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PHYS 242 (Physics II)

Physics. A science that deals with matter and motion and includes mechanics, heat, light, electricity and sound.

Six Major Branches of Physics

1. Mechanics. A branch of physical science that deals with energy and forces and their effect on bodies. This branch of physics deals with motion of bodies, the concept of force, the effect of forces on motion and the form or shape of bodies, energy, momentum, work, and power. Properties of solids, liquids, and gases are likewise studied in this branch.

2. Sound. Its study leads to the consideration of waves and wave motion. This also deals with the different sources of sound, its transmission through various media, acoustics, and hearing.

3. Heat. This branch of physics deals with the temperature scales and measurement, the concept of heat, thermal expansion, heat capacities of substances, changes of state, heat transfer, and thermodynamics which is concerned mainly with the relationship between work and heat.

4. Electricity. This branch focuses on the concepts of electrical changes, the flow of electrical charges - better known as current, the various effects of current, electrical instruments, electrical and magnetic properties of matter, and electronics.

5. Optics. This branch is concerned with the fundamental concepts of electromagnetic waves, absorption and transmission of light, reflection, refraction, optical instruments, and other various phenomena, such as, interference, diffraction and polarization.

6. Atomic & Nuclear Physics. This branch deals with the study of radiation, photoelectric effect, x-rays, structure of the atom, radioactivity, nuclear disintegration, and other properties of nuclei.

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H E A T(Thermal Energy, Temperature, Calorimetry)

This topic explores energy transformations involving the atoms and molecules of which matter is made. This atomic and molecular energy of a system comprises of (1) thermal energy - the random kinetic energy of the atoms and molecules, and (2) the potential energy of these tiny particles resulting from their bonds and interactions with each other. (Ref. #1)

Thermal EnergyThe Thermal Energy in a substance is the kinetic energy associated with the

random motion - translation, collision, or vibration - of all its atoms and molecules. (Ref. #1). It represents the total internal energy of an object: the sum of its molecular kinetic and potential energies.

TemperatureTemperature Defined/Described.

A measure of the average translational kinetic energy of the atoms or molecules in a substance.An indicator of the average thermal energy of the atoms or molecules in a substance.The degree of the hotness or coldness of something (as air, water, or the body) as shown by a thermometer.It is independent on size and measured using a variety of substances that change in a predictable way as their temperatures change.A property which determines the direction of heat flow.

* When temperature increases, so does the average thermal energy of its atoms or molecules. Thermal energy and temperature, then, are closely related quantities - but there is an important distinction. Temperature relies on the average thermal energy per molecule or atom, whereas thermal energy is the total random kinetic energy of all the atoms or molecules in a substance. Take for instance, the river and a cup of water at the same temperature, then the average thermal energy of one water molecule in the river is the same as that of in the cup. Nevertheless, the river contains much more thermal energy because there are so many water molecules in the river than in the cup.

Thermometric Property. A physical property of matter that varies in a predictable way with temperature - metal increases its length as it warms. Other thermometric properties are the increase in the pressure of an enclosed gas as it warms, the increase in the volume of space occupied by a liquid such as mercury as it warms, the change in color of light emitted by a flame as the flame becomes hotter, and the increase in electrical resistance of a metal wire as its temperature increases. Every thermometer uses one of these or some other thermometric property as the basis for measuring temperature.

Thermometric Substances. Substances whose physical properties respond linearly to a change in temperature and hence, are used in setting up a temperature scale.

Thermometer. An instrument which gives a numerical reading of the hotness or coldness of a body. It is temperature-measuring device which includes a thermometric substance.

Galileo. The first man to develop an accurate thermometer. He used a gas for his thermometric substance which was placed in a glass bulb connected to a long, narrow glass tubing. It was inverted in a dish of colored liquid which partly filled the glass tube at ordinary temperature. A temperature rise inside the

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bulb is indicated by a decrease in the height of the column of colored liquid which is due to the expansion of the gas in the bulb. Nonetheless, the said type of thermometer depends to some extent upon the pressure of the air outside.

® Types of Thermometer 1. Constant-volume Gas Thermometer (Galileo's Thermometer). A type which

measures the variation in the pressure of the gas kept at constant volume. This type is more accurate than the liquid-in-glass thermometer.

2. Liquid-in-glass Thermometer. A type of thermometer which makes use of the expansion of a certain liquid confined inside the tube of small bore. The most common liquid used is mercury. A mercury thermometer consists of a long glass tube with a small bore. One end is closed and the other is sealed to a small bulb. The bulb and a portion of the capillary tube are filled with mercury and the rest of the bore above the mercury are evacuated. The walls of the bulb are made very thin for rapid conduction of heat. The stem of the thermometer has a magnifying action on the capillary tube. The advantage of using mercury instead of other liquids is that mercury does not evaporate so easily, and its expansion is more or less uniform with the ordinary range of temperatures.

3. Bimetallic Strip Thermometer (Thermocouple). A type which is composed of a pair of dissimilar conductors with the two junctions kept at different temperatures.

4. Resistance Thermometer. A type which is based on the change in resistance of conductors or semi-conductors with temperature change.

® Common Temperature Scales1. Celsius (Centigrade). Named after the Swedish astronomer, Anders Celsius

(1701-1744), and is used for scientific work in almost all countries. At standard atmospheric pressure (normal atmospheric conditions), water’s ice and steam points are 0ºC and 100ºC, respectively.

2. Fahrenheit. Named after the German physicist, Gabriel Fahrenheit (1686-1736) and is used in civilian life in the U.S. At standard atmospheric pressure, water’s ice and steam points are 32ºF and 212ºF, respectively.

3. Rankine. Named after the British physicist and engineer, William J. M. Rankine (1820-1872). An absolute temperature scale in which each degree equals one degree on the Fahrenheit scale. Absolute zero is at 0ºR (-459.67ºF), and at standard atmospheric pressure, water’s ice and steam points are 491.67ºR and 671.67ºR, respectively. Commonly used in mechanical engineering.

4. Kelvin (Absolute). Invented by the British mathematician and physicist, William Thomson Kelvin (1824-1907). Absolute zero is at 0K (-273.15ºC), and at standard atmospheric pressure, water’s ice and steam points are 273.15K and 373.15K. Used in scientific work and thermodynamics.

5. International (thermodynamic temperature) Scale is based on the property of electrical resistivity, with platinum wire as the standard for temperature between –190 and 660C. Above 660C, to the melting point of gold, 1063C, a standard thermocouple, which is a device that measures temperature by the amount of voltage produced between two wires of different metals, is used; beyond this point temperatures are measured by the so-called optical pyrometer, which uses the intensity of light of a wavelength emitted by a hot body for the purpose.

Fixed Points (Two easily reproducible temperatures in a thermometer)

1. Ice (Freezing) Point. The lower fixed point which is the temperature of a mixture of ice and pure water in equilibrium at standard conditions.

2. Steam (Boiling) Point. The upper fixed point which is the temperature of steam and boiling water in equilibrium at standard conditions.

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Thermal Expansion

All solids (or almost substances) expand with a rise in temperature. The expansion might be very small (perhaps not noticeable with our naked eyes) as in the case of ordinary metals, but these small changes in dimensions usually result in the failure of many engineering structures - that is, if no allowance is made. Due to this nature of solids expansion or contraction joints are provided as in the case of concrete roads and steel rails. The said joints are narrow gaps about an inch wide every 20 to 30 meters in the case of concrete roads, and a few centimeters in the case of steel rails. (Ref. #2)

Architects and engineers must consider thermal expansion and contraction of metal beams when designing skyscrapers and tall bridges. Expansion joints are usually placed in sidewalks, bridges, and highways to prevent their buckling during summer heat. (Ref. #1)

Linear Expansion. The change in length of an object that is warmed or cooled and depends on its (1) change in temperature, (2) original length, and (3) the type of material of which it is made. (Ref. #1)

Consider a rod which has a length Lo at a certain temperature To (refer to the figure that follows). When it is heated to a temperature Tf, which is above its initial temperature (To), its new length Lt is greater than its initial length Lo by an amount ΔL which is proportional to the three factors - the original length Lo, the change in temperature ΔT, and the coefficient of linear expansion α That is, ΔL = Lo α ΔT.

