Physical Science
Temperature and Heat
Slides subject to change
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Scales
Fahrenheit
Celsius
212 °F
100 °C
32 °F
0 °C
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Scales
Celsius
Kelvin
100 °C
373 K
100 units
0 °C
273 K
Absolute zero
0K
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Convert °F to °C

TC = 5/9 (TF – 32) (approx. subtract 32, half
it)
What is normal body temperature (98.6 °F)
in °C ?
 TC = 5/9 (TF – 32)
= 5/9 (98.6 – 32) = 37 °C

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Convert °C to °F
TF = 9/5 TC + 32
 During Napoleon’s famous retreat from
Moscow (Russian War of 1812), the
temperature went as low as –35 °C. What
was that temperature in °F?
 TF = 9/5 TC + 32
= 9/5 (−35) + 32 = –31 °F

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Temperature
Temperature (in Kelvin) is an indication of
average KE of the molecules of a
substance.
 Higher temperature, higher average
speed.

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Temperature Overseas

Moscow

London
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Old Theory of Heat Transfer
Caloric theory: Heat consists of a fluid
called caloric that flows from hotter to
colder bodies.
 The more caloric in a substance, the hotter
it was.
 Heated objects swelled because a big
dose of caloric had just been added.
 Amount of heat in any system would
always stay the same, or “heat is always
conserved.”

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Modern View of Heat Transfer
Net energy passes from one object to
another because of temperature
difference.
 Relates heat energy to kinetic energy.

Molecules
very active,
high T
Molecules
not so active,
low T
Heat energy
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James Prescott Joule
English brewer and
physicist.
 Discovered heat’s
relationship to
mechanical work (1843).


Insulated
barrel of
water
Falling
weight
(work)
Work done by the falling weight is
translated into rotation of the paddles and
increased temperature (energy) of the
liquid in the container.
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Heat Units: Small Calories

One calorie is the heat energy required to
raise the temperature of one gram of water
1 °C.

James Joule found that 4.186 J = 1
calorie.

This is called a “small calorie”
... because a gram is a small mass of
water.
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Large Calories
1 Cal = 1,000 calories = 1 kcal.
 Called a large calorie.
 1 Cal = 4,186 J
 Used to measure energy in food.


Food calories.
 Rich Ice Cream, Vanilla, ½ cup: 266
Calories.
 (266 Cal)(4,186 J/Cal)= 1,110,000J
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Large Calories

One
gram of pure
carbohydrate
Recommended
intake
for an adult person
(example,
sugar)
yieldsCalories/day.
about 4
is about 2,000
- 2,500
Your
Calories
of energy
(16,700
J).
body needs
calories
for energy.

Non-caloric: water, vitamins, minerals,
antioxidants, caffeine, etc.
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Specific Heat

Specific heat c is the heat energy required
to raise the temperature of one kilogram of
a substance 1 °C.
Sometimes the units of c are in
 J/kg-°C
 kcal/kg-°C


We’ll stick with J/kg-°C.
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Heat Capacity Examples
Substance
Aluminum
Glass
Ice
Gold
Steam
Water, liquid
J/kg-°C
920
670
2,100
2,490
2,100
4,186
Specific Heat = Heat energy required to raise the temperature
of one kilogram of a substance 1 °C.
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Heat Capacity
1.0 kg
1.0 kg
Aluminum

Takes 920 J to
raise temperature
1° C
Water

Takes 4,186 J to
raise temperature
1° C
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
Objects that are mostly water (Earth or the
human body) are not as sensitive to
sudden changes in temperature compared
to bare rock.
Temperature Changes
Amount of heat to raise temperature equals
mass times specific heat times temperature
change.
 Let ΔT = Tf – Ti

Q = m c ΔT
Temperature
change, °C
Heat energy, J
Mass, kg
Specific heat,
J/kg-°C
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Cup of Tea
How much heat energy is needed to raise
0.250 kg water (1/4 Liter) from 10 °C to
100 °C (1 cup of tea water)?
Given
Formula
 m = 0.250 kg
Q = mc ΔT
 c = 4,186 J/kg-°C
 ΔT = 100 – 10 = 90 °C
= (0.250)(4,186)(90)
= 94,000 J

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Example
2 kg of carbon steel is heated from 20 °C
to 100 °C. Specific heat of carbon steel
is 0.49 kJ/kg-°C. How much heat required?
Given
Formula
 m = 2 kg
Q = mcΔT
 c = 490 J/kg-°C
 ΔT = (100–20) = 80 °C


