The particles that make up an object can have ordered energy and disordered energy. The kinetic energy of an object due to its motion with velocity v with respect to an observer is an example of ordered energy. The kinetic energy of individual atoms, when they are randomly vibrating about their equilibrium position, is an example of disordered energy. Thermal energy is disordered energy. The temperature is a measure of this internal, disordered energy. The absolute temperature of any substance is proportional to the average kinetic energy associated with the random motion of the substance.
In a gas the individual atoms and molecules are moving randomly. The average kinetic energy of the molecules in a gas is proportional to the absolute temperature T of the gas. In a solid, the atoms can move randomly about their equilibrium positions. In addition, the solid as a whole can move with a given velocity and have ordered kinetic energy. Only the kinetic energy associated with the random motion of the atoms is proportional to the absolute temperature of the solid.
In ideal gases the disordered energy is all kinetic energy, in molecular gases and solids it is a combination of kinetic and potential energy. If we model the atoms in a solid as being held together by tiny springs, then the random internal energy of each atom constantly switches between kinetic energy and elastic potential energy.
In classical physics, zero absolute temperature means zero kinetic energy associated with random motion. The atoms in a substance do not move with respect to each other. (The uncertainty principle in quantum mechanics requires that there is some zero-point energy.) Room temperature is not close to absolute zero temperature. At room temperature the atoms and molecules of all substances have random motion.
In SI units the scale of absolute temperature is Kelvin (K). The Kelvin scale is identical to the Celsius (oC) scale, except it is shifted so that 0 degree Celsius equals 273.15 K. We have
temperature in K = temperature in oC + 273.15.
To convert to temperature in Fahrenheit we can use
temperature in oC = (temperature in oF - 32) ´ 5/9.
|
Two objects with different temperatures can exchange energy, if they are in thermal contact. The energy exchanged between object because they are in thermal contact is called heat. If two objects are in thermal contact and do not exchange heat, then they are in thermal equilibrium.
The zeroth law of thermodynamics states that two object, which are separately in thermal equilibrium with a third object, are in thermal equilibrium with each other. Two objects in thermal equilibrium with each other are at the same temperature.
|
How can we heat things up?
|
|||||
| Atoms in molecules and solids are held together by chemical bonds. Chemical bonds are electromagnetic in origin, but can be modeled well by tiny springs. Two atoms held together by a spring have an equilibrium position. If they are pushed closer together, they repel each other. If they are pulled farther apart, they attract each other. If they are displaced in any way from their equilibrium position and then released, they start vibrating about their equilibrium position. An atom can form different chemical bonds with a variety of other atoms. Different bonds are represented by springs with different spring constants. The stiffer the spring, the more work it takes to pull the atoms apart. If enough work is done, then the spring is stretched too much and it breaks, i.e. the chemical bond breaks. |  |
||||
| At room temperature, gas molecules have
random translational kinetic energy associated with the motion of
their center of mass and random vibrational energy and rotational
kinetic energy associated with the motion about their center of
mass. Collisions continuously transfer energy between the different
degrees of freedom and the average energy in each degree of freedom
is the same. If work is done on the molecules which increases their
vibrational energy, the amplitude of the vibrations increases, and
eventually the chemical bonds break. Most free atoms quickly form
new bonds. If the new bonds are stronger, i.e. if the new springs
are stiffer, then they do more work pulling the atom towards their
new equilibrium positions than was needed to break the old bonds and
the atoms will have more kinetic energy as they pass through the
equilibrium positions. This kinetic energy is quickly shared with
the other degrees of freedom, the energy of all degrees of freedom
increases, i.e. the thermal energy increases. Thermal energy is
released by a chemical reaction.
The temperature increases.
To burn fuel, work must first be done to break the chemical bonds in the fuel. This work provides the activation energy, the energy needed to start the chemical reaction. The free atoms and molecules then bond with oxygen. The new bonds with the oxygen atoms are much stronger than the broken bonds. As the atoms form new bonds, they gain thermal energy. When you strike a match, you first do work against friction to break the chemical bonds in some of the fuel on the head. The free atoms and molecules now combine with oxygen from the air, forming stronger bonds and thus releasing thermal energy. The random kinetic energy of these fast molecules is transferred in collisions to neighboring atoms and molecules, breaking their bonds, etc. Video: Activation Energy |
|||||
If a gas of mass m is confined to a volume V with pressure P and absolute temperature T, the equation of state relates the quantities V, P, and T. For gases at very low densities and pressures the equation of state is found experimentally to be quite simple.
 | Experiments revealed that for a
low-density gas at constant temperature, P is inversely proportional to V.
| ||
| For a low-density gas at constant volume the pressure is proportional to the
temperature.
| ||
| For a low-density gas at constant pressure the volume is proportional to the
temperature.
| ||
| These observation are summarized by the following equation of state:
|
| A tank having a volume of 0.1 m3 contains helium gas at 150 atm.