Figure:Lo

@ To

Lt @ Tf

Based on the preceding figure, change in length is,

ΔL=Lt−Lo

The new length then is,Lt=Lo+ΔL

If we substitute equation 1 in equation 2, the new length could likewise be computed as,

Lt=Lo+Lo α ΔT Where: ΔT =T f−T o

Factoring the right side of the equation, we formed

Lt=Lo[1+α (T f−T o)]

To solve for the coefficient of linear expansion,

α= ΔLLo ΔT

∨α=Lt−Lo

Lo ΔT

The quantity α called the coefficient of linear expansion of the material may hence, be defined as the change in length per unit length of the rod per degree increase in temperature. It is not constant for a particular material but changes slightly with temperature. Nevertheless, because the change is usually small for a certain temperature, we may ignore the ignoble change for a material and use the values in the following table as standard for our discussion.

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Coefficients of Linear Expansion of Common Solids @ 20°CS o l i d α(x10-6 C-1) α (x10-6 F-1)Aluminum (*24) 25 13.888…Brass or Bronze* 19 10.558…Brick and Concrete 7-12 3.888… - 6.666…Copper(*16) 17 9.444…Glass (ordinary) 9 5.000Glass (Pyrex) 3 1.666…Hard Rubber 80 44.444…Ice 51 28.333…Invar Steel* 0.9 0.500Iron (Cast)* 11 6.111…Iron and Steel 12 6.666…Lead 29 16.111…Marble 1.4 - 3.5 0.777… - 1.9444…Quartz 0.4 0.222…Silver 20 11.111…Tungsten* 4.33 2.40555…Zinc 26 14.444…* @ near room temperature

SubstanceLinear Temperature Expansion Coefficient (α)

(10-6 m/m K) (10-6 in/in Fº)ABS (Acrylonitrile Butadiene Styrene) thermoplastic 73.8 41ABS -glass fiber-reinforced 30.4 17Acetal 106.5 59.2Acetal - glass fiber-reinforced 39.4 22Acrylic, sheet, cast 81 45Acrylic, extruded 234 130Alumina 5.4 3.0Aluminum 22.2 12.3Antimony 10.4 5.8Arsenic 4.7 2.6Barium 20.6 11.4Benzocyclobutene 42Beryllium 11.5 6.4Bismuth 13 7.3Brass 18.7 10.4Brick masonry 5.5 3.1Bronze 18.0 10.0Cadmium 30 16.8Calcium 22.3 12.4Cast Iron Gray 10.8 6.0Cellulose acetate (CA) 130 72.2Cellulose acetate butynate (CAB) 80 - 95Cellulose nitrate (CN) 100 55.6Cement 10.0 6.0Cerium 5.2 2.9Chlorinated polyvinylchloride (CPVC) 66.6 37Chromium 6.2 3.4Clay tile structure 5.9 3.3Cobalt 12 6.7

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SubstanceLinear Temperature Expansion Coefficient (α)

(10-6 m/m K) (10-6 in/in Fº)Concrete 14.5 8.0Concrete structure 9.8 5.5Constantan 18.8 10.4Copper 16.6 9.3Copper, Beryllium 25 17.8 9.9Corundum, sintered 6.5 3.6Cupronickel 30% 16.2 9Diamond (Carbon) 1.18 0.66Duralumin 23Dysprosium 9.9 5.5Ebonite 76.6 42.8Epoxy, castings resins & compounds, unfilled 55 31Erbium 12.2 6.8Ethylene ethyl acrylate (EEA) 205 113.9Ethylene vinyl acetate (EVA) 180 100Europium 35 19.4Fluoroethylene propylene (FEP) 135 75Gadolinium 9 5Germanium 6.1 3.4German silver 18.4Glass, hard 5.9 3.3Glass, Pyrex 4.0 2.2Glass, plate 9.0 5.0Gold 14.2 8.2Granite 7.9 4.4Graphite, pure 7.9 4.4Gunmetal 18Hafnium 5.9 3.3Hard alloy K20 6 3.3Hastelloy C 11.3 6.3Holmium 11.2 6.2Ice 51 28.3Inconel 12.6 7.0Indium 33 18.3Invar 1.5 0.8Iridium 6.4 3.6Iron, pure 12.0 6.7Iron, cast 10.4 5.9Iron, forged 11.3 6.3Kapton 20Lanthanum 12.1 6.7Lead 28.0 15.1Limestone 8 4.4Lithium 46 25.6Lutetium 9.9 5.5Macor 9.3Magnesium 25 14Manganese 22 12.3Marble 5.5 - 14.1 3.1 - 7.9Masonry 4.7 - 9.0 2.6 - 5.0Mercury 61

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SubstanceLinear Temperature Expansion Coefficient (α)

(10-6 m/m K) (10-6 in/in Fº)Mica 3 1.7Molybdenum 5 2.8Monel 13.5 7.5Mortar 7.3 - 13.5 4.1-7.5Neodymium 9.6 5.3Nickel 13.0 7.2Niobium (Columbium) 7 3.9Nylon, general purpose 72 40Nylon, Type 11, molding and extruding compound 100 55.6Nylon, Type 12, molding and extruding compound 80.5 44.7Nylon, Type 6, cast 85 47.2Nylon, Type 6/6, molding compound 80 44.4Oak 54Osmium 5 2.8Palladium 11.8 6.6Phenolic resin without fillers 80 44.4Phosphor bronze 16.7Plaster 16.4 9.2Platinum 9.0 5.0Plutonium 54 30.2Polyallomer 91.5 50.8Polyamide (PA) 110 61.1Polybutylene (PB) 72Polycarbonate (PC) 70.2 39Polycarbonate - glass fiber-reinforced 21.5 12Polyester 123.5 69Polyester - glass fiber-reinforced 25 14Polyethylene (PE) 200 111Polyethylene (PE) - High Molecular Weight 60Polyethylene terephthalate (PET) 59.4 33Polyphenylene - glass fiber-reinforced 35.8 20Polypropylene (PP), unfilled 100 - 200 56 - 112Polypropylene - glass fiber-reinforced 32 18Polystyrene (PS) 70 38.9Polysulfone (PSO) 55.8 31Polyurethane (PUR), rigid 57.6 32Polyvinyl chloride (PVC) 50.4 28Polyvinylidene fluoride (PVDF) 127.8 71Porcelain, Industrial 6.5 3.6Potassium 83 46.1Praseodymium 6.7 3.7Promethium 11 6.1Quartz 0.77 - 1.4 0.43 - 0.79Rhenium 6.7 3.7Rhodium 8 4.5Rubber, hard 77 42.8Ruthenium 9.1 5.1Samarium 12.7 7.1Sandstone 11.6 6.5Sapphire 5.3Scandium 10.2 5.7

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SubstanceLinear Temperature Expansion Coefficient (α)

(10-6 m/m K) (10-6 in/in Fº)Selenium 3.8 2.1Silicon 3 1.7Silicon Carbide 2.77Silver 19.5 10.7Sitall 0.15Slate 10.4 5.8Sodium 70 39.1Solder 50 - 50 24.0 13.4Steatitec 8.5 4.7Steel 13.0 7.3Steel Stainless Austenitic (304) 17.3 9.6Steel Stainless Austenitic (310) 14.4 8.0Steel Stainless Austenitic (316) 16.0 8.9Steel Stainless Ferritic (410) 9.9 5.5Strontium 22.5 12.5Tantalum 6.5 3.6Tellurium 36.9 20.5Terbium 10.3 5.7Terne 11.6 6.5Thallium 29.9 16.6Thorium 12 6.7Thulium 13.3 7.4Tin 23.4 13.0Titanium 8.6 4.8Tungsten 4.3 2.4Uranium 13.9 7.7Vanadium 8 4.5Vinyl Ester 16 - 22 8.7 - 12Wood, fir 3.7 2.1Wood, oak parallel to grain 4.9 2.7Wood, oak across to grain 5.4 3.0Wood, pine 5 2.8Ytterbium 26.3 14.6Yttrium 10.6 5.9Zinc 29.7 16.5Zirconium 5.7 3.2

Source: The Engineering Toolbox (www.EngineeringToolBox.comSidelights:

Since the Celsius degree is 9/5 (or 1.8) times as large as the Fahrenheit degree, the coefficient of linear expansion of a given material per C is 9/5 times as large as its coefficient of linear expansion in per F. Or, the α in per F is obtained by multiplying the α in per C by 5/9.Because of its very small coefficient of expansion, Pyrex glass (the new and improved) can be removed from one extreme temperature and dump to another (perhaps from hot to cool or contrariwise) without any danger of cracking. With the same reason, invar steel is used for standard tapes for accurate measurement of distances at ordinary temperatures.Most values at 25ºC (77ºF).

