Transcript Slide 1

This Week
Heat and Temperature
Water and Ice
Our world would be different if water didn’t
expand
Engines
We can’t use all the energy!
Why is a diesel engine more efficient?
Geysers: You have to be faithful
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Physics 214 Fall 2010
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Heat
Heat is a form of energy and any object has internal
energy in the form of kinetic energy of the atoms or
molecules and in potential energy connected with
the molecular structure and the electromagnetic
forces among the constituents.
If we add energy to an object the internal energy
increases.
Energy can be added in many ways. For example
we can do work such as
the frictional forces on a moving object.
electromagnetic radiation from the sun.
a source of heat like a hair dryer.
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First Law of Thermodynamics
The increase in the internal energy of a system is
equal to the amount of heat added to the system
minus the amount of work done by the system.
U = internal energy
Q = heat that is added or removed
Wsystem = is the work done by the system
Looking at the pictures you can see that the force
pushing on the piston does positive work and
therefore is putting energy into the gas.
The work done by the gas is negative because the
force the piston exerts is to the right and the
movement of the piston is to the left.
ENERGY CONSERVATION
ΔU = Q – Wsystem or ΔU = Q + Wexternal
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Temperature
Heat flows from a hot body to a cold body until they reach thermal
equilibrium.
Temperature is the quantity that measures whether one body is
hotter than another and at thermal equilibrium both bodies have
the same temperature.
The simplest picture
is a gas of free molecules where the energy
is kinetic. A higher temperaturemeans higher
velocities and more stored energy.
Increase U and T increases
http://www2.biglobe.ne.jp/~norimari/science/JavaApp/Mole/e-gas.html
http://jersey.uoregon.edu/vlab/Thermodynamics/index.html
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Measuring temperature
Physical properties of an object change with temperature.
For example mercury expands as the temperature
increases and we can use that expansion to measure T.
There are three temperature scales,
Celsius, Fahrenheit and Kelvin
Mixture of ice and water 00C
320F 2730K
Boiling point of water 1000C 2120F 3730K
Tc = 5/9(Tf – 32)
Tf = 9/5Tc + 32
TK = Tc + 273
Note that the temperature at which water boils depends
on the value of the atmospheric pressure so that at
higher altitudes water boils at a lower temperature
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Absolute zero
An ideal gas obeys the law
PV = constant x T
if we keep V constant the pressure is
proportional to T.
If we measure P versus T at constant volume we
find a temperature where the pressure is zero.
This is absolute zero where the stored internal
energy has it’s lowest possible value.
This is Zero degrees Kelvin
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Heat capacity
For a given object at a specific temperature the
total internal energy depends on how big the
object is. If we add a fixed amount of energy
the temperature will rise but the rise will be
smaller the larger the object is. So we define a
quantity called specific heat
c = is the quantity of heat required to raise unit
mass of a substance by one degree.
Q = mcΔT
New unit
1 calorie = heat required to raise 1 gram of
water 10C
1calorie = 4.186 joules
Note to raise 1kg of water 10C requires 1000cal
Note that internal energy can change without a
temperature change. This is due to internal
rearrangement of the molecules
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Ice
Water
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Water and Ice
Water has an unusual behavior as the
temperature is changed. As we lower the
temperature from room temperature it
shrinks but it 40C it starts to expand and
that is why ice floats.
Remember density ρ = mass/volume so
below 40C the volume increases and the
density decreases.
Above 40C the molecules are tightly
packed. Below 40C the molecules form a
crystalline structure that has holes.
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Change of state
We are used to solids, liquids and gases and that most substances
can be in any of the three states depending on temperature and
pressure. Solids have more orderly structure and the
atoms/molecules are tightly joined. In a liquid the atoms/molecules
are loosely joined. In a gas the atoms/molecules are “free”. When
we tear apart something that is joined energy is required and this is
true as we go from solid to liquid and liquid to gas. Taking water as
an example
80 calories/gram are required to melt ice at 0oC to water at 0oC and
540 calories/gram to convert water to gas at 100oC. When water
freezes 80cal/gm has to be removed and when water condenses 540
cals/gm is released.
