Transcript Physics 106P: Lecture 1 Notes

Physics 101: Chapter 12
Temperature & Thermal Expansion

Textbook Sections 12.1 - 12.9
DEMO
UB, Phy101: Chapter 12, Pg 1
Internal Energy and Temperature


All object have “internal energy” (measured in Joules)
random motion of molecules
» kinetic energy
collisions of molecules gives rise to pressure
Amount of internal energy depends on
temperature
» related to average kinetic energy per molecule
how many molecules
» mass
“specific heat”
» related to how many different ways a molecule can
move
translation
 rotation
 vibration

» the more ways it can move, the higher the specific
heat
UB, Phy101: Chapter 12, Pg 2
UB, Phy101: Chapter 12, Pg 3
Temperature Scales
Celcius
Farenheit
Kelvin
212
100
373.15
32
0
273.15
Water boils
Water freezes
DEMO
9
F  C  32
5
K  C  273
5
C  F - 32
9
C  K - 273
NOTE: K=0 is “absolute zero”, meaning (almost) zero KE/molecule
UB, Phy101: Chapter 12, Pg 4
UB, Phy101: Chapter 12, Pg 5
UB, Phy101: Chapter 12, Pg 6
Thermal Expansion


When temperature rises
molecules have more kinetic energy
» they are moving faster, on the average
consequently, things tend to expand
amount of expansion depends on…
change in temperature
original length
coefficient of thermal expansion
» L0 + L = L0 +  L0 T
» L =  L0 T (linear expansion)
» V =  L0 T (volume expansion)
Expansion DEMO
Use expansion
in thermostat
Temp: T
L0
Temp: T+T
L
UB, Phy101: Chapter 12, Pg 7
UB, Phy101: Chapter 12, Pg 8
Not So Useful…
UB, Phy101: Chapter 12, Pg 9
Chapter 12, Preflight
As you heat a block of aluminum from 0 C to 100 C its density
1. Increases
2. Decreases
CORRECT
3. Stays the same
T=0C
M, V0
r0 = M / V0
T = 100 C
M, V100
r100 = M / V100
< r0
UB, Phy101: Chapter 12, Pg 10
Chapter 12, Preflight
Not being a great athlete, and having lots of money to spend, Gill Bates
decides to keep the lake in his back yard at the exact temperature
which will maximize the buoyant force on him when he swims. Which of
the following would be the best choice?
1. 0 C
1000.00
2. 4 C
CORRECT
999.95
999.90
3. 32 C
999.85
4. 100 C
999.80
Density
999.75
5. 212 C
FB = rlVg
999.70
999.65
999.60
999.55
0
2
4
6
8
10
The answer is 4 C, because water has its
greatest density at 4 C. Since bouyant force is
equal to the weight of the displaced fluid or
density*volume, when the density is the largest,
the bouyant force is maximized.
UB, Phy101: Chapter 12, Pg 11
Chapter 12, Preflight
An aluminum plate has a circular hole cut in it. A copper ball (solid
sphere) has exactly the same diameter as the hole when both are at
room temperature, and hence can just barely be pushed through it. If
both the plate and the ball are now heated up to a few hundred degrees
Celsius, how will the ball and the hole fit ?
1. The ball wont fit through the hole any more
2. The ball will fit more easily through the hole
CORRECT
3. Same as at room temperature
The aluminum plate and copper ball both have
different coefficients of thermal expansion.
Aluminum has a higher coefficient than
aluminum which means the aluminum plate hole
will expand to be larger than the copper ball's
expansion and allow more space for the ball to
pass through.
UB, Phy101: Chapter 12, Pg 12
Why does the hole get bigger when the plate expands ???
Imagine a plate made from 9 smaller pieces.
Each piece expands.
If you remove one piece, it will leave an “expanded hole”
Object at temp T
Same object at higer T:
Plate and hole both get larger
UB, Phy101: Chapter 12, Pg 13
UB, Phy101: Chapter 12, Pg 14
UB, Phy101: Chapter 12, Pg 15
UB, Phy101: Chapter 12, Pg 16
UB, Phy101: Chapter 12, Pg 17
Chapter 12, Preflight
You measure your body temperature with a thermometer calibrated in
degrees Kelvin. What do you hope the reading is (assuming you are not
trying to fake some sort of illness) ?
1. 307 K
correct
2. 310 K
3. 313 K
4. 317 K
F  98.6
5
C  (F - 32)  37
9
K  C  273  310
UB, Phy101: Chapter 12, Pg 18
UB, Phy101: Chapter 12, Pg 19
How To Change the Temperature of a System:


Add heat
 Q: heat that flows into a system.
Q = cmT
 Q = amount of heat that must be supplied to raise the temperature
of mass m by an amount T
» Units of Q: Joules or calories


1 cal = 4.186 J
1 kcal = 1 Cal = 4186 J (this is what you read on the label)
 c = specific heat capacity: Heat required to raise 1 kg by 1oC.


