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Peter's Index Physics Home Lecture 4 Course Index Lecture 6
A Semester of First Year Physics with Peter Eyland
Lecture 5 (Expansion and Calorimetry)
In this lecture the following are introduced:
Thermal expansion
Hydrogen bonding
Internal energy
Heat capacity
Specific heat
Heat transfer by mixing
Latent heat
Thermal expansion
The microscopic picture of solids
Solids are three-dimensional arrays of vibrating atoms stacked ~0.1 nm apart.
The atoms are held in place by cohesive forces that are the result of negative potential energy.
In a solid, the negative potential energy of the atoms is much greater than their kinetic energy,
so the atoms are confined down in a potential well and their vibrations do not break the bonds between them
(unless the temperature approaches the melting point).
When the temperature of a body increases the average distance between the atoms usually increases and so the volume
of the body increases.
If a body is hollow it will expand on heating as if it were complete.
It does not expand inwards to fill the empty space - the empty space expands.
Expansion Magic
The expansion of materials when heated was used to good effect in some early Temples.
In the example below, a priest would stand at a burning incense altar and command the doors while throwing
flammable substances on the altar to make the fire rise higher.
When the fire rose, then the doors of the Temple would open apparently by themselves.
The priest could thus command the doors and they would open to him!
Unbeknown to the devotees, the increase in flame fire suddenly heats the hollow bronze altar.
Even though the bronze altar expands momentarily, the air that is trapped inside the altar will expand much more.
The suddenly expanded air increases the pressure on the water below and water is pushed up the pipe so that it
falls into the bucket.
As the bucket fills with water, it pulls on the rope that is connected to a lever arm which opens the Temple door.
Different devotees would perform isolated tasks in re-setting the system (e.g. just empty the bucket)
so that they would not see the whole picture and understand the significance of what they did.
Linear Expansion
The graph on the right gives an expansion relationship between length and temperature. |
In symbols this is:
Replacing the proportionality sign with a constant, this becomes:
The proportionality coefficient, α (alpha), is called the
linear coefficient of thermal expansion.
Expressed this way, the units of α are simply K-1 because the ratio of lengths is dimensionless.
Expanding this:
Note that the slope of the graph is α (alpha) times the reference temperature length.
Example
The Eiffel Tower in Paris is 300.00 m high at 200C and for its steel,
the temperature coefficient of thermal expansion is 1.27 x 10-5 K-1.
Find its height at 400C.
Table of Expansion Coefficients
Material |
Temperature |
α K-1 |
Diamond |
20 0C |
1.0 x 10-6 |
Glass (Pyrex) |
50 0C |
3.2 x 10-6 |
Glass (most) |
50 0C |
8.3 x 10-6 |
Platinum |
20 0C |
8.9 x 10-6 |
Concrete |
20 0C |
12 x 10-6 |
Steel |
20 0C |
13 x 10-6 |
Copper |
20 0C |
17 x 10-6 |
Brass |
20 0C |
19 x 10-6 |
Aluminium |
20 0C |
23 x 10-6 |
Lead |
20 0C |
29 x 10-6 |
Ice |
- 5 0C |
51 x 10-6 |
Celluloid |
50 0C |
109 x 10-6 |
Example
The Moscow to Leningrad/St Petersburg telephone line is 500 m shorter in Winter than Summer.
If the temperature coeficient of linear expansion for copper is 17 x 10-6 K-1,
and the "Winter-to- Summer" temperature variation is 45 0C,
find the Winter-time length of the telephone line.
Example
A steel "seconds" pendulum (i.e. a pendulum with a period of 2 s) is correct at 200 C.
Given that the temperature coefficient of linear expansion for steel is 1.27x10-5 K-1,
find the temperature for which the period is 0.02% larger.
Area and Volume Expansion
Take a square of solid material at the reference temperature with sides of length
.
At a new higher temperature the length will increase by
Δ.
The new area will be given by:
The change in area is:
Similarly for a cube, the volume coefficient of expansion is three time the linear one.
