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Topic31ThermalConceptsNotes-1

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Topic 3: Thermal Physics
Sub Topic 3.1: Thermal Concepts
§ Molecular theory of solids, liquids and gases
§ Temperature and absolute temperature
§ Internal energy
§ Specific heat capacity
§ The topic of thermal physics is a good example of the use of international systems
of measurement that allow scientists to collaborate effectively
§ Phase change
§ Specific latent heat
§ Describing temperature change in terms of internal energy
§ Using Kelvin and Celsius temperature scales and converting between them
§ Applying the calorimetric techniques of specific heat capacity or specific latent
heat experimentally
§ Describing phase change in terms of molecular behaviour
§ Sketching and interpreting phase change graphs
§ Calculating energy changes involving specific heat capacity and specific latent
heat of fusion and vaporization
“Thermal physics deftly demonstrates the links
between the macroscopic measurements essential to
many scientific models with the microscopic
properties that underlie these models.”
The molecular theory of solids, liquids,
and gases
Solids
The molecules (atoms, or ions) are held closely
together by strong forces, usually in regular patterns.
• They can vibrate about their mean positions
• Have fixed shape and volume
The molecular theory of solids, liquids,
and gases
Liquids
• Molecules still vibrate, but some molecules have
enough kinetic energy to overcome the forces,
allowing them to move a little
• Little to no regularity in arrangement
• Constantly changing arrangement
• Fixed volume, but variable shape
The molecular theory of solids, liquids,
and gases
Gases
• Much further apart than solids & gases
• Weak forces betwixt (usually negligible except
collisions)
• Molecules move independently in different
directions/speeds
• Constantly changing
• No fixed shape or volume
Temperature and Absolute Temperature
Temperature:
Scalar Quantity. Measured in degrees
Celsius (oC) or Kelvin (K)
K vs. oC
Celsius (sometimes called Centigrade)
• Based off of when pure water
freezes (0o C) and boils (100o C) at
normal atmospheric pressure
• Values are chosen, not discovered.
• 10o C is not “twice as hot” as 5o C.
Kelvin (Absolute temperature)
• Based off Celsius scale
• Zero reset to Absolute Zero
• Where molecular motion stops
at -273o C
Converting between Celsius and Kelvin:
Temperature (K) = temperature (oC) + 273
• In data booklet under unit conversions, NOT with
thermal equations
• Because increments of both scales are equal, β–³K =
β–³oC
• When using β–³T, you can use either. When using
only one temperature, use K.
• *There is no oK. Only K. Don’t write oK.
Common Temperatures in both Celsius
and Kelvin
Internal Energy
Thermal
Energy:
• Energy transferred from hotter to colder as a result
of temperature difference
• Sometimes called “heat”
• “Heat” can also be used wrongly to describe the
heat inside an object. This is actually called
internal energy.
Internal
Energy:
• The total of the intermolecular potential and random
kinetic energy in a substance.
• Kinetic can be vibrational, translational, or
rotational.
• In gas, potential is negligible, making the energy
almost entirely kinetic.
Internal energy visualized
• All substances are composed of individual
molecules that are in vibration.
• As we heat up a substance its vibrations become
more energetic. This is an increase in the kinetic
energy of the molecules.
• Simultaneously, as heat energy is being added the
molecules are also moving farther apart. This is an
increase in the potential energy of the substance.
• The two energies together are called the internal
energy of the substance. Thus EINT = EK + EP.
• When thermal energy (heat) is added to a substance
it is stored as internal energy.
More on Internal Energy
EINT = EK + EP.
total internal energy
Total Internal Energy
Potential Energy
(due to inter-molecular forces)
Think of Hooke’s law
F = -kx and phase change
Kinetic Energy
(due to
vibration and translation
The thermometer
measures internal
kinetic energy
Thermal equilibrium and heat transfer
Thermal
Equilibrium
• When there is no net flow of thermal energy
between 2 or more objects.
• Thermal energy will flow from hotter (emits
energy) to colder (absorbs energy) until both
temperatures are the same.