Problems: (to be solved on the board)1. A steel beam in a bridge extends 27 meters across a small stream. What is its expanded length

from the winter, when its temperature is 20C, to the summer, when it is 100F?

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2. The aluminum lid of a jar of dill pickles is stuck to the jar. To loosen the lid so it can be opened, hot water is poured over the lid, causing it to expand. If the temperature increase of the lid and glass is 105F, calculate the change in circumference of the lid and of the glass on which it is screwed. The radius of the lid before heating is 4.5 inches.

The different rates at which different metals expand are taken advantage of in bimetallic elements used as thermostat or automatic regulator of temperatures. One form of which is an iron-brass bimetal used in switching devices.

Area and Volume Expansion

Because a material usually expands uniformly in all directions as it becomes warmer, its area and volume increase. As in linear expansion, the change in area and/or volume depends on its change in temperature, its original area and/or volume, and the material of which it is made.

If a thin rectangular sheet of metal is heated, both its length and width will increase in a manner as if they were rods of the same material as the sheet.

Illustration 1: Expansion of a Square Sheet (area)

With Lt=Lo(1+α ΔT ) And since the metal is a square sheet, the lengthEquals width. Thus, the above equation can besquared to determine the new area of the sheet.

¿)]²Lt ²=Lo ²(1+α ΔT )²Lt ²=Lo ²[12+2 (1 ) (α ∆ T )+(α ΔT )2]Lt ²=Lo ²(1+2 α ∆T +α 2 ∆ T2)

Since α is very small, its square becomes much smaller than itself; hence, the last term in the square brackets of the preceding equation may be neglected. The equation may then be simplified as,

Lt ²=Lo ²(1+2α ∆T )

Let Lt ² = At and Lo² = Ao

At=Ao(1+ β ∆ T )

where: Ao = initial or original area at ToAt = new area at a certain temperature Tf

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ß = coefficient of area expansion = 2 α

* Neglecting the last term in the brackets of the previous means neglecting the area of the small square at the lower right corner of the bigger as highlighted in the preceding figure.

Illustration 2: Expansion of a Solid Cube (volume)

With Lt=Lo(1+α ΔT )

And since the metal is a cube, the length equals width equals the thickness. Thus, the above equation can be cubed to determine the new volume of the cube.

[Lt=Lo(1+α ΔT )¿³Lt ³=Lo ³ (1+α ΔT ) ³Lt ³=Lo ³ ¿]Lt ³=Lo ³ ¿)

Since α is very small, its square and cube become much smaller and very much smaller than itself; hence, the third and last terms in the square brackets of the preceding equation may be neglected. The equation may then be simplified as,

Lt ³=Lo ³ (1+3 α ΔT )

Let Lt ² = Vt and Lo² = Vo

Again, employing the reasoning use with the area expansion, the last two terms inside the parentheses - with the square of alpha and cube of alpha, may be neglected without any appreciable error. Therefore, the preceding equation may be modified as,

V t=V o(1+γΔT )

where: γ(gamma) = coefficient of volume expansion = 3 α

The coefficient of volume expansion is three (3) times the coefficient of linear expansion. If all the values in the table of Coefficients of Linear Expansion of Common Solids are multiplied by 3, we will obtain a table of Coefficients of Volume Expansion for the same set of materials.

Coefficients of Volume Expansion (γ ) of Common Fluids (@ 20 or 68F) Substance in per C in per F

Air (& most other gases @ 1 atm) 340 x 10-5 188.888… x 10-5

Ethyl Alcohol (Ethanol C2 H 5 OH) 110 x 10-5 61.111 x 10-5

Methyl Alcohol (Methanol CH 3OH ) 119 x 10-5 66.111 x 10-5

Benzene (C6 H 6) 124 x 10-5 68.889 x 10-5

Gasoline 95 x 10-5 52.778 x 10-5

Glycerin (GlycerolC3 H 8O3) 51 x 10-5 28.333 x 10-5

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Mercury (Hg) 18 x 10-5 10.000 x 10-5

Paraffin Oil 90 x 10-5 50.000 x 10-5

Turpentine 97 x 10-5 53.889 x 10-5

Water, 15C to 100C 21 x 10-5 11.667 x 10-5

Extra Notes/Comments: partially from EncartaBenzene (formerly benzol) is a colorless volatile (evaporating @ a relatively low temperature) toxic liquid with a distinctive odor from petroleum and is used to manufacture dyes, polymers, and industrial chemicals. Turpentine is a colorless, flammable, strong-smelling essential oil used in the manufacture paint solvent and some medicine. It could also mean viscous substance from coniferous trees (e. g., pine trees), or a brownish yellow sticky mixture of essential oil and resin that comes from the terebinth tree. Glycerin is a thick, sweet, odorless, colorless, or pale yellow liquid from fats and oils as a byproduct (incidental product) of soap manufacture, and is used as solvent, antifreeze, plasticizer, manufacture of soaps, cosmetics, lubricants, and dynamite.Ethanol is a colorless liquid with a pleasant smell from fermentation by yeasts and other microorganisms and is used in alcoholic beverages, as solvent, and in the manufacture of other chemicals.Methanol is a colorless volatile poisonous water-soluble liquid, and used as solvent, fuel, and in antifreeze for motor vehicles.Paraffin Oil (or simply paraffin) is a mixture of liquid hydrocarbons obtained from petroleum and used as a domestic heating fuel and as fuel for aircraft. Volatile could also mean: Unstable and potentially dangerous (apt to become suddenly violent and dangerous); upredictable and fickle (changeable in mood, temper, or desire); changing suddenly (characterized by or prone to sudden change); short-lived (continuing for only a short time); or losing data when power is off (describes a computer memory that does not store when the power is off).The coefficient of volume expansion of water is the average value for a certain range of temperature. The expansion of water is sometimes called ANOMALOUS because water contracts when heated from 0C to 4C and then it expands at a non-uniform rate when heated further from 4C to 100C. Thus, the given formula is applicable only when the temperature of water is between 15c and 100C and the result is only approximate.The coefficient of area expansion of liquids are of no use.

The coefficient of volume expansion of liquids is sensitive to pressure on their surface. Generally, except for water, the coefficient of volume of expansion of a given liquid decreases with a rise in pressure on its surface.

Differential Expansion When the container is filled with liquid at a certain temperature and both the

container and its contained liquid are heated to the temperature, the liquid will generally spill. This is due to the fact that liquids have a higher coefficient of volume expansion than solids. The amount of spillage will depend on the differential expansion of the container and its content. (Ref.#2)Problems: (to be solved on the board)

1. During a summer night when the temperature is 68F, your house contains 453 m3 of air. What volume of air leaves the house through an open window if the air warms to 39C on a very hot summer day? Assume that the dimensions of the house experience negligible change and that other conditions are constant. (Ref.#1)

The change in volume of air as it warms and cools has significant effects on our atmosphere. As a gas such as air expands, it becomes less dense; its molecules on the average are spaced farther apart than in a cool gas. For this reason a hot (less dense) will rise and float on a cool (more dense) gas. This property of gases is very important in cleansing the atmosphere surrounding large cities. Air warmed at the earth's surface rises and carries with it air contaminants. Cool, clean air

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sinks down from above to replace the warm air.Occasionally, an inversion occurs. For some reason, the air at the earth's surface remains

cool and is covered by warmer air above. The temperature is said to be inverted. The cool, dense air at the surface does not rise, and pollution accumulates until the inversion breaks.

Another interesting environmental consequence of thermal expansion and contraction is related to the freezing of water at the surface of a lake or pond. Why doesn't the 0C water that is ready to freeze sink to the bottom, allowing slightly warmer water to rise to the surface? This would happen with most substances that expand as they warm and contract as they cool. Water is a notable and important exception. When water cools from 4C to 0C, it expands and becomes less dense than the slightly warmer water; the 0C water has less weight per unit volume than the 4C water. Thus, in a lake or pond, the less dense 0C water floats at the top while the more dense, slightly warmer water sinks to the bottom (refer to the figure that follows). Consequently, a lake freezes the top down rather than from the bottom up. During the winter, fish remains near the bottom of the lake in the slightly warmer water.