Example
A flat thin piece of Aluminium at 20 0C is formed into a circle of radius 340 mm.
The linear temperature coefficient of thermal expansion for Aluminium is 23 x 10-6 K-1.
Find the area at 60 0C.
Water and Hydrogen Bonding
Water normally expands on heating, however when being heated from 00C to 3.98 0C it contracts, reaching a maximum density at 3.98 0C. This is due to Hydrogen bonding. |
Hydrogen bonding arises from a normal bond between a Hydrogen atom and a neighbour.
Since any other atom will bind the electron from the Hydrogen atom more tightly,
the electron will spend more time with the other atom.
This creates a permament dipole with a fraction of the electronic charge that can interact with other dipoles nearby.
The water molecule (DHMO) is the classic situation,
where the Oxygen molecule binds electrons from both Hydrogen atoms more tightly than Hydrogen can.
Hydrogen bonding makes ice less dense than water because the random dipole orientations force
non-compact stacking.
As water cools towards 3.98 0C the molecules become closer together till the dipole forces become significant.
In the region from 00C to 3.98 0C microcrystals of ice are forming so the density drops.
The Hydrogen bond enables life to continue on Earth because in winter,
rivers freeze down from the top and not up from the bottom, enabling plant and animal life to survive.
Internal Energy
Heat is the energy that flows when there is a temperature difference.
When heat flows into or out of a system, conservation of energy says that the energy of that system must change
by the same amount.
The system had an original internal energy and that internal energy has been changed
by the heat flow.
In symbols, Q = U2 - U1, where Q is positive when heat is added to the system.
The internal energy function, U, involves things like pressure, volume and temperature.
It doesn't need to be specified because only a difference is used.
Because heat is a form of energy, the proper S.I. unit is the Joule (J).
Other units such as the calorie are used occasionally: 1 Joule = 4.186 calorie
Note the upper and lower case letters
1 KJ = 4.186 Cal = 4186 cal, where 1 Calorie (used in nutrition) = 1000 calorie
Heat Capacity
When heat is added to a system the temperature normally changes.
The heat capacity, C, of a system is the heat needed to change its temperature by one degree.
The water equivalent of a system is the mass of water with the same heat capacity.
Specific Heat
Heat capacity is an extrinsic property of a system,
it depends on both the mass and type of material of which its parts are made.
The specific heat, c, is an intrinsic property that depends only on the type of material.
Thus, when heat is added to a system with mass m:
The SI unit is J.kg-1.K-1
Table of Specific Heats
Substance |
Specific Heat J.kg-1.K-1 |
Lead |
128 |
Tungsten |
134 |
Mercury |
140 |
Silver |
236 |
Brass |
380 |
Copper |
386 |
Glass |
840 |
Aluminium |
900 |
Ethanol |
2430 |
Sea Water |
3900 |
Water |
4190 |
Molar Specific Heat
The mass of a substance depends on the number of molecules present in it.
To measure the number of molecules, a unit is used called the gram mole.
The number of molecules in a gram mole is called Avogadro's number NA.
The gram mole is a number like a dozen (12), or score (20), or gross (144), but very much bigger.
Avogadro's number is 6.022 × 1023.
The molar specific heat is the specific heat of Avogadro's number of molecules of a substance,
and it has the SI unit of J.mol-1.K-1.
The molar mass is the mass of Avogadro's number of molecules.
Eggsample
An egg has a volume of 50ml.
The mean radius of the Earth is 6370km.
For Avogadro's number of these eggs, find how high they will be stacked.
(Assume perfect stacking and no spaces between eggs.)
Table of Molar Specific Heats
Element |
Molar Specific Heat at room temperature |
Aluminium |
24.4 J.mol-1.K-1 |
Copper |
24.5 J.mol-1.K-1 |
Tungsten |
24.8 J.mol-1.K-1 |
Silver |
25.5 J.mol-1.K-1 |
Lead |
26.5 J.mol-1.K-1 |
Molar Specific Heats and Temperature
|
Calorimetry or Heat Transfer by mixing
When substances are mixed, by being brought into intimate thermal contact, in an isolated system,
where there is no energy loss to the environment outside, then a simple conservation principle applies.
