Transcript Document 7456906

 Thermodynamics


“Thermo” = Study of heat
“dynamics” = Movement of that heat between
objects
 Thermometers

Measure temperature based on physical properties

Mercury based thermometers expand at a predictable rate
with temperature
 Scale of the thermometer measures the amount of
expansion
 Temperature

Kelvin (K), Celsius (C), and Fahrenheit (F)

Temperature values are different on each scale
 Comfortable indoor room temperature



scales
68° F; 20° C; 293.15 K
Water freezes at 0°C; 32°F; 273.15 K
There are 100 degrees between the freezing &
boiling points of water on the Celsius scale, and
100 Kelvins between the freezing and boiling on
the Kelvin scale

Units of these two scales are equal
 Unit of Celsius = Unit of Kelvin
Kelvin temperature
TK  TC  273.15

9 
1C  F
5

 TK
= TC + 273.15
 TC = (5/9)(TF – 32)



TK = Kelvin Temperature
TC = Celsius Temperature
TF = Fahrenheit Temperature
Converting from a Celsius to a Fahrenheit
Temperature
A time and temperature sign on a bank indicates that the outdoor
temperature is -20.0oC. Find the corresponding temperature on
the Fahrenheit scale.
9 F o 
20.0 C 15 Co  36.0 F o
Degrees below ice point
o
32.0 F  36.0 F  4.0 F

ice point
 The
lowest temperature any material could
theoretically reach  As cold as it can get!

Reference point at which molecules are in their
minimum energy state

0 K (-273.15° C; -459.67° F)
 Thermal
energy that flows from one object to
another due to a temperature difference

Energy flows from a higher-temperature object to a lowertemperature object because of the difference in
temperatures.
 Example: In an oven, heat flows from the oven coils to the
air molecules, warming them up, and then to the bread,
warming it as well
 Not
a property of an object
 Represented by the letter Q
 SI Unit = Joules (J)
Internal Energy:
Energy associated with the
molecules and atoms that
make up a system
Heat flows from hot to coldoriginating from the internal
energy of the hot substance.
It is not correct to say that a
substance contains heat.
 Endothermic


Absorb heat from the surrounding area(s)
Cooling effect on the environment
 Exothermic


Processes
Processes
Release heat into the surrounding area(s)
System becomes cooler  Environment becomes
warmer
* Farmers farm oranges in the winter (Oranges could
freeze!)
* Farmer’s Prevention:
- Pour water on the oranges  letting the water
freeze instead!
- Freezing gives off a lot of heat. So, when the water
freezes, it gives the heat to the oranges
- Freezing  Exothermic  The orange absorbs the heat
from the water as the freezing occurs
 http://auto.howstuffworks.com/cooling-
system.htm
HEAT
HEAT
HEAT
HEAT
 If
objects A and B are in thermal equilibrium,
and objects B and C are in thermal
equilibrium, then A and C will be in
equilibrium as well
 Thermal

Expansion
Increase in the length or volume of a material
due to a change in its temperature

Different materials expand at different rates

Thermal Linear Expansion of a Solid

Length an object changes when its temperature changes


Coefficient of linear expansion, is a constant, that specifies
how much a given material expands with a change in
temperature


Measured along one dimension
Represented by greek letter alpha (α)
Linear Expansion Equation

ΔL = Li α ΔT
 L = Length
 α = Coefficient of linear expansion
 
1
 C

C

1
ΔT = Change in temperature (K or °C)
THE BIMETALLIC STRIP
NORMAL SOLIDS
The Buckling of a Sidewalk
A concrete sidewalk is
constructed between two
buildings on a day when the
temperature is 25oC. As the
temperature rises to 38oC,
the slabs expand, but no
space is provided for thermal
expansion. Determine the
distance y in part (b) of the
drawing.
L   LoT

6
 12 10
y
C  3.0 m13 C  0.00047 m
o 1
o
3.00047 m  3.00000 m  0.053 m
2
2
 Thermal

Above 4°C, water contracts and sinks as it cools




expansion of water
You may have experienced this effect if you have
jumped into a lake
 The water is colder the deeper you go
This chilling and sinking of the top layer
continues until the lake is 4° C throughout
From 4°C to 0°C, water expands and stays on
top
At 0°C, water turns into ice and floats (less
dense)

** Crucial for the survival of aquatic life **
Expansion of Water
 Volume
of an object changes when its
temperature changes


ΔV = Vi βΔT




Every substance has a coefficient of volume expansion
(β)  varies by material
V = Volume
Β = Coefficient of volume expansion 
ΔT = Change in temperature
Β ~ 3α

1
o 1
 C 
o
C
An Automobile Radiator
A small plastic container, called the coolant reservoir,
catches the radiator fluid that overflows when an
automobile engine becomes hot. The radiator is made
of copper and the coolant has an
expansion coefficient of
4.0x10-4 (Co)-1. If the radiator
is filled to its 15-quart capacity
when the engine is cold (6oC),
how much overflow will spill into the
reservoir when the coolant reaches its
operating temperature (92oC)?