2. A 1.5-liter Pyrex glass container is filled with turpentine at room temperature of 24C. Determine the amount of turpentine which is expected to spill when the temperature of the said glass container and its turpentine content is raised by 100F.

LiquidVolumetric (Cubical) Coefficient of Expansion

(1/K, 1/C°) (1/F°)Acetic acid 0.00110 0.00061Acetone 0.00143 0.00079Alcohol, ethyl (ethanol) 0.00109 0.00061Alcohol, methyl (methanol) 0.00118 0.00066Ammonia 0.00245 0.00136Aniline 0.00085 0.00047Benzene 0.00125 0.00069Bromine 0.00110 0.00061Carbon disulfide 0.00119 0.00066Carbon tetrachloride 0.00122 0.00068Chloroform 0.00127 0.00071Ether 0.00160 0.00089Ethyl acetate 0.00138 0.00077Ethylene glycol 0.00057 0.00032Freon refrigerant R-12 0.0026 0.00144n-Heptane 0.00124 0.00069Isobutyl alcohol 0.00094 0.00052Gasoline 0.00100 0.00056Glycerine (glycerol) 0.00050 0.00028Kerosene 0.00100 0.00056Mercury 0.00018 0.00010Methyl alcohol 0.00119 0.00066Methyl iodide 0.0012 0.00067n-Octane 0.00114 0.00063Oil (unused engine oil) 0.00070 0.00039Olive oil 0.00070Paraffin oil 0.000764 0.00042Petroleum 0.0010 0.00056n-Pentane 0.00158 0.00088Phenol 0.0009 0.00050Sulphuric acid, concentrated 0.00055 0.00031Toluene 0.00108 0.00060Trichloroethylene 0.001170 0.00065Turpentine 0.001000 0.00056Water 0.000214 0.00012

Source: The Engineering Toolbox (www.EngineeringToolBox.com

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Expansion of Gases

A gas is a form of matter in which the atoms and molecules are relatively far apart. The atoms and molecules are in constant motion and collide frequently and violently with each other. Collisions also occur when gas molecules hit solid and liquid surfaces in their environment. The collisions cause a force and pressure to be exerted on these surfaces.

Gases are described in terms of quantities such as pressure, temperature, the volume of space occupied, and the amount of gas in that space. The said quantities are called state variables because they describe the state or status of a gas - how hot it is, the pressure it exerts on the surface, and so forth. State variables are related to each other by equations called equations of state. Scientists use these equations of state in conjunction with the laws of thermodynamics to study phenomena such as climate changes in the atmosphere, the operation of gas-driven heat engines, and the change in pressure in a person's skull due to clogged sinuses.

Pressure. Force F exerted perpendicular to a surface divided by the area A over which the force is exerted: p= F

A

When molecules of air or of any gas collide with a solid or liquid, they exert force against that surface. The force is of an impulsive nature, much like the force of a tennis ball hitting a practice board. Our skin cannot sense the force of individual collisions, but we can sense the average effect of a large number of the collisions. Each second about 1023

molecules of air strike each square centimeter of our skin. The net effect of these collisions is a force that seems constant in nature.

At sea level, collisions of air molecules with a solid surface can cause an average force of 1.01325 x 105 N to be exerted on each square meter of surface. Thus, the atmospheric pressure at sea level is 1.01325 x 105 N/m2. In the English system of units, the pressure of air at sea level is 14.69594962 lbs/in2 (psi - pounds per square inch). However, the official SI unit of pressure is Pascal abbreviated as Pa and is equivalent to N/m2.

Other hybrid units of pressure based on particular measuring techniques are in common use today. The pressure units millimeter of mercury (mm Hg), also called torr, an inch of mercury (in Hg) evolved because of the use of manometers to measure pressure. A manometer consists of a U-shaped tube containing mercury or water (refer to the figures that follow). If the tube is open on both ends, air pushes down on each column with equal pressure, and the columns on each side of the tube remain at the same elevation (figure a). Nonetheless, if one end of the tube is closed and the air removed from it, creating a vacuum, the air pressure forces the mercury down on the open side and up into the evacuated side (figure b). The height h of the mercury column in the evacuated side above that in the side open to the air is a measure of the pressure of the air. At sea level the mercury column will be about 760 mm or 29.9 in higher on the left side than on the right side of the tube. Thus, we say that the pressure of the earth's atmosphere at sea level is 760 millimeters of mercury (760 mm Hg) or 29.92126 inches of mercury (29.92126 in Hg). Other pressure units occasionally used include the bar (a unit slightly less than one atmosphere).

Vacuum

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When a person blows into the right side right side of an open-tube manometer, the extra pressure on this side forces water down and up into the left side. The absolute pressure,

p=patm+ pgauge

pgauge=ρgh

When a person sucks on the right side of a manometer, the reduced pressure causes the water to rise on this side. The absolute pressure, then is,

p=patm+ pgauge

Equation of State. An equation that interrelates the properties of a substance (volume, pressure, and thermodynamics temperature) of mass m. Ideal (perfect) Gas. A hypothetical (low-density) gas that obeys the gas laws exactly.

LAWS INVOLVING EXPANSION OF GASES

1. Boyle's Law.Robert Boyle(1627-1691), a contemporary of Newton, developed the equation of state that relates the pressure to its volume. He used a manometer closed to one side (refer to figures E and F). A fixed amount of air was trapped in the closed side by a column of mercury, and the pressure of the air could be increased by adding mercury to the open column. He noticed that if the temperature and amount of air in the tube remained constant, the volume of space occupied by the air decreased in inverse proportion to the pressure acting on the air - that is,

V α 1p

Where: V = volume of the gasα = proportionality constant or symbolp = pressure acting on the gas (absolute)

If the pressure is doubled, the volume decreased to one-half its original volume. Boyle described this relation between pressure and volume with the following equation, known as Boyle's Law:

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15

pV=constant (for a fixed amount of gas at constant temperature)

The equation implies that for the product of p and V to be constant, one of the quantities must decrease if the other increases, and vice versa.

Boyle's Law is often used by comparing the product of the pressure and volume in a gas before some process occurs and after it is over. If the amount of gas and the temperature of the gas remain constant, then the product of the final values of pressure and volume must equal the product of the initial values, since both products equal the same constant: (Ref #.1)

Suppose we have a certain amount of gas confined at initial volume Vi, initial pressure Pi, and initial temperature Ti. The gas is then changed to the intermediate conditions of p and V, the temperature remaining constant at Ti.

@ initial conditions: (refer to Figure G)V i=

k i

pi∨k i=p iV i a

@ intermediate conditions:

V= kp∨k=pV b

for gases, ki = k = constant

k = Boltzmann constant = 1.3806504 x10-23 J/K

substituting equations a and b in equation c

pi V i=pV=constant 1

Figure E Figure FManometer

For ideal gases at constant temperature, the volume of the gas is inversely proportional to the absolute pressure. For a real gas, the volume is inversely proportional to the absolute pressure only if the temperature is well above its temperature of condensation. For gases like oxygen, hydrogen, and helium, ordinary room temperature is well above their temperature of condensation.

Example: (to be solved on the board)Suppose that a person's skull is filled with air at a gauge pressure of +7mm

Hg. A skull fracture reduces the volume and, in turn, increases the gauge pressure in the skull to +14mm Hg. Assuming that the air temperature remains constant, compute the ratio of the final skull volume to the initial skull volume.

initial intermediate final

pi V i

T i

p V

T

pf V f

T f

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conditions conditions conditions Figure G. A confined gas at different conditions

2. Charles' Law (Jacques Charles, 1787) At constant pressure, the volume of a confined gas is directly proportional to

the change in its temperature. Temperature should be in the absolute scale.

