Transcript Thermal Processes

Thermal Physics
Energy in Thermal Processes
Energy Transfer
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When two objects of different
temperatures are placed in thermal
contact, the temperature of the warmer
decreases and the temperature of the
cooler increases
The energy exchange ceases when the
objects reach thermal equilibrium
The concept of energy was broadened
from just mechanical to include internal
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Made Conservation of Energy a universal
law of nature
Heat Compared to
Internal Energy
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Important to distinguish between
them
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They are not interchangeable
They mean very different things
when used in physics
Internal Energy
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Internal Energy, U, is the energy
associated with the microscopic
components of the system
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Includes kinetic and potential energy
associated with the random translational,
rotational and vibrational motion of the
atoms or molecules
Also includes any potential energy bonding
the particles together
Thermal Energy
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Thermal Energy is the portion of
the Internal Energy, U, that is
associated with the motion of the
microscopic components of the
system.
Heat
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Heat is the transfer of energy
between a system and its
environment because of a
temperature difference between
them
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The symbol Q is used to represent
the amount of energy transferred by
heat between a system and its
environment
Units of Heat
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Calorie
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An historical unit, before the
connection between thermodynamics
and mechanics was recognized
A calorie is the amount of energy
necessary to raise the temperature of
1 g of water from 14.5° C to 15.5° C .
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A Calorie (food calorie) is 1000 cal
Units of Heat, cont.
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US Customary Unit – BTU
BTU stands for British Thermal Unit
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A BTU is the amount of energy
necessary to raise the temperature of
1 lb of water from 63° F to 64° F
1 cal = 4.186 J
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This is called the Mechanical
Equivalent of Heat
Problem: Working Off
Breakfast
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A student eats breakfast
consisting of two bowls of
cereal and milk, containing a
total of 3.20 x 102 Calories
of energy. He wishes to do
an equivalent amount of
work in the gymnasium by
doing curls with a 25 kg
barbell. How many times
must he raise the weight to
expend that much energy?
Assume that he raises it
through a vertical
displacement of 0.4 m each
time, the distance from his
lap to his upper chest.
h
Problem: Working Off
Breakfast
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Convert his breakfast Calories, E, to
joules:
3

1

10
cal   4.186J 
2
E  (3.2 10 Cal) 

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 1Cal   cal 
 1.34  10 J
6
Problem: Working Off
Breakfast
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Use the work-energy theorem to find the work
necessary to lift the barbell up to its maximum
height.
W  ΔKE  ΔPE  (0  0)  (mgh  0)  mgh
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The student must expend the same amount of
energy lowering the barbell, making 2mgh per
repetition. Multiply this amount by n repetitions
and set it equal to the food energy E:
n(2mgh)  E
Problem: Working Off
Breakfast
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Solve for n, substituting the food energy for E:
E
1.34  10 J
n

2
2mgh
2(25kg)(9. 8m/s )(0.4m)
6
 n  6.84  10 times
3
James Prescott Joule
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1818 – 1889
British physicist
Conservation of
Energy
Relationship
between heat and
other forms of
energy transfer
Specific Heat
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Every substance requires a unique
amount of energy per unit mass to
change the temperature of that
substance by 1° C
The specific heat, c, of a substance
is a measure of this amount
Q
c
m T
Units of Specific Heat
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SI units
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J / kg °C
Historical units
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cal / g °C
Heat and Specific Heat
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Q = m c ΔT
ΔT is always the final temperature
minus the initial temperature
When the temperature increases, ΔT
and ΔQ are considered to be positive
and energy flows into the system
When the temperature decreases, ΔT
and ΔQ are considered to be negative
and energy flows out of the system
A Consequence of
Different Specific Heats
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Water has a high
specific heat
compared to land
On a hot day, the
air above the land
warms faster
The warmer air
flows upward and
cooler air moves
toward the beach
Calorimeter
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One technique for determining the
specific heat of a substance
A calorimeter is a vessel that is a
good insulator which allows a
thermal equilibrium to be achieved
between substances without any
energy loss to the environment
Calorimetry
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Analysis performed using a calorimeter
Conservation of energy applies to the
isolated system
The energy that leaves the warmer
substance equals the energy that enters
the water
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Qcold = -Qhot
Negative sign keeps consistency in the sign
convention of ΔT
Calorimetry with More
Than Two Materials
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In some cases it may be difficult to
determine which materials gain heat
and which materials lose heat
You can start with SQ = 0
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Each Q = m c T
Use Tf – Ti
You don’t have to determine before using
the equation which materials will gain or
lose heat
Phase Changes
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A phase change occurs when the
physical characteristics of the substance
change from one form to another
Common phases changes are
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Solid to liquid – melting
Liquid to gas – boiling
Phases changes involve a change in the
internal energy, but no change in
temperature
Latent Heat
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During a phase change, the amount of
heat is given as
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L is the latent heat of the substance
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Q = ±m L
Latent means hidden
L depends on the substance and the nature
of the phase change
Choose a positive sign if you are adding
energy to the system and a negative
sign if energy is being removed from
the system
Latent Heat, cont.
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SI units of latent heat are J / kg
Latent heat of fusion, Lf, is used
for melting or freezing
Latent heat of vaporization, Lv, is
used for boiling or condensing
Table 11.2 gives the latent heats
for various substances
Problem: Boiling Liquid
Helium
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Liquid helium has a very
low boiling point, 4.2 K,
as well as low latent heat
of vaporization, 2.09 x
104 J/kg. If energy is
transferred to a container
of liquid helium at the
boiling point from an
immersed electric heater
at a rate of 10 W, how
long does it take to boil
away 2 kg of the liquid?
Problem: Boiling Liquid
Helium
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Find the energy needed to vaporize 2 kg of
liquid helium at its boiling point:
Q  mL v  (2kg)(2.09  10 4 J/kg)  4.18  10 4 J
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Divide this result by the power to find the time:
Q mL v 4.18  10 J
Δt  

