PHY 113 C General Physics I
11 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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Review:
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
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1 kg/m3 = 0.001 gm/ml
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Review of equations describing static fluids in terms
of pressure P and density r:
dP
For all fluids near Earth' s surface :
 ρg
dy
For incompressible fluid, r  (constant)  P  P0  ρg ( y  y0 )
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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Bernoulli’s equation:
2
1
1
1
2
rv  rgy  P1  rv  rgy2  P2
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2
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2
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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?
mg/4
mg
P=F/A=(mg/4)/A
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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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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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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
 2 h2  1
2





  Pi Ai   r g h  y wdy  rgw h    rgwh
2 2
i

0
h
Fside
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Webassign questions on fluids (Assignment #17)
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
y
M
r H O A2
2
z
P1  r H 2O g  r Hg gz
 r Hg g h  y  z 
 r H 2O g  r Hg g h  y 
Note that : hA1  yA2
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r H O g  r Hg g h  y 
2
A1h  A2 y

A1 
 r H 2O g  r Hg gh1  
 A2 
h
rH O
2

A1 
r Hg 1  
 A2 
For   0.2 m, A1 / A2  2 :
1000
h  0.2m
 0.0049m
136001  2 
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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.
FB  mg  3.5 N
mg  5.3 N
mg r object

FB
r fluid
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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
FB  mg  0
m  mload  r HeVg
FB  r airVg
mload  r air  r He V
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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 crosssectional area A = 2.40 10-5 m2, and the needle has a crosssectional 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.
1
2
rv  rgy1  P1  rv  rgy2  P2
2
1
1
2
2
2
In this case : y1  y2 ; P1  P2  F / A;
v2   A / a v1  v1
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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.
Not equilibrium:
T1
Equilibrium:
T2
T3
“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.
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Constant temperature “bath”
At equilibrium:
T
T
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Temperature scales
TF=9/5 TC + 32
500
450
400
350
300
250
TF
200
150
100
50
0
-50
Kelvin scale:
-100
T = TC + 273.15o
-150
-200
-100 -50
0
50
100
150 200
TC
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iclicker question:
Suppose you find yourself in a hotel in Europe
or Canada. Which Celsius temperature would
you set the thermostat for comfort?
A. -20oC
B. +20oC
C. +40oC
D. +60oC
E. +80oC
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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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Effects of temperature on matter
Solids and liquids
Model of a solid composed of
atoms and bonds
Thermal exansion:
L = a Li T
L
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Li (equilibrium bond
length
at Ti)
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Typical expansion coefficients at TC = 20o C:
Linear expansion: L = a Li T
a = 11 x 10-6/ oC
Steel:
Concrete:
a = 12 x 10-6/ oC
Volume expansion:
V=L3  V = 3a Vi T = b Vi T
Alcohol:
Air:
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b = 1.12 x 10-4/ oC
b = 3.41 x 10-3/ oC
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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+L
V=L3
V+V=(L+L)3@V(1+3(L/L))
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Brass
Steel
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Switch in thermostat
Modern thermostats use electrical circuits to detect
temperature
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Effects of temperature on materials – continued
strange case of water:
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Effects of temperature on materials – continued -ideal gas “law” (thanks to Robert Boyle (16271691), Jacques Charles (1746-1823), and GayLussac (1778-1850)
8.314 J/(mol K)
PV  nRT
temperature in K
volume in m3 # of moles
pressure in Pascals
1 mole corresponds to 6.022 x 1023 molecules
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PV  nRT
P0 =12.6 atm
T0 =27.5oC
n0
P=?
T=81.0oC
n=n0/3
P0V0  n0 RT0
PV  nRT V  V0
PV
nRT

P0V0 n0 RT0
P
T

P0 3T0
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n  n0 / 3
 354.15 
P  12.6 atm
  4.9atm
 3  300.65 
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Assuming that air behaves like an ideal gas, what is
the density of air at T=0o C and P=1 atm?
PV  nRT
n
P

V RT
m
mass density r  
V
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n M
i
i
i
V
31
Typical composition of air:
url: http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html
n M
i
i
M avg P
m
i
mass density r  

V
V
1000 RT
 1.29 kg/m 3
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