Physics 1
Thermal Physics
Links: http://homepage.mac.com/phyzman/phyz/BOP/2-06ADHT/index.html
and http://www.colorado.edu/physics/2000/index.pl
1
Assignment
•
•
•
•
•
•
•
P&P12:
In class:1(a),3,4(a,b),
6,19,22,23,32,58,59
72oF tooC and -10oC to oF
HW:1(b-d),5,6,8,20,24,37,40,41,61,63,66
Convert 25oF to oC and -210oC to oF
Link:
http://homepage.mac.com/phyzman/phyz/
BOP/2-06ADHT/index.html
2
What do you think? know?
1. Why does popcorn pop?
2. On a camping trip, your friend
tells you that fluffing up a down
sleeping bag before you go to
bed will keep you warmer than
sleeping in the same bag when it
is still crushed. Why?
3
3. Why is it difficult to
build a fire with damp
wood?
4. Why does steam at
o
100 C cause more
severe burns than liquid
o
water does at 100 C?
4
5. Until refrigerators were invented,
many people stored fruits and
vegetables in underground cellars.
Why was this more effective than
keeping them in the open air?
6. In the past, when a baby had a
high fever, the doctor might have
suggested gently sponging off the
baby with rubbing alcohol. Why
would this help?
5
7. Why does water expand when it freezes?
8. Why, during the final construction of the
St. Louis arch, was water sprayed on the
previous sections as the last section was
put in place?
6
Objectives
• 1. Describe thermal energy and compare it
to potential and kinetic energies.
• 2. Describe changes in temperatures of
two objects reaching thermal equilibrium
• 3. Identify various temperature scales, and
convert between them
7
Objectives
• 4. Explain heat as energy transferred
between substances at different
temperatures in one of 3 ways
(conduction, convection, radiation)
• 5. Relate heat and temperature
• 6. Apply principle of energy conservation
to calculate changes in potential, kinetic, &
internal energy
8
Objectives
• 7. Perform calculations with specific heat
capacity
• 8. Interpret the various sections of a
heating curve
9
Objectives
• 9. Recognize that a system can absorb or
release energy as heat in order for work to
be done on or by that system
• 10. Compute work done during
thermodynamic process
• 11. Distinguish between isovolumetric,
isothermal, and adiabatic thermodynamic
processes
10
Objectives
• 12. Illustrate how the first law of
thermodynamics is a statement of
energy conservation
• 13. Calculate heat, work, and change in
internal energy using lst law of T-D
• 14. Apply 1st law of T-D to describe
cyclic processes
• 15. Recognize why 2nd law of T-D
requires 2 bodies at different temps.
11
1.Relate temperature to the kinetic
energy of atoms and molecules
• Temperature scales
• In the USA, the Fahrenheit temperature scale is used.
Most of the rest of the world uses Celsius, and in science
it is often most convenient to use the Kelvin scale.
• The Celsius scale is based on the temperatures at which
water freezes and boils. 0°C is the freezing point of
water, and 100° C is the boiling point. Room temperature
is about 20° C, a hot summer day might be 40° C, and a
cold winter day would be around -20° C.
• To convert between Fahrenheit and Celsius, use these
equations:
•
12
• Convert 72 oF to oC
• C = 5/9 (F-32)
• C = 5/9 (72-32) = 22oC
• Convert -10 oC to oF
• F = 9/5 C + 32
• F = 9/5(-10) + 32 = 14oF
13
Objective 3:
Temperature Scales
Temperature degree scales comparison
14
Celsius to Kelvin:
T = Tc + 273.15
Problem:
1. The lowest outdoor temperature ever
recorded on Earth is -128.6 oF.,
recorded at Vostok Station, Antarctica,
in 1983. What is this temperature on
the Celsius and Kelvin scales?
Answers: -89.22oC, 183.93 K
15
• Obj. #2 - Hotter temperature means
• – more heat present in a substance
• – the faster the molecules of the
substance move.
