fahrenheit scale?
SOLUTION
(a) T C T F 32 68 32 20 C
o 5
9
o
5
9
o
(b) T F T C 32 1800 32 3300 F
o 9
5
o 9
5
o
EXERCISE 2
What’s your normal body temperature? It may not be 98.6F, the oft-quoted
average that was determined in the nineteenth century. What is the difference
between these averages expressed in Celsius degrees?
SOLUTION
The difference between these two averages, expressed in Fahrenheit degrees, is
98.6 F 98.2 F 0.4 F
9
Since 1 C° is equal to F°, we can make the following conversion
5
1 C
(0.4 F) 0.2 C
(9/5) F
3.2 Thermal Expansion
When the temperature of a body increases, the average distance between the
atoms increases and the volume of the body also increases.
The phenomenon is known as thermal expansion.
Thermal expansion is a physical phenomenon in which the increment in
temperature can cause substances to expand.
Linear Expansion
Figure below illustrates the linear expansion of a rod whose length is Lo when
the temperature is To.
Figure shows when the temperature of a rod is raised by T, the length
of the rod increases by L
For linear expansion, an object of length Lo experiences a change L in length
when the temperature changes by T
L L0 T where is the coefficient of linear expansion
OR L – Lo = Lo T
OR L L0 (1 T ) (in terms of final length)
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ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
Thermal stress
For an object held rigidly in place, a thermal stress can occur when the object
attempt to expand or contract.
Thus, the force / stress needed to prevent a solid object from expanding must
be strong enough to counteract any change in length due to a change in
temperature.
F L L T
Stress Y Y 0 YT
A L0 L0
EXERCISE 3
An aluminum bar has the desired length when at 15°C. How much stress is
required to keep it at this length if the temperature increases to 35°C?
SOLUTION
The thermal stress must compensate for the thermal expansion. Y is Young’s
modulus for the aluminum.
Stress F A Y T 25 106 Co 70 10
9
N m2 35 C 15 C 3.5 10
o o 7
N m2
EXERCISE 4
A steel beam is used in the roadbed of a bridge. The beam is mounted between
two concrete supports when the temperature is 23C, with no room provided for
thermal expansion. What compressional stress must the concrete supports apply
to each end of the beam, if they are to keep the beam from expanding when the
temperature rises to 42C?
SOLUTION
F L Lo T
Y Y Y T
Stress = A Lo Lo
YT (2 x1011 )(12 x106 )(19) 4.6 x107 N / m2
The Bimetallic Strip
Each substance has its own characteristic coefficient of expansion.
For example, brass rod expands more than the steel rod because brass has a
larger coefficient of expansion than steel.
A simple device that uses this principle is a bimetallic strip.
A bimetallic strip is made from two thin strips of metal that have different
coefficients of linear expansion as shown in figure below
Figure shows: (a) A bimetallic strip and how it behaves when (b)
heated and (c) cooled
DEPARTMENT OF ENGINEERING 44
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
APPLICATION
Bimetallic strips are frequently used as adjustable automatic switches in
electrical appliances.
Refer figure below.
In part (a), the electricity passes through the heating coil that heat the water.
The electricity can flow because bimetallic strip touches the contact on the
strength adjustment knob, thus providing a continuous path of the electricity.
In part (b), when the bimetallic strip gets hot enough to bend away, the
contacts separate and the electricity stops because it is no longer has a
continuous path along which to flow and the brewing cycle is shut off.
NOTE:
Expansion of holes-Let’s say we have a hole in piece of solid material. When
heated, the material will expand, but what about the hole??? The hole also
expands at the same rate and at the same direction (outwards)-refer to the
diagram.
(a) Unheated
(b) Heated –
dotted line
represents after
heated
Volume Thermal Expansion
For volume expansion, the change V in the volume of an object of volume Vo
is given by V 3V0 T or V V0 T
where is the coefficient of volume expansion and is equal to 3
= 3 (liquids expands more than solids)
DEPARTMENT OF ENGINEERING 45
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
EXERCISE 5
A rod made from a particular alloy is heated from 25 0C to the boiling point of
water. Its length increases by 8.47 10-4 m. The rod is then cooled from 250C to
the freezing point of water. By how much does the rod shrink?
SOLUTION
Assuming that the rod expands linearly with heat, we first calculate the quantity
L/ T using the data given in the problem.