P
P
10W
4
Δt  4.18  10 s  69.7min
3
Sublimation
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Some substances will go directly
from solid to gaseous phase
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Without passing through the liquid
phase
This process is called sublimation
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There will be a latent heat of
sublimation associated with this
phase change
Graph of Ice to Steam
Warming Ice
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Start with one gram of
ice at –30.0º C
During A, the
temperature of the ice
changes from –30.0º C
to 0º C
Use Q = m c ΔT
Will add 62.7 J of
energy
Melting Ice
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Once at 0º C, the
phase change
(melting) starts
The temperature
stays the same
although energy is
still being added
Use Q = m Lf
Needs 333 J of
energy
Warming Water
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Between 0º C and
100º C, the
material is liquid
and no phase
changes take place
Energy added
increases the
temperature
Use Q = m c ΔT
419 J of energy are
added
Boiling Water
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At 100º C, a
phase change
occurs (boiling)
Temperature does
not change
Use Q = m Lv
2 260 J of energy
are needed
Heating Steam
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After all the water is
converted to steam, the
steam will heat up
No phase change occurs
The added energy goes to
increasing the temperature
Use Q = m c ΔT
To raise the temperature of
the steam to 120°, 40.2 J
of energy are needed
Problem Solving Strategies
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Make a table
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A column for each quantity
A row for each phase and/or phase
change
Use a final column for the
combination of quantities
Use consistent units
Problem Solving
Strategies, cont
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Apply Conservation of Energy
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Transfers in energy are given as
Q=mcΔT for processes with no phase
changes
Use Q = m Lf or Q = m Lv if there is a
phase change
In Qcold = - Qhot be careful of sign
ΔT is Tf – Ti
Solve for the unknown
Your Turn
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You start with 250. g of ice at -10
C. How much heat is needed to
raise the temperature to 0 C?
10.5 kJ
How much more heat would be
needed to melt it?
83.3 kJ
Your Turn
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You start with 250. g of ice at -10
C. What will happen if we add 50.
kJ of heat?
10.5 kJ will be used to warm it up
to the MP, and the rest will start
melting the ice.
0.119 kg will be melted
Problem: Partial melting
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A 5 kg block of ice at 0o C is added to an
insulated container partially filled with 10
kg of water at 15 o C.
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(a) Find the temperature, neglecting the heat
capacity of the container.
(b) Find the mass of the ice that was melted.
Problem: Partial melting
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(a) Find the equilibrium temperature.
 First, Compute the amount of energy
necessary to completely melt the ice:
Q melt  mice L f  (5kg)(3.33  10 J/kg)
5
 1.67  10 J
6
Problem: Partial melting
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Next, calculate the maximum energy that can be lost by
the initial mass of liquid water without freezing it:
Qwater  mwatercΔT
 (10kg)(419 0J/kg. o C)(0 o C  15o C)
 6.29  105 J
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This is less than half the energy necessary to melt all
the ice, so the final state of the system is a mixture of
water and ice at the freezing point:
T0 C
o
Problem: Partial melting
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(b) Compute the mass of the ice melted.
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Set the total available energy equal to the heat of fusion
of m grams of ice, mLf:
6.29  10 J  mL f  m(3.33  10 J/kg)
5
5
 m  1.89kg
Final Problem
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100. grams of hot water ( 60. C) is
added to a 1.0 kg iron skillet at
500 C. What is the final
temperature and state of the
mixture?
Final Problem
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16.7 kJ needed to warm water to BP.
226 kJ needed to vaporize water
199.2 kJ will be given up by skillet.
Final temperature will be 100. C
182 kJ of heat from the skillet will be
available to vaporize water
81 grams of water will vaporize.