Temperature and Heat
What's the difference?
Write a definition of each and then compare with
your neighbor.
Temperature vs. Heat



Temperature
Measurement of
average kinetic
energy of molecules in
a substance



Heat
Thermal energy that is
transmitted from one
object to another
Energy in transit
Measurement of
internal thermal
energy
Objects can't contain heat- they can contain thermal energy (measured in temperature) which
is transmitted as heat.
Temperature Scales

Celsius

Kelvin

Water freezes at

Absolute zero

0 degrees

Water freezes at 273

Boils at


100 degrees

To convert from
Fahrenheit:
°F=9/5(°C)+32
To convert from
celsius: K=°C+273
Physical Changes due to heat transfer



When a substance
absorbs or gives off
heat it can

Change temperature

Change phase
BUT NOT BOTH at
same time
Describe what is
happening on this
graph
Temperature Changes


△T depends on

ID of substance

Mass of substance
For example, you
could add 10J to a
textbook and a metal
bar and their temp
change would be
different- thus the
metal pots and pans
Specific Heat
• Specific heat= capacity of a substance to store
heat
• Depends on chemical composition
• Technically= the amount of heat (Joules) required
to raise the temp of a unit mass of the substance
by 1 degree
• Kind of like “thermal inertia”
• SO which do you think has a higher specific heat:
water or aluminum?
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Specific Heat (c)
• Specific heat measured in J/g°C but sometimes
also see it in calories/g°C
• Calorie is a unit of heat based on water: 1 calorie
is heat to raise temp of 1g of water by 1°C
• So specific heat of water is 1cal/g°C which is
equivalent to 4.186J/g°C
• Compare this to aluminum which has a specific
heat of 0.215cal/g°C or 0.900J/g°C
• Aluminum transfers heat much more easily!
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Specific Heat Charts
8
Temperature Changes