80 cals/gm is the latent heat of fusion
540 cals/gm is the latent heat of vaporization
http://jersey.uoregon.edu/vlab/Balloon/
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Gases
Work done by gas = -Fd = -PAd
ΔV = Ad
W = -PΔV
External work done = W = P ΔV
A quantity of gas can be described by three
variables. Pressure, volume and temperature
U = internal energy (kinetic energy of the molecules)
U determines the temperature
Q = heat energy that is put in or taken out
Wgas = work done by the gas when it expands
or is compressed
Wgas is + if it expands – if it is compressed
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Physical Laws for a gas
Energy conservation
ΔU = Q – Wgas
Ideal gas law
PV = NkT (T in degrees Kelvin)
Adiabatic no heat in or out Q = 0
Compression
work is + (ΔU, T increase)
Expansion
work is –
(ΔU, T decrease)
Isothermal T does not change ΔU = 0 Q = W = PΔV
put in heat gas expands take out heat gas must be compressed
Isobaric pressure is kept constant
Put in heat
T increases gas expands (hot air balloon
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Solids and liquids
Q
Temperature depends on the internal energy
Put in heat T rises except at a change of state
We define the specific heat c as the amount of heat
required to raise unit mass of a substance by one degree.
Q = mcΔT
The usual unit is calories/gram/0C
1 cal = 4.186 joules
The specific heat of water is 1cal/gm/0C
Change of state
ice
water
steam
Requires 80 cal/gm to melt ice at 0C to water at 0C
Requires 540cal/gm to make steam at 1000C
To condense steam or freeze water requires the removal of
540 cal/gm or 80cal/gm.
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Heat engines
The use of energy and the conversion of energy is
essential not only in our practical everyday life but is a
requirement for life to exist. The sun puts energy into
the earth and we use such energy sources as gasoline,
coal and nuclear power.
The conversion of one form of energy into another
form of energy is governed by the physical laws. The
generic term used is a heat engine for any system that
takes an energy source and uses it to produce work.
A car engine is a practical example.
Of course there is a practical aspect that an energy
source can become depleted, like oil but we will find
that there is a fundamental law that prevents the
utilization of all the available energy.
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2nd Law of Thermodynamics
Heat engine uses energy QH and produces
work W and releases Qc
Efficiency = ε = W/QH
W = QH – Qc (c = environment)
2nd law
No engine working in a continuous cycle can
take heat at a single temperature and convert
that heat completely to work.
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Carnot Cycle
We can design a “perfect engine”
without friction. The Carnot cycle is is
an ideal engine and an analysis
reveals that
ε=
(TH –Tc)/TH (T in oK)
For example in a gasoline engine TH is the temperature of combustion
and Tc is outside temperature. If we take TH as 1500o and Tc as 300o then
ε = 80%. In practice the efficiency is maybe 35%.
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Car engines
In an engine with spark plugs a gasoline/air mixture is compressed
and the temperature rises and then the spark ignites the mixture
before it reaches the combustion temperature. For increased
efficiency one would like to reach as high a temperature as possible
so fuel injection solves this problem by compressing air and injecting
the fuel after the combustion temperature is reached.
Turbo charged engines force more air into the chamber when the
piston reaches it’s lowest point using power from the engine. Super
chargers do the same thing but with an extra electrical motor. This
causes the engine to run hotter.
Diesel engines use a lower grade of petroleum and also runs at a
much higher temperature
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Review Chapters 10 and 11
W = PΔV for the piston shown
Wexternal = + Wsystem = ΔU = Q – Wsystem or ΔU = Q + Wexternal
Work done on gas = Fd = PAd
ΔV = Ad
W = PΔV
Work done by gas is -W
Temperature scales Celsius, Fahrenheit Kelvin
Mixture of ice and water 00C
320F 2730K
Boiling point of water 1000C 2120F 3730K
Tc = 5/9(Tf – 32)
Tf = 9/5Tc + 32 TK = Tc + 273
c = is the quantity of heat required to raise unit mass of a substance by one degree.