• Obj #5 – Relate heat and temperature
• Heat units: calorie or joule (amount of
heat energy present in a substance)
• Temperature units: degree
(proportional to heat energy present in
a substance)
16
2.Describe changes in temperatures of
two objects reaching thermal equilibrium
• The temperature of the hotter substance will
decrease. The temperature of the colder
substance will increase. Each change will stop
when the temperatures are the same –
thermal equilibrium. In other words, thermal
energy travels from hot to cold.
• Obj. #4 - Heat energy can be transferred
by
– Convection (motion of fluid), conduction
(touching), or radiation (electromagnetic
waves)
17
Convection
Heat transfer in fluids generally takes place via convection.
Convection currents are set up in the fluid because the
hotter part of the fluid is not as dense as the cooler part,
so there is an upward buoyant force on the hotter fluid,
making it rise while the cooler, denser, fluid sinks. Birds and
gliders make use of upward convection currents to rise, and
we also rely on convection to remove ground-level pollution.
Conduction
When heat is transferred via conduction, the
substance itself does not flow; rather, heat is
transferred internally, by vibrations of atoms and
molecules. Electrons can also carry heat, which is the
reason metals are generally very good conductors of
heat.
18
Radiation
The third way to transfer heat,
in addition to convection and
conduction, is by radiation, in
which energy is transferred in
the form of electromagnetic
waves.
More about electromagnetic
waves in a lot more detail in a
later chapter; an e-m wave is
basically an oscillating electric
and magnetic field traveling
through space at the speed of
light.
19
Specific heat
The amount of energy that must be added to
raise the temperature of a unit mass of a
substance by one temperature unit.
• Units: J/kg K
• For Water: 4180 J/kg K
• For Aluminum: 897 J/kg K
• Which one heats faster?
20
A similar table is on page 413 in the Hewitt
book. The units are J/g oC in that table.
21
• Heat Transfer Heat Transfer
Q = mCΔT = mC (Tf – Ti)
• Q = mCΔT = mC (Tf – Ti)
•
•
•
•
Q, quantity of heat in joule
m, mass of substance in kg
c, specific heat for water in 4186 j/kg K
t, temperature in Celsius
22
23
24
See table 12-1 on page 318
Find the amount of heat needed to change the
temperature of 5.0 g of liquid water from 8.0oC to
100oC.
Q = mcDt = .005kg(4186 j/kgoC) (92oC) = 1.9 x 103 j
Again,
specific heat is the amount of
heat necessary to change one kg
of a substance 1 degree Celsius or
Kelvin.
25
• 12/3 When you turn on the hot water to
wash dishes, the water pipes have to heat
up. How much heat is absorbed by a
copper water pipe with a mass of 2.3 kg
when its temperature is raised from 20.0oC
to 80.0oC?
• Q = mcDt
• Q = (2.3kg)(390J/kgoC)(60.0oC)
• Q = 53820 J or 5.4x104J
26
27
Temperature vs Heat
28
12/4b
29
Melting or Freezing
• There is no temperature change. To
determine heat lost of gained use:
• Q = mHf
Hf is heat of fusion
•
3.33x105 J/kg for water
• How much heat must be absorbed to melt
5.0 g of ice at 0.0 oC?
• Q=mHf = 0.005kg x 3.33 105Jkg-1
• Q = 1.7 x 102 J of heat absorbed
30
Evaporation or Condensation
• Again, no temp. change. Use Q = mHv
• For water Hv = 2.26 x 106 J/kg.
• How much heat is lost when 100.g of
steam condenses at 100oC?
• Q = mHv = 0.100kg (2.26 x 106 J/kg)
• Q = 2.26 x 105 J heat lost
31
> One step…
• How much heat is needed to raise the
temperature of 50.0 g of ice at 0oC and change
it to steam at 100.oC?
• Melt + raise temp + evap
• Q = mHf + mcDt + mHv
• Q = 0.050kg[3.33 105Jkg-1 + (4186 j/kg-1oC-1)(100oC)+
•
+(2.26 x 106 J/kg)]
• Q = 3.01 x 106 J of heat absorbed.
32
Your assignment:
• Find the amount of heat involved:
• #1 to heat 75.3g of water from 20.0oC and
change it to steam at 100.oC.