V α T ;V =kT (@ constant pressure P)

@ intermediate conditions: (refer to the preceding fig.)V=kT∨k=V

T a

@ final conditions: (refer to the preceding figure)

V f =k f T f∨k f =V f

T f b

for gases, k = constant; hence,

k = kf c

substituting equations a and b in equation c

VT

=V f

T f=…=constant 2

3. General Gas Law (Pressure-Volume-Temperature Relationship)To derive the General Gas Law, multiply equations 1 & 2

p iV i VT i

=pV f V

T f

dividing both sides by V,p iV i

T i=

p V f

T f

from intermediate to final conditions, pressure is assumed to be constant: that is, P = Pf

p iV i

T i=

p f V f

T f 3

extending equation 3,

p iV i

T i=

p f V f

T f=

pn V n

T n=constant

The product of volume and pressure of a given sample gas, divided by the

corresponding absolute temperature, remains constant.

4. Gay-Lussac's Law (sometimes called Charles' Law)This law was derived published from that of Charles' Law by Gay-Lussac's 15

years after the discovery of Charles' Law. It states that at constant volume, the pressure of a confined gas is directly proportional to the absolute temperature.

p α T∨p=k T

@ initial conditions,

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pi=k iT i∨k i=p i

T i a

@ intermediate conditions,

p=kT∨k= pT b

equating equations a and b,

pi

T i= p

T=…=constant 4

5. Density of a Gas Law

From the General Gas Equation,p iV i

T i=

p f V f

T f

with Density = ratio between the mass of a substance and its volume,

ρ (rho )=mV

∨V =mρ c

substituting equation c in equation 3 we formed,

p i( mi

ρi)

T i=

pf (mf

ρf)

T f

but the mass of a confined gas at the two conditions does not change; hence, mi = mf. Using this premise in the preceding equation, the equation may be simplified as,

p i

ρiT i=

p f

ρf T f=…=constant 5

The preceding observations state that:

* The volume is proportional to the number of moles - that is, doubling the no. of moles while keeping the pressure and temperature constant, doubles the volume.

* The volume varies inversely with the absolute pressure p - that is, if we double the pressure while holding the temperature and the number of moles constant, the gas compresses to one half of its initial volume.

* The pressure is proportional to the absolute temperature - that is, if we double the absolute temperature while keeping the volume and number of moles constant, the pressure doubles.

6. Ideal Gas Equation (the summary of preceding observations)For many gases at temperatures significantly above their condensation

temperature, the energy term of the General Gas equation, pV (pressure-volume) is proportional to the absolute temperature.

pV α T∨pV=NkT ∨Nk= pVT

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Where k=BoltzmannConstant= R

N A

¿number of molesof the gas x the Molar GasConstant¿n RN=number of molecules of a gas∈a volumeof space V¿number of molesof the gas x Avogadro ' s number¿n N A

R=Universal(Molar)Gas Constant¿k N A

¿(1.3806504 x 10−23 J / K)(6.02214179 x1023 /mol )¿8.314472471 J /mol K

pVT

=Nk=n N A k=nR

pV=nRT ∨pV=NkT 6

Experiments on several gases show that, as the pressure approaches zero, the quantity pV/nT approaches the same value of R for all gases. For this reason R is called the Universal Gas constant.

n= mM

M=Molecular (molar )mass

Examples:1. A house whose volume equals 77 m3 contains air at a pressure of 1

atmosphere and at a temperature of 22C. How many molecules of air are in the house? If the average mass of one mole is 4.8 x 10 -26 kg, what is the mass of all the air in the house?

Given: V = 77 m3; P = 1 atm; T = 22Cmaverage per mole = 4.8 x 10-26 kg

Required:a. N b. m

Solutions:a. Based on the available data/information, the Ideal Gas Law is

appropriate to apply. When using this law, temperature must be absolute (that is, in the Kelvin scale) and so units must consistent with other quantities. Thus, we have to convert temperature T into degree Kelvin and pressure P into N/m2.

T=(22+273) K=295 K

p=1atm( 101325 Pa1 atm )=101325 Pa=101325 N

m2

N= pVkT

=101325 N

m2 (77m3 )

1.3086504 x 10−23 J / K (295.15 K )= 7802025 Nm

4.074989656 x10−21J=1.914612222 x 1027

¿1 J=1Nm

Since the average mass of each molecule is 4.8x10-26 kg, the total mass of the air in the house is

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mtotal=N xmaverage mass per molecule

= 1.914612222 x 1027 molecules ( 4.8 x10−26 kg /molecule )

¿91.90138667 kg ( 1 lb0.45359237 kg )=202.607876 lbs

2. A 1.2 cc bubble of water rises from the bottom of lake, where the pressure is 1.75 atm and temperature is 7C, to the lake's surface, where the pressure is 1.0 atm and temperature is 23C. What is the volume of bubble just before it reaches the surface?

Given:Vb = 1.2 cm3; pb = 1.75 atm; Tb = (7 + 273.15)KPs = 1.0 atm; Ts = (23 + 273.15)K

Required: the volume of bubble just before it reaches the surface, Vs

Solution: Relative to the given and required info, the General Gas Equation is applicable with its proportionality technique. This is due to the fact that the number of molecules in the bubble does not change at it rises.

pbV b

Tb=

ps V s

T s=Nk

Rearranging the equation to determine the Vs,

V s=( pb T s

ps T b)V b=( 1.75 atm ∙ 296.15 K

1.0 atm ∙280.15 K )1.2 cc

¿2.219935749∨2.22 cc

Most containers of gas, when near atmospheric pressure, hold huge numbers of atoms or molecules of gas. A cubic centimeter of air contains about 3 x 1019 molecules.

Our lungs hold about 2000 cc of air. One mole of a substance (solid, liquid, or gas) consists of 6.02214179 x1023 atoms or molecules of that

substance. The number 6.02214179 x1023 is called Avogadro's number and is given the symbol NA. One mole of any form of matter contains a mass in grams equal to the atomic or molecular mass of that form of matter. For example, one mole of atomic hydrogen (H) has a mass of 1 gram, one mole of molecular hydrogen (H2) has a mass of 2 g, one mole of helium (He) has a mass of 4 g, one mole of molecular oxygen (O2) has a mass of 32 g, and so on.

One mole of a gas occupies a volume of 22.4 liters or 22.4 x 10-3 m3.

@ S.T.P. (Standard Temperature & Pressure):Standard Temperature, OC = 273.15 KStandard Pressure, 1 atm = 1.01325 x 105 Pa

= 760 mm Hg = 29.92126 in Hg = 14.69595 psi (pounds/in2) = 1.03 x 103 cm H2O

3. A scuba diver breathes 607 cc of air per breath and takes 27 breaths per minute. (a) What volume of air does the diver breathe in two quarters of an hour? (b) The diver is under the water at a depth where the air is inhaled at a pressure of 2.75 atm. The tank from which the air is released is at 240 atm. What must be thee volume of the tank so that the air will last 30 minutes? (c) Compute the mass of the air in the tank at a temperature of 20C (293.15 K). One mole of air has a mass of 29 grams.Solutions:a. With 607 cc as the volume per breath and 27 breaths/min., the volume of air that the

scuba diver takes for two quarters of an hour (30 minutes @ 15 min/quarter of an hour) is

V = (607 cc/breath)(27 breaths/min)(30 min) 1 m3

= 491670 cc = (491670 cc) ------------- 1 x 106 cc

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= 0.49167 m3 (or 491.67 li)

b. The volume of air just computed at a pressure of 2.75 atm must be compressed at a pressure of 240 atm so that its volume Vt (by applying and rearranging Boyle's Law) is,

pV 2.75 atmVt = ----------- = ------------ (0.49167 m3) Pt 240 atm

= 5.63371875 x 10-3 m3

c. To compute for the mass of the air in the tank, it is necessary to know the number of moles of the air in the tank. Thus, we have to apply first the Ideal Gas Law to solve for n.

p tV tn = -------- RT

since the volume is in cubic meter (m3) we have to convert the pressure in terms of N/m2 to achieve consistency of units.

1.01325 x 105 N/m2

Pt = 240 atm --------------------------------- 1 atm

= 2.4318 x 107 N/m2

We find, then that

(2.4318 x 107 N/m2)(5.6337187 x 10-3 m3)n = ------------------------------------------------------- (8.314472 J/mole K) (293.15 K)

= 56.20804014 moles

Since each mole has a mass of 29 grams, 56.20804014 moles has a total mass of

m = (56.20804014 moles) (29 g/mol) = 1630.033164 g or (3.594 lbs)

4. A large cylinder contains 1017 grams of compressed helium gas. How many balloons can be filled with the gas if the volume of each balloon is 0.017 m3 (about 17 cm in diameter), the pressure inside each balloon is 1.25 x 105 N/m2, and the temperature of helium is 27C?Given: p = 1.25 x 105 N/m2

V = 0.017 m3 (per balloon)T = 27C = (27 + 273.15)K = 300.15Kmtotal = 1017 gMHe = 4 4.0 g/mol

Required: No. of balloons that can be filled with the said gas inside the tank.