L 8.47 10–4 m –5
1.13 10 m/C
T 100.0 C – 25.0 C
Therefore, when the rod is cooled from 25.0 °C, it will shrink by
L (1.13 10 –5 m/C) T
(1.13 10 –5 m/C) (0.00 C – 25.0 C) –2.82 10 –4 m
EXERCISE 6
A cylinder of diameter 1.00000cm at 30C is to be slid into a hole in a steel plate.
The hole has a diameter of 0.99970cm at 30C. To what temperature must the
plate be heated? For steel, = 1.2 x 10-5C-1
SOLUTION
L= LoT
L
T=
Lo
L= 1 10-2 – 9.997 10-3 = 3 10-6m
= 1.2 10-5C-1
Lo= 9.997 10-3m
3 106
T= = 25C
1.2 105 (9.997 103 )
The temperature of the plate must be = 30 + 25 = 55C
EXERCISE 7
The brass bar and the aluminum bar in the drawing are attached to an immovable
wall. At 28C the air gap between the rods is 1.3 x 10 -3m. At what temperature
will the gap be closed?
SOLUTION
L = L0T gives for the expansion of the aluminum
LA = ALAT
and for the expansion of the brass
LB = BLBT
Taking the coefficients of thermal expansion for aluminum and brass from table
adding Equations (1) and (2), and solving for T give
DEPARTMENT OF ENGINEERING 46
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
L A L B 1.3 10 3 m
T 21C
A L A BL B 23 10 6 C 1 1.0 m 19 10 6 C 1 2.0 m
The desired temperature is then
T = 28 °C + 21 C° = 49 C
EXERCISE 8
A lead object and a quartz object each have the same initial volume. The volume
of each increases by the same amount because the temperature increases. If the
temperature of the lead object increases by 4.0C., by how much does the
temperature of the quartz increase?
SOLUTION
Recognizing that the lead and quartz objects experience the same change in
volume and expressing that change with Equation 12.3, we have
LeadV0TLead QuartzV0TQuartz
VLead VQuartz
In this result V0 is the initial volume of each object. Solving for TQuartz and
taking values for the coefficients of volume expansion for lead and quartz from
Table gives
1
Lead TLead 87 10 C 4.0 C
6
TQuartz 1
230 C
Quartz 1.5 106 C
EXERCISE 9
A glass is filled “to the mark” with 50.00cm3 of mercury at 18C. If the flask and
its contents are heated to 38C, how much mercury will be above the mark? glass
= 8.5 x 10-6C-1 and βmercury = 182 x 10-6C-1
SOLUTION
Both of glass and mercury will expand
Vmercury= βmercuryVoT
= 182 10-6 (5 10-5) (38 – 18)
= 1.82 10-7m3
Vglass= βglassVoT
= 3glassVoT
= (3 8.5 10-6) (5 10—5) (38 -18) = 2.55 10-8m3
Volume of mercury above mark = Vmercury - Vglass
= 1.82 10-7 – 2.55 10-8 = 1.565 10-7m3
EXERCISE 10
Suppose that the steel gas tank in your car is completely filled when the
temperature is 170C. How many gallons will spill out of the twenty gallon tank
when the temperature rises to 35 0C?
DEPARTMENT OF ENGINEERING 47
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
SOLUTION
Both the gasoline and the tank expand as the temperature increases. The
coefficients of volumetric expansion g and s for gasoline and steel are available
in Table.
The volume expansion of the gasoline is
Vg gV0 T 950 10 –6 C 1 (20.0 gal)(18 C) 0.34 gal
While the volume of the steel tank expands by an amount
V V T 10 C
s s 0 36 –6 1
(20.0 gal)(18 C) 0.013 gal
Vg Vs 0.33 gal
The amount of gasoline which spills out is
3.5 Heat and Internal Energy
Heat is defined as the energy that flows from a higher temperature
object to a lower temperature object because of the difference in
temperatures. (Refer figure)
The SI unit is Joule , J
Heat is sometimes measured with a unit called the kilocalorie (kcal)
The conversion factor between kilocalories and joule is known as the
mechanical equivalent of heat:
1 kcal = 4186 joule or 1 cal = 4.186 joule
Internal energy of a substance is the sum of the kinetic energy (due to the
random motion of the molecules), the potential energy (due to forces that act
between the atoms of the molecule and between molecules) and other kinds
of energy that the molecules of the substance have.