Q = mcΔT
1 calorie = heat required to raise 1 gram of water 10C = 4.186 joules
Change of state (internal energy changes temperature is constant)
80 cals/gm is the latent heat of fusion
540 cals/gm is the latent heat of vaporization
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Gases
Work done by gas = -Fd = -PAd
ΔV = Ad
W = -PΔV
External work done = W =P ΔV
Energy conservation
ΔU = Q – Wgas
Ideal gas law PV = NkT (T in degrees Kelvin)
Adiabatic no heat in or out Q = 0
Compression
work is + (ΔU, T increase)
Expansion
work is –
(ΔU, T decrease)
Isothermal T does not change ΔU = 0 Q = W = PΔV
put in heat gas expands take out heat gas must be compressed
Isobaric pressure is kept constant
Put in heat
T increases gas expands (hot air balloon)
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Heat engines
Efficiency = ε = W/QH
Change in internal energy in one cycle is zero
W = QH – Qc (c = environment)
2nd law
No engine working in a continuous cycle can
take heat at a single temperature and convert
that heat completely to work.
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Geysers (Old Faithful)
Geysers produce a jet of water very often at
equal time intervals. Old Faithful in
Yellowstone erupts every 90 minutes or so.
What is required is a source of heat, a source
of water and a constricted vertical channel
that the water flows into.
As the column of water increases in height the
pressure increases at the bottom of the water
column where the heat is. The increase in
pressure raises the boiling point. This
continues until the water at the bottom starts
to boil.
P = ρgh
This causes an expansion of the column
which reduces the pressure and
immediately takes the temperature of all the
water below the boiling point and there is an
explosive boiling resulting in the eruption.
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Heat
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3E-03 Fire Syringe
Compression and rise in air temperature
What will happen
to the
combustible
material when
the plunger is
rapidly pushed
down ?
Can you guess
the every-day
application of
this
phenomenon ?
This system is analogous
to the combustion cycle
within a diesel engine or
any fuel injected engine.
RAPID COMPRESSION IS ADIABATIC GIVING RAPID RISE OF
AIR TEMPERATURE IN THE CHAMBER WHICH EXCEEDS THE
IGNITION TEMPERATURE OF THE FLAMMABLE MATERIAL.
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Questions Chapter 10
Q1 Is an object that has a temperature of 0°C hotter than, colder
than, or at the same temperature as one that has a temperature of
0°F?
Water freezes at 0o C and 32o F so 0o F is colder
Q2 Which spans a greater range in temperature, a change in
temperature of 10 Fahrenheit degrees or a change of 10
Celsius degrees?
There is 100o C between water freezing and boiling and
180o F so 1oC = 1.8oF so a change of 10oC is larger
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Q4 We sometimes attempt to determine whether another person
has a fever by placing a hand on their forehead. Is this a reliable
procedure? What assumptions do we make in this process?
Body temperatures internally are very similar for all people so
in principle we can tell if a temperature is elevated just like any
other hot object. The assumption is that your hand is at body
temperature.
Q5 Is it possible for a temperature to be lower than 0°C?
Yes. Absolute zero is – 2730 which one can think of as the place
where the molecules of an ideal gas have zero energy
Q6 Is it possible for a temperature to be lower than 0 K on the
Kelvin temperature scale?
Yes. Absolute zero is – 2730K
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Q8 Two objects at different temperatures are placed in contact with
one another but are insulated from the surroundings. Will the
temperature of either object change?
They will exchange heat until they both reach the same temperature
Q10 Two objects of the same mass, but made of different
materials, are initially at the same temperature. Equal amounts of
heat are added to each object. Will the final temperature of the
two objects necessarily be the same?
No. The specific heat, which is the heat energy required to raise
the temperature one degree, is different for each material
Q13 What happens if we add heat to water that is at the
temperature of 100°C? Does the temperature change? Explain.
The water turns into steam at 100o C
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Q21 An ideal gas is compressed without allowing any heat to flow
into or out of the gas. Will the temperature of the gas increase,
decrease, or remain the same in this process?
The temperature will increase ΔU = W
Q23 Heat is added to an ideal gas, and the gas expands in the
process. Is it possible for the temperature to remain constant
in this situation?
Yes. This is an isothermal expansion and W = Q
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Q25 Heat is added to a hot-air balloon causing the air to
expand. Will this increased volume of air cause the balloon to
fall?
Archimedes principle states that the buoyant force is equal to the
weight of liquid displaced. So if the balloon stays the same size
and as the air expands it leaves the balloon it will rise faster
because the weight of the air inside will be less. If no air escapes
but the balloon increases in size it will also rise because the
buoyant force is larger
Q27 A block of wood and a block of metal have been sitting on a
table for a long time. The block of metal feels colder to the touch
than the block of wood. Does this mean that the metal is actually
at a lower temperature than the wood?