Q = cmT : “Cause” = “inertia” x “effect” (just like F=ma)
 cause = Q
 effect = T
 inertia = cm (mass x specific heat capacity)
T = Q/cm (just like a = F/m)
UB, Phy101: Chapter 12, Pg 20
UB, Phy101: Chapter 12, Pg 21
Examples of Specific Heat Capacity
(see Text, Table 12.2)
T = Q/cm
Substance
aluminum
copper
iron
lead
human body
water
ice
c in J/(kg-C)
900
387
452
128
3500
4186
2000
Suppose you have equal masses of aluminum and copper at the same
initial temperature. You add 1000 J of heat to each of them. Which
one ends up at the higher final temperature
a) aluminum
b) copper
c) the same
correct
UB, Phy101: Chapter 12, Pg 22
Example
UB, Phy101: Chapter 12, Pg 23
Chapter 12, Preflight
Suppose you have two insulated buckets containing the same amount of water
at room temperature. You also happen to have two blocks of metal of the same
mass, both at the same temperature, warmer than the water in the buckets.
One block is made of aluminum and one is made of copper. You put the
aluminum block into one bucket of water, and the copper block into the other.
After waiting a while you measure the temperature of the water in both
buckets. Which is warmer?
1. The water in the bucket containing the aluminum block
correct
2. The water in the bucket containing the copper block
3. The water in both buckets will be at the same temperature
Aluminum has a greater specific heat capacity than
copper, so it can retain more heat. Since aluminum has
a greater heat capacity, when added to water, it will
warm up the water more than copper.
UB, Phy101: Chapter 12, Pg 24
UB, Phy101: Chapter 12, Pg 25
Latent Heat
T
100oC
water
temp
rises
water
changes
to steam
(boils)
steam
temp
rises
DEMO
Q added to water
Latent Heat L [J/kg]:
amount of heat needed to add to or remove from a substance
to change the state of that substance
Substance
water
Lf (J/kg) Lv (J/kg)
33.5 x 104 22.6 x 105
These are BIG
numbers !
UB, Phy101: Chapter 12, Pg 26
UB, Phy101: Chapter 12, Pg 27
Example
UB, Phy101: Chapter 12, Pg 28
Chapter 12, Preflight
Summers in Phoenix Arizona are very hot (125 F is not uncommon), and very
dry. If you hop into an outdoor swimming pool on a summer day in Phoenix, you
will probably find that the water is too warm to be very refreshing. However,
when you get out of the pool and let the sun dry you off, you find that you are
quite cold for a few minutes (yes...you will have goose-bumps on a day when the
air temperature is over 120 degrees).
How can you explain this?
The water is evaporating off of your skin. This means that enough heat (or
energy) is entering the water drops to break bonds between water and allow
them to evaporate. Where is this heat coming from? Your body! Heat flows
from your body to the drops of water, making you feel cooler. When the water
is gone, no more heat will flow from your body and you will get hot once again.
However, this whole situation may be avoided by NOT GOING OUTSIDE
WHEN IT IS 125 DEGREES!!!
as the water evaporates off the body, it takes with it the latent heat of
vaporization, which cools you down.
UB, Phy101: Chapter 12, Pg 29
vapor
Vapor Pressure Pvapor
(not on exam)





water
In equilibrium, the “Vapor Pressure” of a liquid depends
only on temperature. (We will discuss partial pressure
more in a week or so when we do Ideal Gas law) .
Boiling occurs when pvap  psurrounding
Boiling water on mountain
boils at T<100C
Boiling water in pressure cooker
1 atm
boils at T>100C
Boiling by cooling
P
“Phase Diagram”
.5 atm
83C 100C
UB, Phy101: Chapter 12, Pg 30
T