How many balloons can the tank blow up, if each filled balloon is a sphere
0.3 m in diameter at an absolute pressure of 1.2 atm?
|
| The mass of a hot air balloon and its cargo (not including the air inside)
is 200 kg. The air outside is at 10 oC and 101 kPa. The
volume of the balloon is 400 m3. To what temperature must
the air in the balloon be heated before the balloon will lift off.
(Air density at 10 oC is 1.25 kg/m3.)
| ||
| If a helium-filled balloon initially at room temperature is placed in a
freezer, will its volume increase, decrease, or remain the same?
|
The earth atmosphere is a gas in the presence of gravity. Gravity prevents the atmosphere from escaping. Near the Earth's surface the acceleration due to gravity is approximately constant (g = 9.8 m/s2). The atmosphere is a fluid, and the pressure in this fluid is a function of the altitude. For a liquid, i.e. an incompressible fluid, we found that P = P0 + rgy, where y is the distance below the reference point. But the atmosphere is a compressible gas. Let us consider a column of gas containing rparticle particles of mass m per unit volume near the Earth's surface.
The pressure times the area (i.e. the force) at height y must exceed that at height y + dy by the weight of the intervening gas. We need
Py+dyA - PyA = -mrparticleVg
= -mrparticlegAdy,
or
dP = -mrparticlegdy.
But from the ideal gas law we know that P = rparticlekT, with k being the Boltzmann constant. We may therefore write
dP = -P(mg/kT)dy, or
dP/P = -(mg/kT)dy.
If the temperature T is constant, then this integrates to
P = P0e-mgy/(kT).
The pressure decreases exponentially with altitude.
Near Earth's surface the temperature T is not constant, it generally decreases with altitude y. In the standard atmosphere T = T0 - ay, with a = 0.0065 K/m. We then have
dP/P = -mgdy/(k(T0 - ay)).
This integrates to
ln(P/P0) =( mg/ka)ln(1 - ay/T0), or
(P/P0)(ka/mg) = (1 - ay/T0).
Since the Earth's atmosphere consist mostly of nitrogen molecules, we have
(P/P0)0.19 = (1 - (0.0065 K/m)y/T0).
T0 and P0 are the temperature (in Kelvin) and the pressure at sea level. The pressure, density, and temperature of the atmosphere decrease as the altitude increases.
Links:
| Standard Atmosphere Calculator |
| The Ideal Atmosphere |
How can we make a balloon rise?
We have to make its weight smaller than the weight of an equal shaped volume of surrounding air. The skin of the balloon and the basket certainly are heavier than an equal volume of the surrounding air. The inside of the skin therefore has to be filled with something appreciably lighter than the surrounding air. Ideally we would empty it out completely. But then the pressure inside would be zero. The air pressure from the outside would collapse the skin. To keep the skin stretched, the inside and outside pressures on each section of skin must be equal, but the weight of the inside material must be less than that of an equal volume of surrounding air. The pressure is proportional to the particle density and the temperature, while the weight is proportional to the particle density and the weight of each particle.
P = krparticleT,
weight = rparticle ´ volume ´ weight per particle.
To get the weight down and keep the pressure the same we have two options.
| Option 1: We can decrease the weight per particle. Air is composed of 77% nitrogen, 21% oxygen and 2% other gases by weight. The density of air at room temperature and atmospheric pressure is approximately 1.3 kg/m3. If we replace the air with Helium gas at the same temperature, the same particle density will yield the same pressure. But the mass density of Helium gas at room temperature and atmospheric pressure is only approximately 0.18 kg/m3. If we make the balloon big enough, the total volume of air displaced by the balloon will weigh more than the balloon itself, and the net force on the balloon will be upward. Since the density of air decreases with altitude, the weight of the volume of air displaced by the balloon will decrease as the Helium balloon rises. Eventually it will equal the weight of the balloon and the net force on the balloon will be zero. As the balloon is rising it is also acted on by the drag force. The direction of the drag force is opposite the direction of the balloon's velocity. Because the drag force the the acceleration of the balloon will quickly decrease to close to zero and the balloon will rise with nearly constant velocity. You can explore this behavior in an exercise associated with this module. |
| Option 2: We can increase the temperature of the air in the balloon. Then we can obtain the same pressure with a smaller particle density. Again the weight of the balloon decreases and the net force on the balloon is upward. The hot air balloon will rise. |
Links
| How a helium balloon works |
| Experiments with a helium balloon |
Exercise (You can earn up to 5 points extra credit by completing this exercise.)