• #2 to cool 124g of water from 100.oC and
change it to ice at OoC.
33
Specific Heat Capacities
Which
one is
greatest
that you
use
everyday?
OR
J/kgoC
34
Law of Heat Exchange
• Heat lost = heat gained within a closed
system.
35
Law of Heat Exchange
Q lost = Q gained
36
• 0.300 kg of coffee, at a temperature of 95 °C, is
poured into a room-temperature (20.oC) steel
mug, of mass 0.125 kg. Assuming no energy is
lost to the surroundings, what does the
temperature of the mug filled with coffee come
to?
• Applying conservation of energy, the total
change in energy of the system must be zero.
So, we can just add up the individual energy
changes (the Q's) and set the sum equal to zero.
The subscript c refers to the coffee, and m to the
mug.
•
37
Note that room temperature in Celsius is about
20°. Re-arranging the equation to solve for the
final temperature gives:
38
39
The temperature of the coffee doesn't drop by
much because the specific heat of water (or
coffee) is so much larger than that of steel.
This is too hot to drink, but if you wait, heat will
be transferred to the surroundings and the
coffee will cool.
40
Latent Heat is energy transferred
during phase changes
• Latent Heat
• Crystalline materials change phase -- melt and freeze or vaporize
and condense -- at a single, fixed temperature.
Energy is required for a change of phase of a substance. The ratio
of the energy to the mass of the substance involved is called the
latent heat of the substance. This is much like the specific heat we
have just discussed. It is called latent heat because there is no
change or difference in temperature.
•
Latent heat of fusion Hf describes the heat necessary to melt (or
freeze) a unit mass of a substance.
Q = m Hf
• Latent heat of vaporization Hv describes the heat necessary to
vaporize (or condense) a unit mass of a substance.
• Q = m Hv
•
41
Temperature vs Heat
42
Formulas
Temperature change use: Q = mc D t
Melting or freezing: Q = m Hf
Evaporation or Condensation: Q = mHv
Hf is latent heat of fusion, 3.33 x 105 J/kg
Hv is latent heat of vaporization, 2.26 x 106
J/kg
• These values are for water.
•
•
•
•
•
43
44
Problem
• A jar of tea is placed in sunlight until it
reaches an equilibrium temperature of
32oC. In an attempt to cool the liquid to
0oC, which has a mass of l80 g, how much
ice at 0oC is needed? Assume the specific
heat capacity of the tea to be that of pure
liquid water.
45
m tea = 180gm ice = ? which is the mass of the water that
has melted
c tea = c water = 4186 J/kgoC
H f = 3.33 x 105 J/kg and t final = 32oC
Q lost = Q gained tea loses and water gains only melting
the ice
(mcDt)tea = (mHf)ice
m ice
= (mcDt)tea
Hf ice
m ice
= (.180kg)(4186 J/kgoC)(32oC)
3.33x105 J/kg
= 7.2 x 10-2kg
46
1st Law of Thermodynamics
DU = Q - W
The change in thermal energy of
an object is equal to the heat
added to the object minus the
work done by the object.
See Fig 12-11 on
Page 326
47
A heat engine
Transforms heat at high
temperature into mechanical energy
and low-temperature waste heat
A heat pump (refrigerator) absorbs
heat from the cold reservoir and
gives off heat to the hot reservoir.
48
49
50
2nd Law of Thermodynamics
• In the 19th Century French engineer
Sadi Carnot studied the ability of
engines to convert thermal energy into
mechanical energy.
• He developed a logical proof that even
an ideal engine would generate some
waste heat.
• Carnot’s result is best described by the
term entropy which is the measure of
the disorder in a system.
51
The change is entropy, D S, is shown by the
equation:
DS = Q / T
Units: J/K
The change in entropy of an object is equal to
the heat added to the object divided by the
temperature of the object.
Natural processes occur in a direction that
increases the entropy of the universe. All
processes tend toward disorder unless some
action occurs to keep them ordered. i.e., heat
can flow only from hot to cold naturally.
52
Practice Problems
53
12/32 answer
54
55
56
57
•The End
www.internationalhero.co.uk
www.federalinfrared.com
58