Q = (2)(0.49)(100 – 20) = 78,400 J
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Phases of Matter

Phase changes may occur as we add heat
energy:
phase change
phase change
Solid → Liquid → Gas
Solid
Particles have fixed positions
in space relative to each
other.
Gas
Liquid
Particles without a definite
shape or volume.
Particles free to move, but
they form a discrete surface.
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Ice – Water Transition
Start with ice, at a temperature
lower than the melting point.
 Add heat to raise ice to melting
point.
 At 0 °C, keep adding heat,
temperature is constant.
 Finally converts to water, and
temperature rises again.

100 °C
0 °C
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Latent Heat: Phase Changes

Heat required for Solid → Liquid transition
called Latent Heat of Fusion Lf.
Q = m Lf

Heat required for Liquid → Gas transition
called Latent Heat of Vaporization Lv.
Q = m Lv
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Latent Heat

The following table shows the latent heats and change of
phase temperatures of some common fluids and gases.

From Wikipedia
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Water
cice = 2,100 J/kg-°C
cwater = 4,186 J/kg-°C
Lf = 335,000 J/kg

How much heat is required to raise 1.0 kg of
ice from –10 °C to liquid at 25 °C?
Three Steps
1. Heat ice from –10 °C to 0°C (Q1 = mciΔT)
2. Melt solid ice to water at 0 °C (Q2 = m Lf)
3. Heat water from 0 to 25 °C (Q3 = mcwΔT)
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Water
cice = 2,100 J/kg-°C
cwater = 4,186 J/kg-°C
Lf = 335,000 J/kg
1. Heat solid ice from –10 °C to 0 °C:
Q1 = mciΔT = (1.0)(2,100)(10) = 21,000 J
2. Melt solid ice to water at 0 °C :
Q2 = m Lf = (1.0)(335,000) = 335,000 J
3. Heat liquid water from 0 °C to 25 °C:
Q3 = mcwΔT = (1.0)(4,186)(25) = 105,000 J
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Water
cice = 2,100 J/kg-°C
cwater = 4,186 J/kg-°C
Lf = 335,000 J/kg

How much heat is required to raise 1.0 kg of
ice from –10 °C to liquid at 25 °C?
Three Steps
1. Heat ice from –10 °C to 0°C (Q1 = 21,000 J)
2. Melt solid ice to water at 0 °C (Q2 = 335,000 J)
3. Heat water from 0 to 25 °C (Q3 = 105,000 J)
Total Heat Energy = Q1 + Q2 + Q3 = 461,000 J
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The Letter “C”
Note we now have several physical heat
characteristics containing the letter “C.”
1. °C – degree Celsius – A measure of
temperature.
2. cal – small calorie – A measure of heat
energy – 4.186 joules.
3. Cal – large Calorie – A larger measure
of heat energy – 4,186 joules or 1000 cal.
4. c – specific heat – capacity of a material
to hold heat energy – units are J/kg-°C.

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Boiling Point



Boiling point of water is
affected by pressure.
High-pressure, “pressure
cooker,” liquids (water,
shortening) boil at higher
temperature, food cooks
faster.
Col. Sanders
uses pressure
cooker ca. 1939
Low pressure, water boils at lower
temperature. Denver 94.4 °C (202 °F).
Takes longer to make a hard-boiled egg.
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Thermal Expansion

Matter tends to change in volume in
response to a change in temperature.

When heated, molecules move more
vigorously and maintain a greater average
separation.

Thus, normally, objects expand when
heated, shrink when cooled.
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Road Expansion Joints
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Heat Buckling
32
a solution to the
problem—rails slide by
each other
Bi-Metallic Strip
brass
steel
Thermostat
bi-metallic strip
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Water – The Exception
An exception is water. It
expands as it solidifies.
 densities:
 water: 1.00 g/cc
 ice: 0.92 g/cc


Enables ice to float. Density
decreases because it’s the
same mass, but more volume.
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
How does this happen?
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Heat Transfer
Conduction is the transfer of heat
by molecular collisions. Within a
mass.
 Convection: Movement of a
substance from one position to
another. Air flow.


