Download Boddeker`s Ch 16 Temperature and Heat (PHY122)

Thermodynamics
Ch 16 – Temperature and Heat
16.1 Temperature and the Zeroth Law of Thermodynamics
The Zeroth Law of Thermodynamics
If two objects are separately in
thermal equilibrium with a 3rd object,
then the first two objects are in
thermal equilibrium with each other.
Thermal equilibrium: If two objects are in thermal
equilibrium then neither object will exchange energy by heat
or electromagnetic radiation if they are placed in thermal
contact.
Temperature is a property that determines whether an
object is in thermal equilibrium.
Thermal contact: If in thermal contact two objects will
exchange energy between them if a temperature difference.
16.2 Temperature Scales
You shouldn’t ever have to memorize formulas in Physics. Take for example the formula from or to
Celsius to/from Fahrenheit.
WATER
Freezing
Boiling
Difference
Fahrenheit
32°
212°
180°
Celsius
0°
100°
100°
Let’s try using this logic…
Take the ratio of
180/100 = 9/5 & 100 / 180 = 5/9
98.6 °F
Where is the “zero” (with respect to water) for the
Let’s bring the Fahrenheit scale down to a
Celsius scale?
more logical “zero” point.
Ans: 0 °C
 98.6 °F - 32 °F = 66.6 F°
 (take note of the units, 66.6 F°)
For the Fahrenheit scale?
Ans:32 °F
We now need to bring this “BIG” scale down
Now, which scale has the most number between
to the “small” scale by multiplying by 5/9
freezing and boiling?
 66.6 F° * 5/9 = 37 °C
Fahrenheit (we’ll call this the “BIG” scale; conversely Celsius is
the “small” scale)
Let’s go in reverse…
So we say the “zero” point for the Fahrenheit scale is
32 °F…
37 °C uses the “small” scale
 I think most people will agree this is a weird
 let’s go to the “BIG” scale
“zero” point
 37 °C * 9/5 = 66.6 F°
 How can we bring this “zero” point to zero?
o Ans: Subtract 32 °F
But the “zero” point of the Fahrenheit scale
 We also recognize that the Fahrenheit scale is the isn’t “zero”…it’s 32 °F
“BIG” scale…so how can we bring it to a small
 66.6 F° + 32 F° = 98.6 °F
scale?
o Ans: Multiply by a small ratio, 5/9
The Kelvin scale is the proper scale in most situations (PV = nRT, etc). When
in doubt…just use the Kelvin scale. Why?
1 Kelvin = 1 °C and
T = TC + 273.15
So if the temps are given in °C then ΔT will be the same
whether in the Kelvin or Celsius scales.
If you notice
T  temperature in the Kelvin scale
TC  temperature in Celsius
TF  temperature in Fahrenheit
The Kelvin scale is an absolute scale which means you
can not go below zero.
At 0 K (do not use the degree sign) all movement (i.e. in an atom) stops. As
you can see, we can not achieve the Absolute Zero point, but we can achieve
(optional)
Thermometers make use of
thermometric properties.
Types of thermometers
 volume expansion (Hg
column)

constant volume gas (Δ
gas pressure)

thermocouples (Voltage
differences of two
different metals)

electrical resistance
(Platinum wire)
temps very close.

thermographic (emitted
radiation)
Galileo’s Thermometer Demo: TH-E-GT
16.3 Thermal Expansion
(of Solids and Liquids)
Let’s pretend we have a 10 cm steel ruler an a 2
meter stick (also steel)
Now let’s pretend we have two 10 cm steel rulers,
A and B.
If we heat both of the steel sticks…which will
expand more a greater amount?
10 cm steel ruler, A, undergoes ΔTC = 10 C°
10 cm steel ruler, B, undergoes ΔTC = 100 C°
Which will expand more?
Most of you will say the long one will expand
more and you’ll be RIGHT!
Some may say they will expand the same and
you’ll be CORRECT also (almost)!
Explanation:
Since they are both steel, they will expand by
the same proportion; but not the same amount…
(We know the longer one will expand by the
greater total length since 5% of 2 meters is
much greater than 5% of 10 centimeters)
B, will expand a factor of 10 greater A
Most of you already knew these results.
These results also lead to one conclusion:
ΔL = α L ΔT
Which states that the change of length of an
object is dependant on
 Type of material
 Initial Length of material
 Change of temperature
Application of Thermal Expansion/Contraction
You are attempting to
What happens to the coupling?
form the best junction
 it expands, including the hole
between two pipes.
in the middle
What happens to the pipes?
Question: What would happen if the
coupling and the pipes were made of
different materials and the entire new
system became very hot or very cold?
Use a coupling between
pipes
 heat the coupling
(torch)
 cool the ends of the
pipes (ice bath or oil
bath cooled with
liquid nitrogen)
 they contract, become
smaller
Ans: Different materials have
different coefficients of linear
expansion, so if the coupling is a
Slide the pipes into the
different material, the coupling would
coupling…let achieve room temp.
either crack (contract too much) or leak
You now have a coupling as strong
(expand too much)
as the pipe itself.
This is the same principle as the
bimetallic strip discussed in your book.
Volume Thermal Expansion
One of the notable exceptions to
First NOTE:
β equals 3 * α
Coef of Volume
Expansion
= 3*(Coef of Linear Exp)