How do we measure temperature?
| Assume we have a gas in a container with a movable
piston under atmospheric pressure. As we increase the temperature
of the gas it expands, its volume increases. The piston moves out.
As we cool the gas the piston moves in. The ideal gas law gives us
the volume of the gas as a function of temperature.
PV = NkT, V = (Nk/P)T. The volume is proportional to the absolute temperature of the gas. Assume the temperature change from 0 oC to 10 oC. We have (DV/V) = (DT/T) = 10/273 = 0.037 = 3.7%. By monitoring the volume of the gas we can monitor its temperature. We have a gas thermometer. |
 |
||||||||||||
| Solids and liquids also expand as the temperature
increases. For most solids we can define an
average coefficient of linear expansion,
a. The change in length Δl of a
solid is proportional to its length l and the change in
temperature ΔT. The proportional constant is a.
Δl = alΔT. The average volume expansion coefficient
b is defined through
For some applications it is important to choose materials that have a small coefficient of linear expansion. (Temperature variations can destroy a precision alignment.) For other applications it is important that all materials have similar coefficients of linear expansion. (When different parts of a structure expand by different amounts, the structure buckles.) |
Linear expansion coefficients a: (per oC)
|
||||||||||||
| If we want to build a thermometer, we want to use a
material with a large coefficient of volume expansion. For most
liquids, the coefficient of volume expansion is on the order of 10-4.
For each degree Celsius temperature change, the volume changes by (ΔV/V)
= b = 10-4 = 0.0001 = 0.01%.
If we tried to observe this change by monitoring the height of water
in a glass, we would probably be unsuccessful.
Assume you have 1 liter = 1000 cm3 of liquid with b = 10-4 in a container with bottom area A = 100 cm2. The height of the liquid is 1 cm. (Volume = area * height.) You increase the temperature of the liquid by 20 oC. ΔV = bVΔT = 10-4 * 1000 cm3 ´ 20 = 2 cm3. The change in height is Δh = ΔV/A = 2 cm3/(100 cm2) = 0.02 cm = 0.2 mm. If, however, you put a tight lid on the container at a height of 10 cm with a hole connected to a fine capillary, then the liquid will rise in the capillary to a much greater height. Assume you connected a tube with a cross sectional area of 1 cm2. Then Δh = ΔV/A = 2 cm3/(1 cm2) = 2 cm. You have constructed a usable thermometer. |
 |
||||||||||||
| Thermometers can also be constructed
from bimetallic strips.
Two strips of metal, one of steel and one of aluminum, that have the same length at 0
oC,
will have different length a 200 oC. Assume the strips are 10
cm long at 0 oC.
The change in length of the steel strip at 200 oC is Dl
= alDT = 11´10-6
10 cm 200 = 0.022 cm = 0.22 mm
and the change in length of the aluminum strip at 200 oC is Dl
= alDT = 23´10-6
10 cm 200 = 0.046 cm = 0.46 mm.
If the two strips are bonded together, the bimetallic strip will buckle or curl.
Long
bimetallic coils will uncoil slightly as the temperature rises if the inner strip has the
larger a. If one end is fixed and a pointer is attached to the
other end the pointer will move. The position of the pointer indicates the temperature. Some liquid crystals change color when their temperature changes. They selectively only reflect one color of the white light falling onto them. The color that is reflected back depends on the temperature and can be used as a temperature indicator. Link:
|
|||||||||||||
| In a constant volume gas thermometer the pressure at 20 oC is
0.98 atm. (a) What is the pressure at 45o C? (b) What is the temperature if the pressure is 0.5 atm?
| ||||||
| Liquid nitrogen has a boiling point of -195.81 oC at atmospheric
pressure. Express this temperature in (a) degrees Fahrenheit and (b) Kelvin.
| ||||||
| A copper telephone wire has essentially no sag between poles 35 m apart on
a winter day when the temperature is -20 oC. How much longer
is the wire on a summer day, when the temperature is 35 oC?
|
| A square hole 8 cm along each side is cut in a sheet of copper.
Calculate the change in the area of the hole if the temperature of the sheet
is increased by 50 K.
|
Link to other Web material:
| How Thermometers Work |
| Temperature and Thermal Expansion |
| Standard Atmospheric Computations |
Please complete assignment 3 now. For the User ID use your registered password.
You can submit the assignment up to three times. Each time the computer will tell you your score.