Radiation: transfer
energy by
electromagnetic waves.
Tanning salon.
Infrared Heater
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Heat Conduction



Heat conduction is the
transfer of thermal energy
between neighboring
molecules due to a
temperature difference.
It always takes place from
a region of higher
temperature to a region of
lower temperature
Acts to equalize the
temperature differences.
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Thermal Conductivity

Property of a material that
indicates its ability to conduct
heat.
Material
W/m-K
Air
0.025
Soil
Glass
Aluminum
Stainless Steel
Copper
0.15
1.1
237
12-45
401
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Convection
Convective
heat transfer is
the transfer of
heat from one
place to
another by the
movement of
gas or liquids.
 Convection
oven has a fan
inside.

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Radiant Heater
Radiant heating is a
technology for
heating indoor and
outdoor areas.
 Radiant heating
consists of radiant
energy being
emitted from a heat
source.

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Kinetic Theory of Gases
Molecules
 All directions at high speeds.
 Collide with each other and walls of the
container.
 Pressure results from
collisions with the walls.
 More collisions/second =
higher pressure.

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Ideal Gas Law

Let
Important! Kelvin!
 P = pressure
 T = temperature (in K)
 N = number of molecules
 V = volume

Ideal gas law,
 P proportional to

NT
V
pressure is proportional to number of molecules times
temperature (Kelvin) divided by volume.
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Two Results
Example: Hold N and T constant,
Pα 1
or P1/P2 = V2/V1
V

Hold N and V constant
Pα T
or P1/P2 = T1/T2

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Example, N and V Constant
Temperature of a cylinder of gas at T1=
20 °C heated to T2=1,000 °C (a campfire).
 What is the new pressure relative to initial
pressure?
 Convert to Kelvin K ! (TK = TC +273)
 T1= 20 °C = 293 K
 T2= 400 °C = 1,273 K
 P2/P1 = T2/T1 = (1,273/293) = 4.3
 P2 = 4.3 P1 new pressure is 4.3 times the

initial pressure.
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Solar Radiation


Radiation from the Sun heats Earth surface.
Some absorbed by atmosphere.
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Solar Radiation


Of the radiation that reaches Earth’s surface.
70% absorbed by surface, 30% reflected.
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Global Warming

Infrared absorption warms the atmosphere –
both directly from Sun and reflection from
Earth surface.

Atmosphere transfers the energy it receives
both into space and back to the surface.
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Greenhouse Gases
Radiation absorption in infrared spectrum
is limited to greenhouse gases.
 The more greenhouse gases, the more
infrared absorption.

Molecules
Percent Greenhouse
Gases in Atmosphere
Water vapor
~60%
Carbon dioxide
~20%
Methane (CH4), Nitrous oxide
(N2O), ozone (O3),
~20%
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Water Vapor

Water vapor contributes
significantly to the
greenhouse effect ensuring a
warm habitable planet.
Water vapor also forms clouds that reflect
sunlight, which moderates global warming.
 Amount of water vapor is not affected by
human activity.

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Increasing Carbon Dioxide
Level of CO2 is affected 5−10% by human
activity, but especially by increasing
temperatures. Oceans give up CO2.
 Increased CO2 usually follows increased
temperatures by 200-800 years, based on
Antarctic ice core studies.

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“Chartsmanship”
CO2 Concentration Tasmania
370
400
365
350
360
300
355
250
PPM
PPM
CO2 Concentration Tasmania
350
200
345
150
340
100
335
50
330
1975
1980
1985
1990
1995
2000
2005
0
1975
1980
Year
 In
official
reports.
1985
1990
1995
2000
2005
Year
 To
scale.
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2008
Goddard Institute for Space Studies
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Sunspots
Sunspot intensity cycles every 10-11 years.
 Amount of emitted radiant energy varies with
sunspot activity. Significant influence on
Earth’s climate.

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1990 IPCC Graph


Warmer during Medieval Warm Period 1000−
1300 A.D. Colonization of Greenland by the
Vikings early in the millennium.
Colder during Little Ice Age (1550−1650 A.D).
Greenland abandoned, Thames River froze over.
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Global Warming Summary

Over last 100 years, average temperature has
increased by approximately 0.7 °C.
Solar activity, orbital variations closely
associated with climate.
 CO2 and N2O concentrations are increasing.
 Overall sea levels difficult to measure.
 In last 10 years satellites measure very
minor increase in sea levels.
 97% of Antarctic is colder, ice is growing.


See data (+ 3 mm/yr) http://nsidc.org/sotc/sea_level.html
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