PHYS1001

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PHYS1001 Physics 1 REGULAR Module 2 Thermal
Physics HEAT CAPACITY LATENT HEAT
What is cooking all about?
PHYSICS FUN EXCITING SIMPLE
ptC_heat .ppt
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• Overview of Thermal Physics Module
• Thermodynamic SystemsWork, Heat, Internal
  Energy0th, 1st and 2nd Law of Thermodynamics
• Thermal Expansion
• Heat Capacity, Latent Heat
• Methods of Heat TransferConduction, Convection,
  Radiation
• Ideal Gases, Kinetic Theory Model
• Second Law of ThermodynamicsEntropy and Disorder
• Heat Engines, Refrigerators

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? HEAT CAPACITY LATENT HEAT 17.5
p582 Heat Heat capacity (specific heat
capacity) Phase changes Conservation of energy
calorimetry
References University Physics 12th ed Young
Freedman
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What happens when we heat a substance? How does
the temperature change? When does the state of
matter change (phase changes)?
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Phases of matter
Gas - very weak intermolecular forces, rapid
random motion
high temp low pressure
Liquid - intermolecular forces bind closest
neighbours
low temp high pressure
Solid - strong intermolecular forces
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Heating 1.0 kg ice to steam (-20 oC to 120 oC)
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Simple model for heating a substance at a
constant rate
T(t)
boiling point
g
l /g
melting point
l
s / l
s
Q(t)
mcsDT
mclDT
mcgDT
mLf
mLv
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Q m LS
? Phase changes at constant temperature
deposition
LS latent heat of sublimation or heat of
sublimation
sublimation
freezing
condensation
gas
liquid
solid
evaporation
melting
Q m Lf
Q m LV
At melting point Lf latent heat of fusion or
heat of fusion
At boiling point LV latent heat of vaporization
or heat of vaporization
Q gt 0 energy absorbed by substance during phase
change Q lt 0 energy released by substance
during phase change
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? Specific heat or specific heat capacity, c
Tf
DT Tf - Ti
Ti
Mass of object m Specific heat (capacity) c
heat Q
NO phase change during temperature change
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Specific heat
Latent heats
Latent heat phase change (formation or breakage
of chemical bonds requires or releases
energy) Water - large values of latent heats at
atmospheric pressure Lf 3.34x105 J.kg-1 (273
K) Lv 2.26x106 J.kg-1(373 K)
Substance c (J.kg-1.K-1) Aluminum
910 Copper 390 Ice
2100 Water 4190 Steam
2010 Air 1000 Soils /
sand 500
You can be badly scolded by steam more
dangerous than an equivalent amount of boiling
water WHY?
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Latent heat and phase changes As a liquid
evaporates it extracts energy from its
surroundings and hence the surroundings are
cooled. When a gas condenses energy is released
into the surroundings. Steam heating systems
are used in buildings. A boiler produces steam
and energy is given out as the steam condenses in
radiators located in rooms of the building.
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Evaporation and cooling Evaporation rates
increase with temperature, volatility of
substance, area and lower humidity. You feel
uncomfortable on hot humid days because
perspiration on the skin surface does not
evaporate and the body can't cool itself
effectively. The circulation of air from a fan
pushes water molecules away from the skin more
rapidly helping evaporation and hence cooling.
Evaporative cooling is used to cool buildings.
Why do dogs pant? When ether is placed on the
skin it evaporates so quickly that the skin feels
frozen. Ethyl chloride when sprayed on the skin
evaporates so rapidly the skin is "frozen" and
local surgery can be performed.
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Problem C.1 The energy released when water
condenses during a thunderstorm can be very
large. Calculate the energy released into the
atmosphere for a typical small storm.
Where did the water come from?
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Solution Identify / Setup Assume 10 mm of rain
falls over a circular area of radius 1 km h 10
mm 10-2 m r 1 km 103 m volume of water
V ? r2 h ? (106)(10-2) 3?104 m3 mass of
water m ? kg density of water ? 103
kg.m-3 m ? V (103)(3?104) 3?107 kg Latent
heat change of phase Q m L Lv
2.26?106 J .kg-1 Energy released in atmosphere
due to condensation of water vapour
r
h
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Execute Q m LV (3?107)(2.26?106) J 7?1013
J Evaluate The energy released into the
atmosphere by condensation for a small thunder
storm is more than 10 times greater then the
energy released by one of the atomic bombs
dropped on Japan in WW2. This calculation gives
an indication of the enormous energy
transformations that occur in atmospheric
processes.