Solution: We have to determine first the no. of moles needed for each balloon. This is achieved by applying the Ideal Gas Law:

pV (1.25 x105 N/m2)(0.017 m3)n = ---------- = --------------------------------------------- RT (8.314472 J/molK)(300.15 K)

= 0.851502469 mole (per balloon)

Next, we have to determine the mass of helium required for each balloon, and it is equal to

m = n M = (0.851502469 mol)(4.0 g/mol) = 3.406009876 g (per balloon)

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Finally, the no. of balloons that can be filled by a 1017-gram helium gas is,

1017 g / 3.406009876 g = 298.5898565 balloons or 299 balloons

Supplementary Problems: (to be solved on the board)1. The density of nitrogen is 1.37 kg/m3 at S.T.P. Find the density at 45C and

0.997 x 105 N/m2.2. What is the volume occupied by a 7.25 grams O2 at S.T.P. The molecular

mass of oxygen is 32 g/mol.3. A steel tank contains carbon dioxide @ 0C and a pressure of 1.777 atm.

Determine the internal gas pressure when the tank is heated to the boiling point of water at normal conditions.

4. A mass of H2 occupies 37 m3 at 17C and 1.75 atm. Find its volume at 35C below zero and 3.75 atm.

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CALORIMETRY

Calorimetry. The science which involves measurements of the quantity of heat transferred between bodies (Ref.#2). It is an experimental technique used to measure the specific heat capacity of the substance. It is likewise used to measure the energy required to melt or boil a substance, to measure the metabolic rates of animals, and to calculate the chemical energy released during chemical reactions and even during explosions (Ref.#1).

Calorimetry measurements are made using a device calorimeter. It is used to measure the specific heat capacity of a solid material (refer to the figure that follows). Calorimeters vary considerably in construction, depending on the type of measurement to be made. A typical calorimeter consists of a can in an insulated container, and a sealable top cover or cap. The insulation around the can (i.e., Styrofoam or a vacuum jacket) prohibits the transfer of heat into or out of the can.

Heat Quantities.James Joule and Count Rumford proved that heat is a form of energy. It causes a body to

rise in temperature, to fuse, to evaporate, or to expand. It is the thermal energy absorbed or released during a temperature change. It is a measure of the total molecular energy of the body and its quantity is dependent on the body's size: That is, if two bodies with the same temperature differ in size (mass), one is bigger than the other, the one with the bigger size is likely to contain more heat than the other. Heat is energy transferred to or from an object and that of some other object it contacts in its environment due to a temperature difference. It is likewise defined as a quantity, which when added to a body rises its temperature, or when removed from a body decreases its temperature; provided, the state of the body does not change. It is given the symbol Q (nevertheless, some authors use the symbol H) and expressed in terms of joules, calories, or Btu.

Conversion Factors:1 Joule = 0.2390057 cal = 9.478172 x 10-4 BTU1 cal = 4.184 Joules = 3.965667 x 10-3 Btu1 BTUint = 252.1644 cal = 1055.056 Joules1 calint = 4.1868 Joules = 4.186109 Joulesint

One calorie is the quantity of heat required to change the temperature of 1 gram of water by 1C.One kilocalorie is the quantity of heat required to change the temperature of 1 kilogram of water by

1C.One BTU (commonly used in engineering) is the quantity of heat required to change the

temperature of 1 standard pound of water by 1F. For accurate work, the temperature change in the definition must be centered around 59F.

To convert BTU to calories,

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1 BTU = heat required to change 1 lb of water by 1Fbut 1 lb = 453.59237 grams and 1F = 5/9C

1 BTU = 5/9 (453.59237) = 252 calories

If the state or phase of a body does not change, there is a simple relationship between the of heat lost or gained by the body and its change in temperature. The temperature change of a body, in the same state, is proportional to the heat change (Ref.#2).

Specific Heat Capacity and Thermal CapacityThe specific heat capacity of a substance is the amount of heat that must be added to a unit

mass of that substance to raise its temperature by a degree. It is a characteristic of a substance, is given a symbol c, and expressed as energy per unit mass per degree temperature - for example, cal/gC, J/kgC, Btu/slugC, Btu/lbF. However, some writers define specific heat as a ratio - that is, it is a ratio of the thermal capacity of a certain substance to that of water. Defined as a ratio, it is unitless and hence, its numerical value is the same in any system of units.

If we know the specific heat capacity of a substance, we can determine the amount of heat required to raise the temperature of a mass m of a certain substance by an amount ÄT.

Q=mass x specific heat xchange∈temp .Q=mc ΔT

The heat (thermal) capacity of a substance is the amount of heat required to raise the temperature of that substance by a degree. It is given a symbol C, and expressed in J/K, kcal/C, Btu/F, or cal/C.

If the mass of a body is m (in kg) and its specific heat capacity is c (in kcal/kgC), then its heat capacity is mass x specific heat: C = (m x c) kcal/C

C=m c∨C=∆ Q∆T

Method of Mixtures. A fundamental principle that if several bodies at different temperatures are placed inside an insulated enclosure, the bodies will come to a common final temperature. If two bodies at different temperatures are placed in thermal contact, and we assume that the interchange of heat is between the two only, the amount of heat gained by the colder body is equal to the amount of heat given up by the other body. Or, if several bodies at different temperatures are mixed, in coming to a final common temperature, the amount of heat absorbed by some of the bodies is equal to the amount of heat given up by the other bodies.

Heat Gained+Heat Lost=0or,

Heat Gained=Heat Lost

Note: Heat Lost indicates a negative value.

Mean specific and molar heat capacities at constant pressure (P = 1.0 atm) and temperature range 0C to 100C Specific heat (c)

__________________________ M Molar (C) Substance J/kgC cal/gC g/mole J/moleCAlcohol (Ethyl) 2428 0.580 46.00 112.0Aluminum 900 0.215 27.00 24.6Beryllium 1820 0.435 9.01 17.7Benzene 1720 0.411 Brass 390 0.094 Brick 840 0.201 Cadmium 230 0.055* Carbon 507 0.121 12.00 6.1

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Concrete 880 0.210 Copper 387 0.0924 63.50 24.8Dry Air 1000 0.239 Ethylene Glycol 2386 0.570 62.00 148.0Glass 837 0.200 Gold 129 0.0308Human Body 3470 0.829 Ice(-25C to 0C) 2090 0.500 18.00 36.5Iron or steel 448 0.107 55.90 26.3Lead 130 0.031 207.00 26.9Marble (CaCO3) 879 0.210 100.00 87.9Mercury 138 0.033 201.00 27.7Methanol 2550 0.609 Nitrogen 1040 0.249 Oxygen 910 0.217 Protein 1675 0.400Salt 879 0.210 58.50 51.4Sand 820 0.196 Silicon 703 0.168Silver 234 0.056 108.00 25.3Steam 2010 0.480Steel 480 0.114Sodium Chloride 880 0.210 * Tungsten 134 0.032 184.00 24.8Turpentine 1800 0.42Wood 1680 0.401Water 4186 1.000 18.00 75.4 Zinc 390 0.092

* at room temperature* the values of c in cal/gC is numerically equal in Btu/lbF

Examples:1. A lady wishes to take a bath but the water is cold enough that it requires her to

warm it first by 20C. How much heat is required if she needs to warm 150 kg of water?

Given: cwater = 4185 J/kgC (refer to the table) mwater = 150 kg ΔT = Tf - To = 20C

Req'd: Q to be added to a 150 kg waterSol'n: Q = mwatercwaterΔT = (150 kg)(4185 J/kgC)(20C)

= 1.2555 x 107 J (answer)

2. How much heat must be removed from a 55-kg body of a woman at 40C to lower her temperature to 33C?Given: chuman body = 3470 J/kgC

mwoman = 55 kgTo = 40CTf = 33C

Req'd: Q to be removed from a woman's bodySol'n: Q = mhuman body cwoman(Tf - To)

= (55 kg)(3470 J/kgC)(33 - 40)C = -1.33595 x 106 J (answer)

* The negative sign reminds us that heat was transferred out of the woman's body.