materials contracting as they get
colder.
WATER!!! Water’s highest density
is at 4 °C
This is very important for life, if
you don’t see why…ask in CLASS!!!
Derivation below
And the formula follows
similarly to linear
expansion ΔL = α L ΔT
ΔV = β V ΔT
Vo + V
Vo + V
Vo + V
=
(l + l)
(h + h) (w + w)
= (l + lT) (h + h T) (w + w T)
= l*h*w (1 + T)3
Derivation:
 Remember Young’s Modulus…we now have
another application
Young’s Modulus
= Stress / Strain;
where stress F/A and strain is ΔL/L
(Y)
(Y) ΔL/L
Stress
Stress
Stress
= F/A
/ ΔL/L
= F/A (= stress)
= (Y) ΔL
/L
; where ΔL = α L ΔT
= (Y) α L ΔT / L
= Y α ΔT
So if you know Young’s Modulus and the coef of
linear expansion, you can calculate Stress.
Special Properties of water
Vo + V
V/Vo
= Vo (1 + 3T + 3(T)2 + (T)3)
= 3T + 3(T)2 + (T)3
We also know T <<1 for most materials…so
V/Vo = 3T
V = Vo3T
Thermostat Demo: TH-A-DT
16.4 Heat and Mechanical Work
Units of heat
calorie
Heat is defined as the
transfer of energy across
the boundary of a system
due to a temperature
difference between the
system and its surroundings.
the amount of energy transfer required to raise the
temp of 1 gram of water from 14.5 °C to 15.5 °C
joule
The SI unit: 1 cal = 4.186 Joules
1 kcal = 1 Calorie = 4186 Joules
BTU
English unit: (Q required to raise 1 lb, 1 °F)
1 Btu = 1054 Joules
If work is done on a system, then the system will transfer energy across its boundary.
Work on system
The dentists drill causes your teeth to get hot…so
cold water is sprayed on your teeth to cool down the
teeth. The cold water is external to the system and
its temperature raises when in contact with the hot
tooth.
16.5 Specific Heats
Work by system
If energy is added to a system, work may be
done by the system.
Burn coal in a steam engine, and the steam
engine is able to move the entire train.
Heat Capacity, C : Q = C ΔT
C, the heat capacity is defined as the energy needed to raise
the temperature of a sample by 1°C
Specific Heat: Q = m c ΔT c, the
heat capacity of a sample per unit
mass
Question: Which would require more energy to bring to
boil if both are initially at room temperature?
A) A cup of water
B) A swimming pool full of water
Question: Which would require more energy to
rise from 20 °C to 50 °C?
A) A cup of water
B) A cup of alcohol
Ans: I wouldn’t want to have to pay the energy bill
required to raise a swimming pool full of water to its
boiling point.
Ans: This one is harder…but I believe many
know raising the temperature of alcohol requires
less heat than water.
Conclusion: The greater the mass…the more energy
required to bring to boiling point.
Question: Which would require more energy to bring to
boil?
A) A cup of water at 20 °C
B) A cup of water at 90 °C
Conclusion: The material being heater makes a
bit difference, so materials like water require
much more energy to change its temperature
compared to materials like alcohol and sand.
Question: Which would require more energy to
bring to 20 °C?
C) A cup of ice at 0 °C
D) A cup of water at 0 °C
Ans: This one is easy…the one at 90 °C is almost at boil
already, thus the energy input required to bring it to boil
will way less than the cup at 20°C
Ans: This one is easy…imagine adding 0°C water
to your Coke or Pepsi to cool it down. 0°C ice
requires much energy to bring it to 0°C water.
Conclusion: The greater the ΔT…the greater amount of
energy was required.
Conclusion: The change of state requires a large
change of energy.
SI units are generally preferable, but I’m presenting this section in cgs units
Question: A large block of ice, 1000 g, is at -20 °C. How much energy is required to bring it to 100 °C steam?
Ans:
Q
Q
Q
Q
Q
Q
v  vaporization
f  fusion
units of c  calories / (gram C°)
= cice m ΔT + m Lf
+ cwaterm ΔT + m Lv
= ½ (1000)(20) + 1000(80)
+ 1 (1000)(100) + 1000 (539)
= 10,000 cal
+ 80,000 cal
+ 100,000 cal + 539,000 cal
= 729,000 calories
(1 Calorie = 1 kilocalorie)
= 729 Calories (4.186 J / Cal)
= 3052 Joules
(a good exercise for you is to work the problem out using
16.6 Conduction, Convection, Radiation
Conduction
P=Q/t
P = kA ΔT/Δx
P = kA ΔT/Δx
P = A ΔT/(Δx/k)
P = A ΔT/(ΣRi)
Δx:
The thicker the wall…more isolated
A:
More area…
the more energy flows through the wall
ΔT: The greater the temperature difference…
the more energy flow through the wall
k:
And different materials will have
different coefficients for heat flow
R  ft2 °F hr/Btu
1 ft2 °F hr/Btu (0.305m/ft)2 (5°C/9°F)(3600s/hr)(1Btu/1054J)
1 ft2 °F hr/Btu = 0.1765 m2 °F sec / J
Standard 2x4 insulation is R-11 English units
In metric this would be R-2
Radiation
Convection
Stefan’s Law: P = σAeT4
SI units)
Energy transferred by the movement
of a warm medium
Convection Currents in Air: TH-B-CC
σ = 5.669 x 10-8 W/m2K4,
e is emissivity (0 to 1)
Ideal absorber, e = 1 (Black Body)
Ideal reflector, e = 0
Demo: Light the Match: TH-B-LM
Crooke’s Radiometer Demo: TH-D-CR