Figure shows how heat as energy is transit from hot to cold.
(a) Heat flows from the hotter coffee to the colder hand.
(b) Heat flows from the warmer hand to the colder glass of ice water.
DEPARTMENT OF ENGINEERING 48
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
3.6 Heat and Temperature Change: Specific Heat Capacity
Solids and Liquids
Greater amounts of heat are needed to raise the temperature of solids or
liquids to higher values
Similar concepts apply when the temperature is lowered, except that heat
must be removed.
The heat, Q that must be applied or removed to change the temperature of a
substance of mass, m by an amount T is
Q mcT
where c is the specific heat capacity of the substance.
Unit for specific heat capacity: J / (kg C)
Calorimetry
To determine the specific heat of a solid or liquid, we conduct the following
steps:
i. raise the temperature of the substance to some value
ii. place it in a vessel containing cold water of known mass and
temperature
iii. measure the temperature of the combination after the equilibrium is
reached.
Vessels having this property are called calorimeters and analysis performed
using such vessels is called calorimetry.
The principle of conservation of energy for perfectly insulated container
requires that heat lost by warmer materials equals heat gained by
cooler materials. (That is if there is no heat loss to the external
surroundings)
From the principle of conservation of energy for the isolated system,
heat loss by the hotter object = heat gained by the cooler object.
Q 0
Q gained Qlost
[mc (T )] gained [mc (T )] lost
**When calculating heat contributions, always write any temperature
changes as the higher minus the lower temperature.
EXERCISE 11
A 0.4kg iron horseshoe that is initially at 500C is dropped into a bucket
containing 20kg of water at 22C. What is the final equilibrium temperature?
Neglect any energy transfer to or from the surroundings
SOLUTION
- Qhot = Qcold
(mCT)iron = (mCT)water
(0.4) (452) (500 – Tf) = (20) (4186) (Tf – 22)
(90400 – 180.8Tf) = (837201Tf – 1341840)
Tf= 39.3C
DEPARTMENT OF ENGINEERING 49
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
EXERCISE 12
An aluminum cup contains 225g of water and a 40g copper stirrer, all at 27C. A
400g sample of silver at an initial temperature of 87C is placed in the water. The
stirrer is used to stir the mixture gently until it reaches its final equilibrium
temperature of 32. Calculate the mass of the aluminum cup.
SOLUTION
The total energy absorbed by the cup, stirrer and water equals the energy given
up by the silver sample. Thus,
(mcup cal + mstirrer ccopper + mwater cwater) (Twater) = (mcT)silver
Tsilver
mcup = [ ( msilver csilver - mstirrerccopper - mwater ) ]
Twater
1 87 32
mcup = [ (0.4) (235) - 0.04 (387) – 0.225
9 10 32 27
2
mcup = 0.085kg
EXERCISE 13
An engineer wishes to determine the specific heat of a new alloy. A 0.15kg
sample of the alloy is heated to 540C. It is then quickly placed in 400g of water
at 10.0C, which is contained in a 200g aluminum calorimeter cup. The final
temperature of the mixture is 30.5C. Calculate the specific heat capacity of the
alloy. (The specific heats of water and aluminum are 4186J/kg C and 900J/kgC)
SOLUTION
(heat lost by sample) = (heat gained by water) + (heat gained by calorimeter
cup)
ms cs Ts mwcw Tw mcal ccal Tcal
(0.15)cs (540 30.5) (0.4)(4186)(30.5 10) (0.2)(900)(30.5 10)
76.4cs 34300 3700
cs 500 J / kg C
EXERCISE 14
A thermometer has a mass of 31.0 g, a specific heat capacity, c = 815 J/kgC and
a temperature of 12.0C. It is immersed in 119 g of water and the final
temperature of the water and thermometer is 41.5C. What was the temperature
of the water before the insertion of the thermometer?
SOLUTION
Since there is no heat lost or gained by the system, the heat lost by the water in
cooling down must be equal to the heat gained by the thermometer in warming
up. The heat Q lost or gained by a substance is given by Equation as Q = cmT,
where c is the specific heat capacity, m is the mass, and T is the change in
temperature. Thus, we have that
cH O mH O TH O ctherm mtherm Ttherm
2 2 2
Heat lost by water Heat gained by thermometer
We can use this equation to find the temperature of the water before the
insertion of the thermometer.