4
Vcoolant  4.10 10

C
Vradiator  5110 C
6

o 1
15 quarts 86 Co 0.53 quarts
 15 quarts 86 C  0.066 quarts
o 1
o
Vspill  0.53 quarts  0.066 quarts  0.46 quarts
 Property

of a material
A constant that tells how much the temperature
of a mass of material changes when a particular
amount of heat is transferred

A material with a large specific heat requires more
heat per kilogram to a produce a given change in
temperature than one with a smaller specific heat
Q = cmΔT
Q = Heat (J or calories)
Q = Negative, heat energy is removed
Q = Positive, heat energy is added
c = Specific heat (J/kg x K) or (cal/g°C)
m = mass
ΔT = Temperature change in C° or K
(ΔT = Tf – Ti)
A Jogger
In a half-hour, a 65-kg jogger can generate 8.0x105 J of
heat. This heat is removed from the body by a variety of
means, including the body’s own temperature-regulating
mechanisms. If the heat were not removed, how much
would the body temperature increase?
Q  mcT
Q
8.0 10 J
o
T 


3.5
C
mc 65 kg 3500 J kg  C o
5


CALORIMETRY
If there is no heat loss to
the surroundings, the heat
lost by the hotter object
equals the heat gained by
the cooler ones.
Measuring Specific Heat Capacity
The calorimeter is made of 0.15 kg of aluminum
and contains 0.20 kg of water. Initially, the
water and cup have the same temperature
of 18.0oC. A 0.040 kg mass of unknown
material is heated to a temperature of
97.0oC and then added to the water.
After thermal equilibrium is reached, the
temperature of the water, the cup, and the
material is 22.0oC. Ignoring the small amount
of heat gained by the thermometer, find
the specific heat capacity of the
unknown material.
mcTAl  mcTwater  mcTunknown


c unknown 
mcT Al  mcT water
mT unknown
9.00 10 J kg  C 0.15 kg4.0 C  4186 J kg  C 0.20 kg4.0 C 


2
o
o
0.040 kg75.0 C o
 1300 J kg  C o
o
o

A constant that tells how much the temperature of a
particular number of moles of a material changes when a
particular amount of heat is transferred

A mole of a substance is a measure of quantity based
on the number of particles making up an object


1 mole = 6.022 x 1023 molecules or atoms
Q = knΔT




Q = Heat (J)
k = Molar specific heat (J/mol x K)
n = Number of moles
ΔT = Temperature change in C° or K
 Transformation
between solid and liquid,
liquid and gas, or solid and gas

All require the addition of energy
 When
water freezes into ice or steam
condenses into water  Energy is released

* Although energy is being added or released
during a phase change  Temperature of the
substance remains constant
THE PHASES OF MATTER
During a phase change, the temperature of the mixture
does not change (provided the system is in thermal
equilibrium).
 Energy
required per kilogram to cause a
phase change in a given material

Latent heat of fusion


Transforming from solid to liquid, or liquid to solid
Latent heat of vaporization

Transforming from liquid to gas, or vice-versa
 Latent
heat values  Same in either
“direction” of phase change
 Latent


Heat Equations
Q = Lfm
Q = Lvm




Q = Heat (J)
m = Mass (kg)
Lf = Latent heat of fusion (J/kg)
Lv = Latent heat of vaporization (J/kg)
Ice-cold Lemonade
Ice at 0oC is placed in a Styrofoam cup containing 0.32 kg of
lemonade at 27oC. The specific heat capacity of lemonade is
virtually the same as that of water. After the ice and lemonade
reach an equilibrium temperature, some ice still remains. Assume
that mass of the cup is so small that it absorbs a negligible amount
of heat.

14mL2 43  1cmT
44 2 4 43
f ice
Heat gained
by ice

lemonade
Heat lost by
lemonade

14mL2 43  1cmT
44 2 4 43
f ice
Heat gained
by ice
lemonade
Heat lost by
lemonade
cmT lemonade

m 
ice
Lf
4186 J kg  C 0.32 kg27 C  0 C


 0.11 kg
o
3.35 10 5 J kg
o
o
 Flow
of thermal energy directly through a
material without motion of the material
itself


Example: Cast iron pan on stove  Handle
eventually gets hot
Thermal Conductors:


Materials that conduct heat well
Thermal Insulators:

Materials that conduct heat poorly
 Process
in which heat is carried from one
place to another by the bulk movement of a
fluid

Heat transfer through a gas or liquid caused by
movement of the fluid
Hot Water Baseboard Heating and
Refrigerators
Hot water baseboard heating units are mounted on the
wall next to the floor. The cooling coil in a refrigerator is
mounted near the top of the refrigerator. Each location is
designed to maximize the production of convection
currents.
 Radiation


Heat transfer by electromagnetic waves
Material that is a good absorber is also a good
emitter
 Global
Warming & the Greenhouse Effect