11/07/2013phy 113 c fall 2013 -- lecture 201 phy 113 c general physics i 11 am - 12:15 pm tr olin...
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![Page 1: 11/07/2013PHY 113 C Fall 2013 -- Lecture 201 PHY 113 C General Physics I 11 AM - 12:15 PM TR Olin 101 Plan for Lecture 20: Chapter 19: The notion of temperature](https://reader035.vdocuments.site/reader035/viewer/2022062712/56649c7f5503460f94935eaa/html5/thumbnails/1.jpg)
PHY 113 C Fall 2013 -- Lecture 20 111/07/2013
PHY 113 C General Physics I11 AM - 12:15 PM TR Olin 101
Plan for Lecture 20:
Chapter 19: The notion of temperature
1. Review of fluid physics
2. Temperature equilibrium
3. Temperature scales
4. Temperature in ideal gases
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PHY 113 C Fall 2013 -- Lecture 20 211/07/2013
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PHY 113 C Fall 2013 -- Lecture 20 311/07/2013
The physics of fluids.
•Fluids include liquids (usually “incompressible) and gases (highly “compressible”).
•Fluids obey Newton’s equations of motion, but because they move within their containers, the application of Newton’s laws to fluids introduces some new forms.
Pressure: P=force/area
1 (N/m2) = 1 Pascal
Density: r =mass/volume 1 kg/m3 = 0.001 gm/ml
Review:
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PHY 113 C Fall 2013 -- Lecture 20 411/07/2013
)(ρ (constant) fluid, ibleincompressFor 00 yygPP
Review of equations describing static fluids in terms of pressure P and density :r
gdy
dPρ :surface sEarth'near fluids allFor
Note that for compressible fluids (such as air), the relationship between pressure and density is more complicated.
Buoyant force for fluid acting on a solid – net force due to volume Vdisplaced being displaced in fluid:
FB=rfluidVdisplacedg
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PHY 113 C Fall 2013 -- Lecture 20 511/07/2013
Bernoulli’s equation:
22222
111
212
1 PgyvPgyv
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PHY 113 C Fall 2013 -- Lecture 20 611/07/2013
Webassign questions on fluids (Assignment #17)
A large man sits on a four-legged chair with his feet off the floor. The combined mass of the man and chair is 95.0 kg. If the chair legs are circular and have a radius of 0.500 cm at the bottom, what pressure does each leg exert on the floor?
mgmg/4
P=F/A=(mg/4)/A
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PHY 113 C Fall 2013 -- Lecture 20 711/07/2013
Webassign questions on fluids (Assignment #17)
A swimming pool has dimensions 32.0 m ✕ 7.0 m and a flat bottom. The pool is filled to a depth of 2.50 m with fresh water. (a) What is the force exerted by the water on the bottom?
(b) What is the force exerted by the water on each end? (The ends are 7.0 m.)
(c) What is the force exerted by the water on each side? (The sides are 32.0 m.)
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PHY 113 C Fall 2013 -- Lecture 20 811/07/2013
Webassign questions on fluids (Assignment #17)
A swimming pool has dimensions 32.0 m ✕ 7.0 m and a flat bottom. The pool is filled to a depth of 2.50 m with fresh water. (a) What is the force exerted by the water on the bottom?
h=2.5m
Fbottom=PA=rghA
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PHY 113 C Fall 2013 -- Lecture 20 911/07/2013
Webassign questions on fluids (Assignment #17)
A swimming pool has dimensions 32.0 m ✕ 7.0 m and a flat bottom. The pool is filled to a depth of 2.50 m with fresh water.
(b) What is the force exerted by the water on each end? (The ends are 7.0 m.)
h=2.5m
w=7.0m
22
2
0 2
1
2 gwh
hhgwwdyyhgAPF
i
h
iiside
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PHY 113 C Fall 2013 -- Lecture 20 1011/07/2013
Webassign questions on fluids (Assignment #17)
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PHY 113 C Fall 2013 -- Lecture 20 1111/07/2013
21
1
: thatNote
2
2
yAhA
yhgg
zyhg
zggP
HgOH
Hg
HgOH
22A
M
OH
Dy
Dz
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PHY 113 C Fall 2013 -- Lecture 20 1211/07/2013
2
1
2
1
21
1
1
2
2
2
AA
h
A
Aghg
yAhA
yhgg
Hg
OH
HgOH
HgOH
m0049.02113600
1000m2.0
:2/ m, 0.2For 21
h
AA
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PHY 113 C Fall 2013 -- Lecture 20 1311/07/2013
Webassign questions on fluids (Assignment #17)
The gravitational force exerted on a solid object is 5.30 N. When the object is suspended from a spring scale and submerged in water, the scale reads 3.50 N (figure). Find the density of the object.
fluid
object
B
B
F
mg
Nmg
NmgF
3.5
5.3
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PHY 113 C Fall 2013 -- Lecture 20 1411/07/2013
Webassign questions on fluids (Assignment #17)
A light balloon is filled with 373 m3 of helium at atmospheric pressure. (a) At 0°C, the balloon can lift a payload of what mass? Note: rair = 2.9 kg/m3 : rHe = 0.179 kg/m3
Vm
VgF
Vgmm
mgF
Heairload
airB
Heload
B
0
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PHY 113 C Fall 2013 -- Lecture 20 1511/07/2013
Webassign questions on fluids (Assignment #17)
A hypodermic syringe contains a medicine with the density of water (see figure below). The barrel of the syringe has a cross-sectional area A = 2.40 10-5 m2, and the needle has a cross-sectional area a = 1.00 10-8 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force of magnitude 2.65 N acts on the plunger, making medicine squirt horizontally from the needle. Determine the speed of the medicine as it leaves the needle's tip.