The heat lost by the hot substance equals the heat gained by the cold substance.
Example
An insulated vertical Copper pipe of mass 0.25 kg is initially at 150C.
The bottom end is capped and 0.3 kg of water at 940C is poured into it.
The specific heat of Copper is 386 J.kg-1.K-1.
The specific heat of Water is 4190 J.kg-1.K-1.
Assuming no heat loss to the environment, find the final temperature.
Apply the general principle is conservation of energy, i.e.
Heat lost by water = Heat gained by copper
Example
A 0.15 kg sample of a new alloy is heated to 5400C.
It is then quickly placed in 0.40 kg of water at 100C which is contained in a 0.20 kg
aluminum calorimeter cup at the same temperature.
The final temperature of the mixture is 350C.
The specific heat of Water is 4190 J.kg-1.K-1.
The specific heat of Aluminium is 900 J.kg-1.K-1.
Assume no heat is lost to the surroundings.
Calculate the specific heat of the sample.
Latent Heat
The word "latent" means "hidden". While a substance is boiling the temperature remains constant.
Heat is added but the temperature doesn't change, so the heat to boil it completely away into a gas is called a
latent heat.
The latent heat of vapourisation is the heat absorbed (at constant temperature) to boil a substance into a gas:
or the heat released (at constant temperature) to condense a gas into a liquid.
The latent heat of fusion is the heat absorbed (at constant temperature) to melt a substance into a liquid:
or the heat released (at constant temperature) to freeze a liquid into a solid.
In symbols:
Table of Latent Heats
Substance |
Fusion kJ.kg-1 |
Vapourisation kJ.kg-1 |
Mercury |
11.4 |
296 |
Nitrogen |
25.5 |
201 |
Oxygen |
13.9 |
213 |
Gold |
64.4 |
1580 |
Silver |
105 |
2336 |
Water |
333 |
2256 |
Copper |
207 |
4730 |
Example
Find how much energy a refrigerator has to remove from 1.5 kg of water at 200C,
to make ice at -120C. Assume no energy loss to the environment.
The specific heat of Water is 4.19 kJ.kg-1.K-1.
The specific heat of Ice is 2.1 kJ.kg-1.K-1.
The latent heat of fusion of Water is 333 kJ.kg-1.
The general principle is conservation of energy, that is,
Heat lost by water = Heat to cool water to 00C + heat to freeze water at 00C +
heat to cool ice to -120C
Example
A 0.50 kg chunk of ice at -100C is placed in 3.0 kg of water at 200C.
Find the temperature and phase of the final mixture. Assume no energy loss to the environment.
Heat to raise ice to 00C:
Heat to melt ice at 00C:
Heat to cool water to 00C:
The heat available from the warm water to cool down to 00C is more than sufficient
to raise the ice to 00C and melt it, (Q3> >Q1+Q2)
so the result will be water at a temperature above 00C.
Example
When 1.0 kg of solid mercury at its melting point of -390C is placed in a 0.05 kg of Aluminium container
filled with 1.2 kg of water (both at 200C), the final temperature of the mixture is found to be
16.20C.
The specific heat of Mercury is 140 J.kg-1.K-1.
The specific heat of Aluminium is 900 J.kg-1.K-1.
Assume no heat is lost to the surrounds.
Find the latent heat of fusion of Mercury.
Summarising:
In the temperature range between 200C and 1000C there is a reasonable
linear relationship between expansion and temperature.
Hydrogen bonding arises when an electron spends more time with the other atom thus creating a dipole (i.e. a partly exposed proton).
When heat flows into or out of a system the internal energy changes, Q = U2 - U1
The heat capacity of a system is the heat needed to change its temperature by one degree.
The specific heat is an intrinsic property of the material:
The molar specific heat is the specific heat of one mole of a substance
When substances are mixed, then the heat lost by the hot substance equals the heat gained by the cold substance.
The latent heat is the heat absorbed or released (at constant temperature) to change the phase of a substance.
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