Specific Heat Capacity
Specific Heat • The energy transferred to 1 kg of the substance
causing its temperature to increase by 1 K.
Capacity (c)
-1 -1
• Unit: J kg K
• 𝑐=
#
$βˆ†&
or more commonly:
𝑄 = π‘šπ‘βˆ†π‘‡
• Q: amount of energy supplied to mass (J)
• m: mass (kg)
• βˆ†T: Change in temperature (K)
• (*This is the same as βˆ†oC)
• c: Specific heat capacity (J kg-1K-1)
In other words…
Objects with a higher heat capacity heat up (or cool down) slowly
compared to an equal mass of substance with a lower specific heat
capacity.
An Iron Kiln: To watch a guy use primitive technology to create an iron kiln from scratch, check
out this link.
Practice Problem:
A piece of iron of mass 0.133 kg is placed in a kiln until it reaches the
temperature 𝛉 of the kiln. The iron is then quickly transferred to 0.467 kg
of water held in a thermally insulated container. The water is stirred until it
reaches a steady temperature. The following data are available.
Specific heat capacity of iron = 450 J kg-1 K-1
Specific heat capacity of water = 4.2 x 103 J kg-1 K-1
Initial temperature of the water = 16oC
Final temperature of the water = 45oC
The specific heat capacity of the container and the insulation is negligible.
a) State an expression, in terms of 𝛉 and the above data, for the energy
transfer of the iron in cooling from the temperature of the kiln to the
final temperature of the water.
Energy Transfer: Q
Mass of iron: 0.133 kg
Initial Temp of Iron: 𝛉
Final temp of water (and iron): 45oC
Specific Heat Capacity of Iron: 450 J kg-1 K-1
𝑄 = π‘šπ‘βˆ†π‘‡
𝑄 = 0.133 × 450 × 45 − πœƒ
𝑸 = πŸ”πŸŽ ×(πŸ’πŸ“ − 𝜽)
b) Calculate the increase in internal energy of the water as the iron cools in the water.
Increase in Internal Energy: Q
Mass of water: 0.476 kg
Initial Temp of Water: 16 oC
Final temp of water: 45 oC
Specific Heat Capacity of Iron: 4.2 x 103 J kg-1 K-1
𝑄 = π‘šπ‘βˆ†π‘‡
𝑄 = 0.476 × 4200 × 45 − 16
𝑸 = πŸ“. πŸ•×πŸπŸŽπŸ’ J (using 2 s.f.)
c) Use your answers from b) and c) to determine 𝛉
*Energy absorbed by the water (Q in part b) is equal to the
energy transferred by the iron (Q in part a). The negative
indicates that energy was emitted by the iron.
𝑄 = 60 ×(45 − πœƒ)
−5.8×10D = 60×(45 − πœƒ)
−5.8×10D = 2700 − 60πœƒ
−6.1×10D = −60πœƒ
1010 = πœƒ
𝛉 = 1010oC to 3 s.f.
EXAMPLE: Air has a density of about r = 1.2 kg m-3. How much heat, in
joules, is needed to raise the temperature of the air in a 3.0 m by 4.0 m
by 5.0 m room by 5°C?
•
•
•
From the previous table we see that c = 1000 J kg-1 K-1 .
The change in temperature is given: DT = 5°C.
We get the mass from r = m / V or
m = r rV = (1.2)(3.0 x 4.0 x 5.0) = 72 kg.
Q = mcDT = (72)(1000)(5) = 360,000 J or 360 kJ.
Let’s try another example:
After a long day of yelling at kids, Mr. Christofi needs a (boiling) hot cup of
tea to relax during an afterschool meeting, which starts in 10 minutes. The
teachers lounge kettle has a power rating of 500 W and Mr. Christofi’s tea cup
holds 500g of water (22oC starting temperature). If it takes 1 minute to walk to
the teachers lounge, and 2 minutes to walk to the meeting (slowly with a hot
tea!), will Mr. Christofi be late to the meeting?