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3. How much heat is needed to boil a 5-lb water initially at 75F? Assume normal conditions.

Given: mwater = 5 lbscwater = 1 Btu/lbFTo = 75FTf = 212F

Req'd: Q to boil 5 pounds of water

Sol'n: Q = mwatercwater(Tf - To) = (5 lb)(1 Btu/lbF)(212 - 75) F = 685 Btu (answer)

4. The temperature of a 357-gram metal specimen is raised to 99C. It is plunged into a 107-g silver calorimeter containing 500 grams of water at 23C. If the final temperature of the mixture is 27C, what is the approximate specific heat capacity of the said specimen?

Given: mmetal = 357 gmcalorimeter = 107 gmwater = 500 gTo = 99C (of the metal specimen)To (of calorimeter) = To (of water) = 23CTf (of the mixture) = 27Ccsilver calorimeter = 0.055 cal/gCcwater = 1.0 cal/gC

Req'd: cmetal specimen

Sol'n: Applying the Method of Mixture

Heat given up by the specimen + Heat absorbed by water and the calorimeter = 0

Qmetal + (Qwater + Qcalorimeter ) = 0 ------> a

But: Qmetal = mmetalcmetal (Tf - To)metal = 357g cmetal (27 - 99) C

Qmetal = -25704 gC cmetal ------> b

Qwater = mwatercwater (Tf - To)water = 500 g (1.0 cal/gC)(27 - 23) C

Qwater = 2000 cal ------> c

Qcalor = mcalorccalor (Tf - To)calor = 107 g (0.055 cal/gC)(27 - 23) C

Qcalor = 23.54 cal ------> d

Substituting equations b, c, and d in equation a,

-25704 gC cmetal + 2000 cal + 23.54 cal = 0-25704 gC cmetal + 2023.54 cal = 0

Transposing terms to solve for cmetal,

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-25704 gC cmetal = -2023.54 cal

-2023.54 calcmetal = --------------- = 0.079 cal/gC

25704 gCStates of Matter

All matter consists of some distribution of atoms or molecules..1. Solid. Atoms or molecules are usually bonded together in a well-defined structure (limited

region). They can vibrate about an equilibrium position but are unable to rotate or move to new positions in the solid. This is caused by the mutual attraction among molecules. Solids have definite shape.a. Crystalline Solid (atoms have ordered structure)b. Amorphous Solid (atoms are arranged randomly as in a glass)

2. Liquid. Atoms or molecules are far apart because molecules break away from their neighbors and move about more freely. Mutual forces of attraction exist but they are not enough to prevent translational motion. Liquids have no definite shape but dependent on their containers. Like solids, liquids offer strong repulsive atomic forces acting internally to resist compression - that is, when an attempt is made to compress them

3. Gas. Molecules or atoms are loosely bonded together. They are in constant random motion and exert only weak forces on each other. Gases have no definite shape and size, no fixed volume, and much distance between adjacent particles.

4. Plasma. A highly ionized substance containing equal amounts of positive and negative charges: That is, a collection of free, electrically charged particles - negatively charged electrons and positively charged ions. This is happening when matter is heated to high temperatures causing many electrons surrounding each atom to get freed from the nucleus. Plasmas exist inside the stars.

5. Bose-Einstein Condensate. This state of matter was the only one created while you were alive. In 1995, two scientists, Cornell and Weiman, finally created this new state of matter. Two other scientists, Satyendra Bose and Albert Einstein, had predicted it in the 1920. They didn't have the equipment and facilities to make it happen in the 20s. Now we do. If plasmas are super hot and super excited atoms, the atoms in a Bose-Einstein condensate (BEC) are total opposites. They are super-unexcited and super-cold atoms.

The BEC happens at super low temperatures. At zero Kelvin all molecular motion stops. Scientists have figured out a way to get a temperature only a few billionths of a degree above absolute zero. When temperatures get that low, you can create a BEC with a few special elements. Cornell and Weiman did it with Rubidium.

So it's cold. A cold ice cube is still a solid. When you get to a temperature near absolute zero something special happens. Atoms begin to clump. The whole process happens at temperatures within a few billionths of a degree so you won't see this at home. The result of this clumping is the BEC. A group of atoms takes up the same place, creating a "super atom." There are no longer thousands of separate atoms. They all take on the same qualities and for our purposes become one blob.

6. Fermionic Condensate. Scientists have created a new form of matter, which they say could lead to new ways of transmitting electricity. The fermionic condensate is a cloud of cold potassium atoms forced into a state where they behave strangely.

The new matter is the sixth known form of matter after solids, liquids, gases, plasma and a Bose-Einstein condensate, created only in 1995.

To make the condensate the researchers cooled potassium gas to a billionth of a degree above absolute zero - the temperature at which matter stops moving.

They confined the gas in a vacuum chamber and used magnetic fields and laser light to manipulate the potassium atoms into pairing up and forming the fermionic condensate.

Jin pointed out that her team worked with a supercooled gas, which provides little opportunity for everyday application. But the way the potassium atoms acted suggested there should be a way to turn it into a room-temperature solid.

It could be a step closer to an everyday, usable superconductor - a material that conducts electricity without losing any of its energy.

"If you had a superconductor you could transmit electricity with no losses," Jin said. "Right now something like 10% of all electricity we produce in the United States is lost. It

heats up wires. It doesn't do anybody any good." Superconductors could allow for the development of magnetically levitated trains. Free of

friction they could glide along at high speeds using a fraction of the energy trains now use.

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Phase Transitions and GasesMatter exists in various forms, or phases. If the temperature and/or pressure of a sample of matter is adjusted, the matter may undergo a phase transition. During a phase transition, matter shifts between its three states: solid, liquid, and gas. © Microsoft Corporation. All Rights Reserved.

Changes of States (Phase Transitions)

1. Fusion/Melting. The process of transforming the solid form of a certain substance into its liquid form without necessarily changing the temperature of the said substance.

Solid LiquidIce H2O0C 0C

If heat is added to a solid such as ice, its temperature increases. The atoms and molecules vibrate with greater amplitude. As more and more heat is added, eventually the vibrations become so violent that the atoms or molecules break away from their neighbors and move about more freely.

The amount of heat required to break the crystalline structure of a solid to form a liquid also differs for different substances.

The heat that is being added to a substance at melting point or at boiling point is not manifested by any change in temperature. Hence, it is to be latent heat or hidden heat. Latent heats are pressure-

dependent

The Latent Heat of Fusion (Lf) of a solid is the amount of energy needed to melt a unit mass of a solid at its melting temperature. The melting temperature varies from substance to substance

The amount of heat required for the process of fusion is,

Q=m Lf

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The change of state and change of temperature do not occur simultaneously in the same substance.

Heats of Fusion and Vaporization Substance Melting Lf Boiling Lv

Point J/kg Point J/kg

Aluminum 658.70 398400 2300.00 10500000Antimony 630.50 165000 1440 561000Copper 1083.00 134000 1187.00 5069000Ethyl Alcohol -114.00 1042000 351.00 854000Gold 1063.00 64500 2660 1578000Helium -268.93 20900Hydrogen (H2) -259.31 58600 -252.89 452000Iron 1530.00 272100 2500.00 6364000Lead 327.3 24500 1750 871000Nitrogen (N2) -209.97 25500 -195.80 201000Oxygen (O2) -218.79 13800 -183.00 213000Silver 960.80 88300 2193 2336000Sulfur 119.00 38100 444.60 326000Water 0.00 334000 100.00 2256000

Fixed Point Temperatures and Latent Heats of Fusion and Vaporization at a pressure of 1 atm

Substance Melting Lf Boiling LvPoint (ºC) kcal/kg Point (ºC) kcal/kg

Aluminum 660 93.0 2056 2000Ammonia -77 108.0 -34 327Copper 1083 50.6 2595 1760Ethyl Alcohol -114 25.0 78 204Gold 1063 16.1 2966 446Helium -269 5.97Hydrogen -218.8 15.0 -253 106.70Iron 1539 65.0 2470 1620Lead 327 6.3 1744 222Mercury -39 2.8 358 71Nitrogen -210 6.1 -196 48Oxygen 219 3.3 -183 51Platinum 1774 27.1 4407 640Silver 960 243.0 2212 552Tin 232 14.4 2270 650Tungsten 3400 44.0 5927 1180Water 0 80.0 100 540

* The amount of energy needed to melt larger or smaller amounts of matter is directly proportional to the mass of the matter involved.