Solving the equation above for TH 2 O and using the value of cH O from Table we
2
have
DEPARTMENT OF ENGINEERING 50
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
ctherm mtherm Ttherm
TH O
2
cH O mH O
2 2
815 J/ kg C 31.0 g 41.5 C 12.0 C
1.50 C
4186 J/ kg C 119 g
The temperature of the water before the insertion of the thermometer was
T 41.5 C 1.50 C = 43.0 C
3.7 Heat and Phase Change: Latent Heat
In some situations, the transfer of energy does not result in a change in
temperature.
This occurred when the substance change from one form to another, which
referred to as a phase change.
Figure below summarizes the three familiar phases of matter (solid, liquid and
gas) and the phase changes that can occur between any two of them
melting evaporating
solid liquid gas
freezing condensing
subliming
solid gas
condensing
The heat, Q that must be supplied or removed to change the phase of a
mass, m of a substance is
Q mL
where L is the latent heat of the substance. (Unit: J/kg)SI unit of L=J / kg
Heat of fusion: the heat required to change 1.0 kg of substance from the
solid to the liquid state.
Heat of vaporization: the heat required to change a substance from the
liquid to the vapor phase.
Figure below displays a graph that indicates what typically happens when
heat is added to a material that changes phase.
DEPARTMENT OF ENGINEERING 51
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
The graph shows the way temperature of water changes as heat is added,
starting with ice at -40C. The pressure is atmospheric pressure.
a) From -40C to 0C: no phase change occurs. The temperature of
the ice changes as heat is added. Amount of energy added is Q =
mciceT.
b) At 0C: no temperature change occurs. the ice-water mixture
remains at this temperature even though energy is being added until
all the ice melts (phase change occurs) to become water at 0C. The
energy required to melt is Q = mLf
c) From 0C to 100C: no phase change occurs. The energy added to
the water is used to increase its temperature. The amount of energy
necessary to increase the temperature from 0C to 100C is Q =
mcwaterT.
d) At 100C: another phase change occurs as the water changes to
steam. The water-steam mixture remains at constant temperature,
100C even though energy is being added until all the liquid has been
converted to steam. The energy required to convert water to steam at
100C is Q = mLv
e) Above 100C: during this portion of the curve, no phase change
occurs, so all the energy added is used to increase the temperature of
the steam. The energy that must be added to raise the temperature of
the steam is Q = mcsteam T.
EXERCISE 15
Suppose that the amount of heat removed when 3.0 kg of water freezes at 00C
were removed from ethyl alcohol at its freezing/melting point of -1140C. How
many kilograms of ethyl alcohol would freeze?
SOLUTION
We want the same amount of heat removed from the water as from the ethyl
alcohol, i.e., Qwater = Qalcohol
or (mLf)water = (mLf)alcohol
DEPARTMENT OF ENGINEERING 52
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
Lf water 33.5 10 4 J/kg
m alcohol m water 3.0 kg 9.3 kg
L f alcohol 10.8 10 4 J/kg
EXERCISE 16
A woman finds the front windshield of her car covered with ice at -120C. The ice
has a thickness of 4.5 x 10-4 m, and the windshield has an area of 1.25 m 2. The
density of ice is 917 kg/m3. How much heat is required to melt the ice?
SOLUTION
The heat required is Q = mLf + cmT, where m = V.
Thus, Q = VLf + cVT
–4 4 3
Q =(917)(4.50 10 )(1.25){33.5 10 +[2.00 10 ](12.0}
1.85 10 5 J
EXERCISE 17
Find the mass of water that vaporizes when 2.1 kg of mercury at 205 C is added
to 0.11 kg of water at 80.0C
SOLUTION
From the conservation of energy, the heat lost by the mercury is equal to the
heat gained by the water. As the mercury loses heat, its temperature decreases;
as the water gains heat, its temperature rises to its boiling point. Any remaining
heat gained by the water will then be used to vaporize the water.
According to Equation, the heat lost by the mercury is Qmercury (cmT )mercury .
The heat required to vaporize the water is, from Equation, Qvap (m vap L v )water .