The amount of heat added relates to the
change in temperature:
Q=mc△T


Where Q= amount of heat coming in or going
out (+ for increase in temp, - for decrease in
temp)
c is specific heat of substance
Note relationships- more mass means less △T
Higher specific heat means less △T for same Q
Heat Transfer and Phase
Changes
• If you add or subtract heat at a phase
change, the phase change will occur
instead of temperature change- remember
that these cannot occur together! Once
the phase change is complete, the
temperature can change again
• Q=mL
– Q= amount of heat
– L=latent heat of fusion or vaporization
Problem Solving: Heat Transfer
• In a half hour, a 65kg jogger can generate
8.0x105 J of heat. The heat is removed from
the joggers body through natural
mechanisms. If this heat were not removed,
how much would the body temperature
increase? Specific heat capacity of the
human body is 3500J/kg°C.
• Is heat transfer involved in phase change or
temperature change? Which equation is
used?
Problem Solution
• Q=cmΔT
• ΔT=Q/cm=8.0x105 J
/(3500J/kg°C)(65kg)=3.5°C
Problem solving with phase
changes
• Phase changes are reversible
• Note which phase change- are you using
latent heat of vaporization or fusion?
• Are you increasing temperature outside of
phase change? If so, you need to solve
both parts separately
• Try # 10 on HW
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Heat Transfer and Thermal
Expansion
• When substance changes temperature it also
changes size
• Most things expand as they increase temperature
but not…
• Water between4°C-0°C
• Each material has a coefficient of linear
expansion=α and change in length equals:
• ΔL=αLiΔT
• Using this formula, what is the unit then for the
coefficient of thermal expansion, α?
• 1/°C
Problem Solving: Thermal
Expansion
• A metal ball has a diameter that is slightly greater
than the diameter of a hole that has been cut into
a metal plate. The coefficient of linear expansion
for the metal in the ball is greater than that for the
plate. Which one (or more) of the following
procedures can be used to make the ball pass
through the hole?
• A: raise the temperatures of the ball and plate by
the same amount
• B: lower the temperatures of the ball and plate by
the same amount
• C: heat the ball and cool the plate
• D: cool the ball and heat the plate
Solution: thermal expansion
• B and D
• Since the coefficient of linear expansion of
the ball is greater than the plate, it will
shrink more per change in temperature as
the temperature of both is lowered. Also,
by cooling the ball you will decrease its
size and by heating the plate you will
increase the size of the hole.
Thermal Expansion
• Thermal expansion is
a property of the
material
• different materials
expand differently
• Engineers need to
take this into account
in their designs:
expansion joints in
bridges
• bi-metal strip demo
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Thermal expansion
• This is how
thermostats workbimetallic strips in
refrigerators, ovens,
etc. open and close a
switch as the
bimetallic strip bends
one way or the other
due to temp changes
• If you have a metal lid
on a glass jar that is
stuck on too tight,
how can you use this
to get the lid off?
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Volume Thermal Expansion
• Similar to linear expansion, the expansion
of volume with an increase in temp is
related to the initial volume, the change in
temperature, and a proportionality
constant
• This is primarily useful with liquids since
they don’t expand linearly
• The coefficient of volume expansion = β
• ΔV=βVoΔT
Problem solving- thermal
expansion
• A concrete sidewalk is constructed between two
buildings on a day when the temperature is 25 degrees
celsius. The sidewalk consists of 2 slabs, each 3m in
length and negligible thickness. As the temperature
rises to 38 degrees, the slabs expand but no space is
provided for thermal expansion. Concrete has a
coefficient of linear expansion of 12 x 10-6 The buildings
don’t move so the slabs buckle upward. Determine the
vertical distance (y) that the slaps stick up after buckling.
• Hint- DRAW!
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Solution
• The slabs expand linearly according to
ΔL=αLiΔT
• ΔL=(12 x 10-6)(3m)(13)=0.00047m
• The new length of the slab is then 3.00047 which
gives us the hypotenuse of the triangle. The
height, y, is found through the pythagorean
theorem. The x is the original length, 3m.
• Y=sqrt [(3.00047)2+(3)2]= 0.053m
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How does heat transfer?
• You’ve heard this before…
• Conduction, convection, radiation
• Remember them?
Convection
• Think convection
currents!
• The different temp
molecules actually
move
• Thus the “hot air
rises”
phenomenon
Conduction
• Heat transfer through
collisions of moleculesenergy transferred from
higher temp to lower
temp
• Thermal conductors
transfer heat well
• Thermal insulators do
not
Factors Affecting Conduction
• Amount of heat (Q) conducted along a bar
depends on what factors?
t =Time
T= Temperature difference
A=Thickness or cross section area
L=Length (inversely proportional)
The material involved (thermal conductivity,
k, of that material)
Conduction of Heat through a
material
• Q=(kAΔT)t/L
• So try one: When excessive heat is
produced in the body, it must be transferred
to the skin and dispersed to maintain a
constant 37.0°C. One possible mechanism
for transfer is through body fat which has a
thermal conductivity of 0.20J/sm°C.
Suppose the heat travels through 0.030m of
fat to reach the skin which has a surface area
of 1.7m2 and a temperature of 34.0 °C. Find
the amount of heat that reaches the skin in a
half hour.
Problem Solving- Conduction
• How much heat would flow through an 8m2 area of an uninsulated concrete block
(k = 0.8 W/Km) house in 24 hours if the
temperature is 37ºC on the one side and
22ºC on the other. The block is 25 cm
thick.
Solution- Conduction
• ΔQ = kA(T2-T1)t/ L
• ΔQ = (0.8)(8)(37-22)[(24)(3600)]/0.25
• ΔQ = 3.32 x 107 J
Radiation
• Energy transfer
without a mediumthrough
electromagnetic
waves
• We will do lots
more on radiation
later!
1st Law of Thermodynamics
• Energy Conservation ΔU+ΔK+ΔQ=0
• When you add energy to a system it can
do 2 things:
– Increase the internal energy of system if it
stays in the system (measured by increased
temp)
– Do work if it leaves the system
30
1st Law of Thermodynamics
• ΔU = Q + W
• ΔU represents the net change in the
internal energy of the gas
• Q represents the net heat added (+) or
removed (-) from a confined gas
• W is work done by the confined gas (-) or
on the confined gas (+)
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Problem:
• A 5-kg aluminum block slides from rest
down a 1-meter long, 37º incline. When it
arrives at the base of the incline it's speed
is only 3 m/sec.
• How much energy is lost to frictional heat?
32
Solution
• PEtop = mgh = 5(10)(1 sin 37º) = 5(10)(.6)
= 30 J
• KEbottom = ½mv2 = ½(5)(3)2 = 22.5 J
• Energy lost = 7.5 J
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One step further
• If 880 J of heat are needed to raise the
temperature of 1 kg of aluminum by 1 Cº that is, aluminum has a specific heat of
880 J/kg Cº - how much did the
temperature of our block increase after
sliding down the incline?
34
solution
• 7.5 = mc ΔT
• 7.5 = (5)(880) ΔT
• ΔT = 0.0017 Cº
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nd
2
rd
3
and
Laws of
Thermodynamics
• Heat flows from hot to cold
• No system can reach absolute zero
36
Heat Engines
• Transfer internal
energy into
mechanical work
• Heat flows from hot
reservoir (burning
fuel) to cold reservoir
(exhaust) and some
of that energy is
removed from system
as work (drives
piston)
•
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Carnot Efficiency
• Heat Engines
CANNOT be 100%
efficient: always some
heat exhaust
• Ideal efficiency
depends on T
differences
• =Thot-Tcold
Thot
• Heat engines are
limited by this- they
must strive fro huge
differences in temp
• In cars, fuel cells and
electric motors are
NOT heat engines so
they can achieve
higher eficiency
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Four Gas Processes
Isothermal
• Constant Temp
Isobaric
• Constant Pressure
• P1V1=P2V2
• V1/T1=V2/T2
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Four Gas Processes
Isochroic
• Constant Volume
• P1/T1=P2/T2
Adiabatic
• Constant Heat- accomplish
this by performing rapidly
(pumping up a tire) or
isolating the system
• When you pump up a tire,
you compress the gas
causing the temperature to
rise even though you have
not added heat energy!
• Opposite true for
expansion- air expanding
40
rapidly drops in temp