112
2121
22222
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212
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;/ ; :case In this
vvaAv
AFPPyy
PgyvPgyv
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PHY 113 C Fall 2013 -- Lecture 20 1611/07/2013
Dictionary definition: temperature – a measure of the the warmth or coldness of an object or substance with reference to some standard value. The temperature of two systems is the same when the systems are in thermal equilibrium.
“Zeroth” law of thermodynamics:
If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other.
T1 T2 T3
Not equilibrium: Equilibrium:
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PHY 113 C Fall 2013 -- Lecture 20 1711/07/2013
Constant temperature “bath”
T
T
At equilibrium:
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PHY 113 C Fall 2013 -- Lecture 20 1811/07/2013
-100 -50 0 50 100 150 200-200
-150
-100
-50
0
50
100
150
200
250
300
350
400
450
500
T
F
TC
Temperature scales TF=9/5 TC + 32
Kelvin scale:
T = TC + 273.15o
T 0
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PHY 113 C Fall 2013 -- Lecture 20 1911/07/2013
iclicker question:Suppose you find yourself in a hotel in Europe or Canada. Which Celsius temperature would you set the thermostat for comfort?
A. -20oCB. +20oCC. +40oCD. +60oCE. +80oC
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PHY 113 C Fall 2013 -- Lecture 20 2011/07/2013
There is a lowest temperature:
T0 = -273.15o C = 0 K
Kelvin (“absolute temperature”) scale
TC = -273.15 + TK
Example –
Room temperature = 68o F = 20o C = 293.15 K
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PHY 113 C Fall 2013 -- Lecture 20 2111/07/2013
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PHY 113 C Fall 2013 -- Lecture 20 2211/07/2013
Effects of temperature on matter
Solids and liquids
Li (equilibrium bond length at Ti)
Model of a solid composed of atoms and bonds
DL
Thermal exansion:
DL = a Li DT
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PHY 113 C Fall 2013 -- Lecture 20 2311/07/2013
Typical expansion coefficients at TC = 20o C:
Linear expansion: DL = a Li DT
Steel: a = 11 x 10-6/ oC
Concrete: a = 12 x 10-6/ oC
Volume expansion:
V=L3 DV = 3a Vi DT = b Vi DT
Alcohol: b = 1.12 x 10-4/ oC
Air: b = 3.41 x 10-3/ oC
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PHY 113 C Fall 2013 -- Lecture 20 2411/07/2013
iclicker question
On the last slide – we suggest that b=3a. Is this result
A. One of those mysteries of physics that has no explanation?
B. A result that we can derive?
L L+DL
V=L3
V+DV=(L+DL)3@V(1+3(DL/L))
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PHY 113 C Fall 2013 -- Lecture 20 2511/07/2013
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PHY 113 C Fall 2013 -- Lecture 20 2611/07/2013
Brass
Steel
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PHY 113 C Fall 2013 -- Lecture 20 2711/07/2013
Switch in thermostat
Modern thermostats use electrical circuits to detect temperature
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PHY 113 C Fall 2013 -- Lecture 20 2811/07/2013
Effects of temperature on materials – continued strange case of water:
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PHY 113 C Fall 2013 -- Lecture 20 2911/07/2013
Effects of temperature on materials – continued -- ideal gas “law” (thanks to Robert Boyle (1627-1691), Jacques Charles (1746-1823), and Gay-Lussac (1778-1850)
nRTPV
pressure in Pascals
volume in m3 # of moles
temperature in K
8.314 J/(mol K)
1 mole corresponds to 6.022 x 1023 molecules
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PHY 113 C Fall 2013 -- Lecture 20 3011/07/2013
nRTPV
P0 =12.6 atmT0 =27.5oCn0
P=?T=81.0oCn=n0/3
atm.
. atm. P
T
T
P
P
RTn
nRT
VP
PV
/n nVVnRTPVRTnVP
9.4653003
15354612
3
3
000000
000000
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PHY 113 C Fall 2013 -- Lecture 20 3111/07/2013
Assuming that air behaves like an ideal gas, what is the density of air at T=0o C and P=1 atm?
V
Mn
V
m
RT
P
V
n
nRTPV
iii
density mass
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PHY 113 C Fall 2013 -- Lecture 20 3211/07/2013
Typical composition of air: url: http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html
3kg/m 1.29
1000density mass
RT
PM
V
Mn
V
m avgiii