Thus, the total amount of heat gained by the water is
Qwater (cmT )water (mvap Lv ) water .
Qlost by Qgained by
mercury water
(cmT ) mercury (cmT ) water ( mvap Lv ) water
where Tmercury (205 C 100.0 C) and T water (100.0 C 80.0 C)
The specific heats of mercury and water are given in Table, and the latent heat of
vaporization of water is given in Table. Solving for the mass of the water that
vaporizes gives
T
c m e r cur y m e r cur y m e r cur y c wa te rmwa te rTwa te r
m
m vap
( Lv ) wa te r
[139 J/(kg C)](2.10 kg)(105 C) [4186 J/(kg C)](0.110 kg)(20.0 C )
22.6 10 5 J/kg
9.49 10 –3 kg
EXERCISE 18
A cube of ice is taken from the freezer at 8.5º C and placed in a 95-g aluminum
calorimeter filled with 310 g of water at room temperature of 20.0°C. The final
situation is observed to be all water at 17.0°C. What was the mass of the ice
cube?
DEPARTMENT OF ENGINEERING 53
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
SOLUTION
The heat lost by the aluminum and 310 g of liquid water must be equal to the
heat gained by the ice in warming in the solid state, melting, and warming in the
liquid state.
mAl cAl Ti Al Teq mH2 O cH2 O Ti H2 O Teq mice cice Tmelt Ti ice Lfusion cH2 O Teq Tmelt
0.095 kg 900 J kg Cº 3.0 Cº 0.31 kg 4186 J kg Cº 3.0 Cº
mice 9.90 10 3 kg
2100 J kg Cº 8.5Cº 3.3 10 5 J kg 4186 J kg Cº 17 Cº
EXERCISE 19
A 0.2 kg piece of aluminum that has a temperature of -155C is added to 1.5 kg
of water that has a temperature of 3.0C. At equilibrium the temperature is 0.0C.
Ignoring the container and assuming that the heat exchanged with the
surroundings is negligible, determine the mass of water that has been frozen into
ice.
SOLUTION
Using the energy-conservation principle
cAluminum mAluminum TAluminum cWater mWater TWater mIce Lf, Water
Heat gained by aluminum Heat lost by water Heat lost by water that freezes
Solving for mIce, taking values for the specific heat capacities from Table, and
taking the latent heat for water from Table , we find that
cAluminum mAluminum TAluminum cWater mWater TWater
mIce
Lf, Water
9.00 102 J/ kg C 0.200 kg 0.0 C 155 C
33.5 10 J/kg
4
4186J/ kg C 1.5 kg 3.0 C 0.0 C
0.027 kg
33.5 104 J/kg
EXERCISE 20
An unknown material has a normal melting/freezing point of -25C and the liquid
phase has specific heat capacity of 160 J/kgC. One –tenth of a kilogram of the
solid at -25C is put into a 0.15kg aluminum calorimeter cup that contains 0.1kg
of glycerin. The temperature of the cup and the glycerin is initially 27C. All the
unknown material melts, and the final temperature at equilibrium is 20.0C. The
calorimeter neither losses energy nor gains energy from the external
environment. What is the latent heat of fusion of the unknown material?
SOLUTION
The system is comprised of the unknown material, the glycerin, and the
aluminum calorimeter. From the principle of energy conservation, the heat
gained by the unknown material is equal to the heat lost by the glycerin and the
calorimeter. The heat gained by the unknown material is used to melt the
material and then raise its temperature from the initial value of –25.0 °C to the
final equilibrium temperature of Teq 20.0 C .
DEPARTMENT OF ENGINEERING 54
ESM 1224, SEMESTER 3, 2006/2007 CHAPTER 3
Qgained by Qlost by Qlost by
unknow n gly cerine calorim eter
m u Lf c u mu Tu c gl m gl Tgl c alm al Tal
Taking values for the specific heat capacities of glycerin and aluminum from
Table, we have
(0.10 kg)Lf [160 J/(kg C)](0.10 kg)(45.0 C) [2410 J/(kg C)](0.100 kg)(7.0 C )
[9.0 10 2 J/(kg C)](0.150 kg)(7.0 C )
Lf 1.9 10 4 J/kg
Solving for Lf yields,
DEPARTMENT OF ENGINEERING 55