P= 500 W
m = .500 kg
Ti = 22oC
Tf = 100oC
c = 4180 J kg-1 K-1
t=?
𝑄 = 𝑃𝑑
𝑄 = π‘šπ‘βˆ†π‘‡
𝑃𝑑 = π‘šπ‘βˆ†π‘‡
500𝑑 = .500(4180)(100 − 22)
𝑑 = 326.04 𝑠
𝑑stsuv
326.04 𝑠
=
+1+3=
60𝑠
8.43 min
Phase Changes
Phase
Changes
PHASE CHANGE PROCESS
EXAMPLE
solid ¾® liquid
melting
ice to water
liquid ¾® solid
freezing
water to ice
liquid ¾® gas
gas ¾® liquid
boiling
condensing
water to steam
steam to droplets
solid ¾® gas
gas ¾® solid
sublimation
deposition
frost evaporation
frost
What is temperature?
Temperature
Measure of the average kinetic energy of the particles
in a substance.
As temperature is increased, particles
move faster, which increases the kinetic
energy. This causes the substance to
expand.
Thermometers Expansion causes the liquid in the
bulb to expand, rising up into the tube.
A Thermostat uses 2 strips of metal bound together. As the
temperature rises, the metals expand at different rates,
causing the strips to bend.
Specific Latent Heat
Specific Latent
Heat (L)
• The energy required to achieve a change of
phase
• “Latent” means “hidden”
Q = mL
unit: J kg-1
Q = Thermal Energy (J)
m = Mass (kg)
L = Specific Latent Heat (J kg-1)
Specific Latent Heat
Specific Latent
Heat of Fusion
(Lf)
(melting)
• Energy required to change the phase of 1 kg of
substance from solid to liquid (without DT)
Specific Latent
Heat of
vaporization
(Lv)
(boiling)
• Energy required to change the phase of 1 kg of
substance from liquid to gas (without DT)
Phase Change Graphs
Latent heat of
vaporization
Calculating Thermal Energy
Temperature Change: Use Q = mcDT
Latent
heat of
fusion
Phase Change: Use Q = mL
Assumptions and Info about the graph:
• Assumes energy is supplied by constant power
source.
• “Energy” could be replaced with time axis
and it would produce the same graph (Q=Pt)
Additional Information:
• The graph remains flat until ALL of the substance has been converted
(to solid, liquid, or gas)
• Notice that the flat part of the graph has no change in temperature,
therefore, the Latent Heat equation also does not contain a temperature
variable.
• One substance will have different Specific Latent Heat values (L) for
fusion and vaporization. Make sure you use the correct one.
Example Problem: How much energy is needed in order to
change a 3 kg block of ice (starting temperature -40oC into 3 kg
of water vapor at 160oC?
2014 Physics HL Paper 2 Question (7 marks)
This question is about energy.
a) At its melting temperature, molten zinc is poured into an iron
mould. The molten zinc becomes a solid without changing
temperature.(a) Outline why a given mass of molten zinc has a greater
internal energy than the same mass of solid zinc at the same
temperature. (3 Marks)
b) The zinc is allowed to cool in the mould. The temperature of the iron
mould was 20°C before the molten zinc, at its melting temperature, was
poured into it. The final temperature of the iron mould and the solidified
zinc is 89°C. The following data are available.
Mass of iron mould =12kg
Mass of zinc =1.5kg
Specific heat capacity of iron =440Jkg–1K–1
Specific latent heat of fusion of zinc =113 kJ kg–1
Melting temperature of zinc =420°C
Using the data, determine the specific heat capacity of zinc. (4 marks)
Essential Idea
“Thermal physics deftly demonstrates the links
between the macroscopic measurements essential
to many scientific models with the microscopic
properties that underlie these models.”
At a macroscopic level, we experience temperature; we heat things up, and observe ice
melting. We can measure the pressure of a gas and discover how it relates to its
temperature. But, to explain all this, we need to invent a microscopic world, where
unseen entities interact according to their own rules; where entities give rise to our
everyday experiences through a series of correspondencies.
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