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Problem C.2 For a 70 kg person (specific heat
3500 J.kg-1.K-1), how much extra released energy
would be required to raise the temperature from
37 C to 40 C? Solution Identify / Setup m 70
kg c 3500 J.kg-1.K-1 ?T (40 37) C 3
C Specific heat capacity Q m c ?T
Execute Q m c ?T
(70)(3500)(3) J 7.4?105 J 0.74
MJ Evaluate
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Water has a very large specific heat capacity
compared to other substances cwater 4190
J.kg-1.K-1 The large heat capacity of water
makes it a good temperature regulator, since a
great amount of energy is transferred for a given
change in temperature. Why is there a bigger
difference between the max and min daily
temperatures at Campbelltown compared to
Bondi? Why is water a good substance to use in a
hot water bottle? Why is the high water content
of our bodies (c 3500 J.kg-1.K-1) important in
relation to the maintenance of a constant core
body temperature?
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CALORIMETRY PROBLEMS This is a very common type
of problem based upon the conservation of energy.
It involved changes in temperature and phase
changes due to heat exchanges. Setup All
quantities are taken as positive in this
method. Identify the heat exchanges (gained or
lost), phase changes and temperature changes.
Conservation of energy energy gained
energy lost
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Problem C.3 How much ice at 10.0 C must be
added to 4.00 kg of water at 20.0 C to cause the
resulting mixture to reach thermal equilibrium at
5.0 C. Sketch two graphs showing the change in
temperature of the ice and the temperature of the
water as functions of time. Assume no energy
transfer to the surrounding environment, so that
energy transfer occurs only between the water and
ice.
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Solution Identify / Setup
ICE gains energy from the water
Ice -10 oC
Ice/water 0 oC
Water 5 oC
WATER losses energy to the ice
Water 20 oC
Water 5 oC
Heat gained by ice Qice heat lost by water
Qwater
Q m c ?T Q m L
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mice ? kg mwater 4.00 kg temperature
rise for ice to melt ?Tice1 0 (10) C
10 C temperature rise of melted ice ?Tice2
(5 0) C 5 C temperature fall for water
?Twater (20 5) C 15 C cice
2100 J.kg-1.K-1 cwater 4190 J.kg-1.K-1
Lf 3.34?105 J.kg-1
NB all temperature changes are positive
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Execute
T
ice
0 oC
t
equilibrium temperature reached
T
water
0 oC
t
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heat gained by ice
Qice mice cice ?Tice1 mice Lf mice cwater
?Tice2 Qwater mwater cwater ?Twater Qice
Qwater mice cice ?Tice1 mice Lf mice cwater
?Tice2 mwater cwater ?Twater
heat lost by water
conservation of energy
alternatively can calculate each term
Evaluate
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Problem C.4 June 2007 Exam Question (5
mark) A sample of liquid water A and a sample of
ice B of identical masses, are placed in a
thermally isolated container and allowed to come
to thermal equilibrium. The diagram below is a
sketch of the temperature T of the samples verses
time t. Answer each of the following questions
and justify your answer in each case. 1 Is the
equilibrium temperature above, below or at the
freezing point of water? 2 Has the liquid
water partly frozen, fully frozen, or not at
all? 3 Does the ice partly melt, or does it
undergo no melting?
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Solution Identify / Setup
Phase change temperature remains constant Q ?
m L Ice melts at 0 oC and liquid water freezes
at 0 oC Temperature change no change in
phase Q m c ?T Ice warms and liquid water
cools Energy lost by liquid water (drop in
temperature) Energy gained by ice (rise in
temperature phase change)
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Execute
(1) Ice increase in temperature initially and
then remains constant when there is a change in
phase. Therefore, the equilibrium temperature
reached is the freezing point. (2) The ice
reaches the freezing point first and the then the
temperature remains constant. As the water cools,
the ice melts. The temperature never rises above
the freezing point, therefore, only part of the
ice melts. (3) The temperature of the water
falls to its freezing point and never falls below
this and hence it is most likely that no liquid
freezes.
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YF Example 17.10 Whats Cooking
Cu pot m 2.0 kg T 150 oC
Water added m 0.10 kg T 25 oC
Final temperature of water and pot Tf? energy
lost by pot energy gained by water 3 possible
outcomes 1 none of water boils 25 oC lt Tf lt
100 oC 2 some of the water boils Tf 100 oC 3
all water boils to steam 100 lt Tf lt 150 oC
have to becareful with such problems
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How do we measure a persons metabolic rate?
Santorio Santorio weighed himself before and
after a meal, conducting the first controlled
test of metabolism, AD 1614.
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A fever represents a large amount of extra energy
released. The metabolic rate depends to a large
extent on the temperature of the body. The rate
of chemical reactions are very sensitive to
temperature and even a small increase in the
body's core temperature can increase the
metabolic rate quite significantly. If there is
an increase of about 1 C then the metabolic rate
can increase by as much as 10. Therefore, an
increase in core temperature of 3 can produce a
30 increase in metabolic rate. If the body's
temperature drops by 3 C the metabolic rate and
oxygen consumption decrease by about 30. This
is why animals hibernating have a low body
temperature. During heart operations, the
person's temperature maybe lowered.