Transcript Chapter 12: Temperature and Heat

Chapter 12: Temperature and Heat
 Now, we move to a new area and take up the
study of Thermodynamics
 Thermodynamics deals with the mechanics of a
(large) collection of particles (gas, liquid, solid) and
how these particles interact (on average) with their
environment
 First, we need some definitions (see Chap. 11,
sections 1-3):
m
 kg 
  , units of  3 
- Density
V
m 
(or mass density)
-Specific Gravity
SG 

 water (4 C )

 water (4 C)  1000kg/m

-Pressure
Force F
P

Area A
3
N
units of  2   Pascal  [Pa]
m 
 Consider a swimming pool of surface area A and
depth h which is filled with water (total mass m)
A
h
FBD
Ft=PtA
h
mg
There is a pressure Pt on top of the water
due to the column of air above
Fb=PbA
 Pt is equal to the atmospheric pressure
= 1.01x105 Pa (at sea level)
= 1 atmosphere
 To determine pressure at the bottom of the
pool, apply Newton’s 2nd Law
F
 Fb  Ft  m g  0
Pb A  Pt A  m g
y
 The density of the water and the volume of
the water are
m
  , V  hA  m   V   hA
V
Pb A  Pt A  (  hA) g  Pb  Pt   gh
 Therefore, the change in pressure from the
bottom to top is
P  Pb  Pt   gh
- Temperature
Boiling point of
water
Freezing point
of water
(at sea level
and 1
atmosphere)
°F
K
°C
212
100
372.15
32
0
273.15
-459.7
-273.15
0
 The Kelvin (K) scale is an absolute temperature
scale – 0 K is absolute minimum temperature (no
negative values)
 In fact, 0 K can never be reached. This is known
as the Third Law of Thermodynamics (Hopefully, we
will have time to discuss the 0th-2nd Laws)
 However, experiments have been performed in
which a gas has been cooled to < 0.001 K.
 Converting between scales:
T (K)  T (  C)  273.15, T (K)  T (  C)
T (  C)  95 (T (  F) - 32.0)
 What is temperature? It is a measure of the
internal energy of a substance (aggregate random
motion of all its atoms and/or molecules)
 This energy is proportional to the temperature.
3
For an ideal gas
2
E  kT
Effect of Temperature Changes on
Liquids and Solids
 Consider a thin rod of length Lo at initial
temperature To.
 If it is heated to a warmer temperature Tf, its
length will increase to a new length Lf  Linear
Thermal Expansion of a Solid
L f  Lo  L   Lo T
whereT  T f  To
Lo
Lf
Lf
Tf>To
Tf<To
 = coefficient of linear expansion
(solids only), depends on the
material (see Table 12.1), has
units of (C°)-1
 Can also be applied to a hole in a material.
Consider a thin board with a circular hole of
diameter Do which is heated from To to Tf
D   Do T
To
Tf
 Can not be applied to liquids or gases as they
have no fixed shape
 But for liquids and solids, we have another effect
called the Volume Thermal Expansion. It can not
be applied to gases as they are compressible.
 Consider a substance of volume Vo and To
To,Vo
Tf>To
 For solids, =3
V   Vo T
=coefficient of volume
expansion, depends on
substance (see Table 12.1),
units of (C°)-1
Example: Problem 12.30
Many hot-water heating systems have a reservoir
tank connected directly to the pipeline, so as to
allow for expansion when the water becomes hot.
The heating system of a house has 76 m of copper
pipe whose inside radius is 9.5x10-3 m. When the
water and pipe are heated from 24 to 78 °C, what
must be the minimum volume of the tank to hold
the overflow of the water?
Solution:
Given: Lo=76 m, ro=9.5x10-3 m, T0=24°C, Tf=78°C
Method: Need to know initial volume of pipe interior
which is also the initial volume of water. When
heated both water and pipe volume expanded. The
difference is the volume needed for the expansion
tank.
From Table 12.1 get coefficients of volume
expansion
copper c  51x10 C , water  w  207x10 C
-6
-1
What is the volume of the pipe interior?
Ao   r , Vo  Lo Ao  Lo r
2
o
2
o
-6
-1
For volume expansion use
V   Vo T   Lo r T
2
o
Where the change in temperature
T  T f  To  78  24  54 C

Vc   c Lo r T
-6
o -1
-3
2
o
 (51x10 C )(76m) (9.5x10 m) (54 C)
 5.9 x10-5 m 3
2
Vw   w Lo ro T
-6
o -1
-3
2
o
 (207x10 C )(76m) (9.5x10 m) (54 C)
-4
3
 2.4 x10 m
Vtan k  Vw  Vc  1.8x10-4 m 3
2
o