The mechanical equivalent of heat to work foot-pound is,1 B.T.U. = 252.0 cal = 777.9 foot-pound 1 cal = 4.1868 J

2. Freezing/Solidification. The process of changing the liquid form of a certain substance into its solid form with out any change in temperature. This is the reverse process of melting.

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Liquid SolidH2O Ice0 C 0 C

If energy is removed from a liquid, its temperature decreases; the vibrational, rotational, and translational motions of the atoms and molecules in the liquid are reduced. When the liquid reaches what is called its freezing temperature, the random motion of its atoms and molecules is slow enough that they start to fuse or bond with each other; the solid state begins to form. This transition from the liquid to the solid state is called freezing and occurs at the temperature at which the solid melts. As atoms or molecules fuse together into the structure of the solid state, energy is released. The amount of energy released equals the energy needed to break the bonds when the solid melts. The amount of heat required for freezing is,

Q=−m Lf

* The negative sign indicates that the liquid loses or releases energy when it freezes.

* Heat Reservoir. Any substance which does not change in temperature upon the removal, or upon the addition of heat (e.g., a big body of water).

Example: A 10 million kilograms of ice in a small pond has an average temperature of -7.0C during the middle of winter. How much heat must be added to the said ice to convert it to water at 30C for summer swimming?

Solution: The conversion of the ice to water in this case takes place in three phases.

Phase 1: The ice should be warmed from -7.0C to 0C, its melting temperature. Heat required for this process is,

Q(to warm ice) = mciceΔt= 10000000 kg x 2090 J/kgC x [0-(-7)C]=1.463 x 1011 J

Phase 2: The ice at 0C should be melted to form an equal amount of water at the same temperature (0C). Heat required for the said process is,

Q(ice to water) = mLf (melting)= 10000000 kg x 3.35 x 105 J/kg= 3.35 x 1012 J

Phase 3: Finally, the water should be warmed to 30C. Heat required for this is,

Q(warm water to 30C) = mcwaterΔt= 10000000 kg x 4180 J/kgC x (30 - 0)C= 1.254 x 1012 J

Hence, heat required for the conversion process is the sum of the three results,Q(required) = 4.7503 x 1012 J

3. Vaporization. The process of changing the liquid form of a certain substance into its gaseous form without any change in the temperature of the substance. This process takes place at the boiling point of the said substance.

Liquid GasH2O Steam100C (212F) 100C (212F)

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Heat required for this process is,Q=m Lv

where: Lv is the Heat of Vaporization of the substance

The Latent Heat of Vaporization of a substance is the energy required to convert a unit mass of the said substance from liquid form to gaseous form at the liquid's boiling point.

The boiling temperature of a liquid is the temperature at which any heat added to the liquid causes the liquid to vaporize but does not cause an increase in temperature. Like freezing temperature, boiling temperature varies from substance to substance.

The energy required to boil a liquid also depends on its mass - the greater the mass, the more energy is required to boil it.

Example: How much heat is needed to vaporize a 35-gram ethyl alcohol?

Given: Lv of ethanol is 8.54 x 105J/kgm of ethanol is 35g or 0.035 kg

Required:Q to vaporize ethanol

Solution:Q = m Lv = 0.035 kg x 8.54x105 J/kg

= 29750 J

4. Condensation. The reverse process of vaporization. This process occurs when gas molecules are cooled to their boiling temperature. The further removal of energy causes the gas molecules to come together (that is, condense to form a liquid. The condensation temperature is the same as the boiling temperature.

Gas LiquidSteam H2O100C (212F) 100C (212F)

Heat required for this process is,Q=−m Lv

The (latent) Heat of Condensation is the amount of heat that must be given up by a unit mass of a (gaseous) substance to form a liquid at its condensation (or boiling) point. The heat of condensation is equal to the heat of vaporization.

The temperature at which a liquid boils or a gas condenses depends on the pressure of the gas above the liquid. As the pressure of air pressure above the liquid (say, water) decreases, so does its boiling temperature (refer to the following table).

Boiling Temperature of Water as a Function of Pressure

Pressure (mm Hg) Boiling Temp. (C)

1500 120.31000 108.0 760 (atmospheric pressure at sea level) 100.0

700 97.8

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600 93.6 400 83.0 100 51.6 50 38.1

20 22.2 10 11.3 4.6 0.0 2.0 -10.8

5. Sublimation. The process by which a material passes directly from the solid state to the gaseous state without becoming a liquid. The solid is said to sublime only at its sublimation point. It is an endothermic phase transition that occurs at temperatures and pressures below a substance's triple point in its phase diagram. (Applicable only to some materials, e.g., dry ice, iodine (produces fumes on gentle heating), naphthalene, arsenic (@ high temperatures) and ammonium chloride (decomposes into hydrogen chloride and ammonia when heated).

Solid GasDry Ice (Card Ice or Frozen CO2) CO2@ 1 atm, -78.5°C (194.65 K, −109.3 °F, 350.37 °R)@ 5.2 atm, -56.4°C (216.75 K, -69.52 °F, 390.15 °R).

The heat required to sublime a unit mass of a (sublimable) substance is called the Heat of Sublimation (Ls). It is analogous to heat of vaporization and its value depends on the pressure under which sublimation occurs. Sublimation temperature corresponds to the boiling point and is sensitive to pressure for the same reason the boiling point is. Liquid carbon dioxide cannot exist at a pressure lower than about 5 x 105 Pa (about 5 atm), and “dry ice” (solid carbon dioxide) sublimes at atmospheric pressure. Sublimation of water from frozen food causes freezer burn. Sublimation is a technique used by chemists to purify compounds.

6. Deposition. The process by which a material passes directly from gaseous state to the solid state skipping the liquid phase. This is the reverse of sublimation. It occurs when frost forms on cold bodies such as refrigerator cooling coils.

The changes in chemical composition are not changes of state. The melting point of any substance is the temperature at which equilibrium will exist

between solid and liquid phases of the substance at normal pressure (760 mm Hg). This temperature may also be termed as freezing point.

At any temperature, there will be a pressure of the gaseous phase of a substance such that the liquid and gaseous phases are in equilibrium. This is the vapor pressure of the liquid at that temperature.

The temperature at which the vapor pressure of a liquid is equal to normal atmospheric pressure is the boiling point of the liquid, which could likewise be called its condensation temperature.

Very pure water can be cooled several degrees below the freezing temperature without freezing; the resulting unstable state is described as supercooled. When a small ice crystal is dropped in or the water is agitated, it crystallizes within a second or less. Supercooled water vapor condenses quickly into fog droplets when a disturbance, such as dust particles or ionizing radiation, is introduced. The principle is used in “seeding” clouds, which often contain supercooled water vapor, to cause condensation and rain.

In thermodynamics, the word endothermic ("within-heating") describes a process or reaction in which the system absorbs energy from the surroundings in the form of heat. Its etymology stems from the prefix endo- (derived from the Greek word ένδον, endon, "within") and the Greek word thermasi, (meaning “to heat”). The opposite of an endothermic process is an exothermic process, one that releases energy in the form of heat. The term endothermic was coined by Marcellin Berthelot (25 October 1827 – 18 March 1907). - http://en.wikipedia.org/wiki/Endothermic

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The enthalpy of sublimation (also called heat of sublimation) can be calculated as the enthalpy of fusion plus the enthalpy of vaporization. The reverse process of sublimation is deposition. The formation of frost is an example of meteorological deposition. The enthalpy of sublimation of dry ice is 571 kJ/kg (25.2 kJ/mol).

Snow and ice sublime, although more slowly, below the melting point temperature. This allows a wet cloth to be hung outdoors in freezing weather and retrieved later in a dry state. In freeze-drying, the material to be dehydrated is frozen and its water is allowed to sublime under reduced pressure or vacuum. The loss of snow from a snowfield during a cold spell is often caused by sunshine acting directly on the upper layers of the snow. Ablation is a process that includes sublimation and erosive wear of glacier ice.

Lc= 46,000 J/g = 4.6 x 107 J/kg (Heat of combustion of gasoline)