No. What you feel is heat flow and the thermal conductivity of metal is much
bigger than that of wood. This is also why you feel much colder when the
air is damp and why trapped dry air in fiberglass is used for insulation. It is
also because room temperature is lower than body temperature
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Ch 10 E 2
Temperature is 14° F.
What is the temperature in Celsius?
Tc= 5/9 (TF – 32) = 5/9 (14 - 32)=5/9 (-18) = -10° C
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Ch 10 E6
How much heat does it take to raise the temperature
of 70 g of H2O from 20°C to 80°C? C= 1 cal/gram/oC.
Q = mcT=(70)(1)(80-20) = 4200 cal
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Ch 10 E 12
Add 600 J to 50 g of H2O initially at 20°C
a) How many calories?
b) What is the final temperature of the H2O
a) 600 /4.186 = 143.3 cal = Q
b) Q=mcT
T=Q/mc=143.3/(50)(1)=2.87°C
TF=22.87°C
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Ch 10 E 16
Add 500 cal of heat to gas. Gas does 500 J of
work on surroundings.
What is the change in internal energy of gas?
U = Q-W
Q = 500/ 4.186 = 2093 J
W=500 J
U = 2093 – 500 = 1593 J
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Ch 10 CP 2
A student’s temperature scale = 0°s = ice point of H2O; 50° s = boiling
point of H2O. The student then measures a temperature of 15°s.
a) What is this temperature in degrees Celsius?
b) What is this temperature in degrees Farenheit?
c) What is this temperature in Kelvins?
d) Is the temperature range spanned by 1°s larger or smaller than spanned by
1°C?
a) Ice point H2O = 0°s=0°C
Boiling point H2O = 50°s=100°C
Obvious relationship: Ts=1/2 Tc, Tc=2Ts
Tc=2Ts=2(15) = 30°C
b) TF = 9/5 Tc+ 32 = 9/5(30) + 32 = 86°F
c) Tk = Tc + 273.2 = 30 + 273.2 = 303.2K
Ts
50°s
0°s
Tc
100°c
0°c
d) 1°s spans 2°C, so that the range spanned by 1°S is LARGER than that
spanned by 1°C.
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Ch 10 CP 4
150 g of metal at 120°C is dropped in a beaker containing 100 g H2O
at 20°C. (Ignore the beaker). The final temperature of metal and water
is 35°C.
a) How much heat is transferred to the H2O?
b) What is the specific heat capacity of metal?
c) Use the same experimental setup – how much metal at 120°C to have final
temperature of metal and water = 70°C?
a) Q = mcT = (100)(1)(35-20) = 1500 cal
b) Q = mcT → c = Q/mT =
c = -1500/(150)(35-120) = 0.12 cal/gram/°C
c) Q H2O = mcT = (100)(1)(70-20) = 5000 cal
Q metal = - Q H2O = - 5000
Q metal = mcT; m = Q metal/cT = -5000/(0.12)(70-120)
m = 833 g
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Ch 11 E 6
A Carnot engine takes in heat at 650 K and releases
heat to a reservoir at 350K.
What is the efficiency?
ec = (TH – Tc) / TH = 650-350 / 650 = 0.46
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Ch 11 E 8
Carnot engine operates b/w 600 K and 400 K and
does 200 J of work in each cycle.
a) What is the efficiency?
b) How much heat does it take in from the high-temp
reservoir during each cycle?
a) εc= (TH- Tc)/TH = (600-400)/600 = 0.33
b) ε = W/QH , QH = W/ε = 200/0.33 = 600J
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Ch 11 CP 2
Carnot engine operates b/w 500° C and 150° C and does
30 J of work in each cycle.
a) What is the efficiency?
b) How much heat is taking in from the high-temp reservoir each cycle?
c) How much heat is released to low-temp reservoir each cycle?
d) What is the change, if any, in the internal energy of gas each cycle?
a) εc= (TH-Tc)/ TH= (500+273.3)-(150+273.3)/(500+273.3) = .045
b) ε = W/QH , QH = W/ε = 30/0.45 = 66.3 J
c) W = QH-Qc, Qc = QH-W = 66.3J - 30J = 36.3J
d) U = Q-W, W = 30J
Q = QH - Qc = 66.3 – 36.3 = 30 J
U = 30-30 = 0
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