Lecture16 - Temperature Temperature and the Zeroth...

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Unformatted text preview: Temperature Temperature and the Zeroth Law When you put something into the refrigerator, it will eventually be the same temperature as the refrigerator. Why ? When two object are in contact, energy can flow from the hoCer object to the colder one unDl they are in thermal equilibrium. Chapter 19 If bodies A and B are each in thermal equilibrium with a third body C, then A and B are in thermal equilibrium with each other. This statement is called the Zeroth Law of Thermodynamics. This law the basis for defining temperature. Because of the zeroth law, temperature is well defined, independent of the object you measure it on. How to measure temperature If we have a system that changes some measurable property as a funcDon of how ‘hot’ it is, we have a thermometer. Most thermometers are based on the fact that many materials: gases, liquids and solids expand when their temperature increases. Historically, three different temperature scales have been used: Fahrenheit: A non ­scienDfic scale, now only used in the United States. Celsius: A non ­scienDfic scale, that is used everywhere else. Kelvin: The preferred scienDfic scale, used in science worldwide. Kelvin: 0 K is defined as absolute zero, which is the lowest theoreDcal temperature (no transfer of energy). 273.16 K is defined by the triple point of water (temperature at which ice, liquid & water vapor can coexits). Celsius: 0oC is defined by the freezing point of water and 100oC is defined by the boiling point of water. Fahrenheit: 32oF is is defined by the freezing point of water and 212oF is defined by the boiling point of water. In this scale normal human body temperature is about 100oF (98.6oF). Conversions Celsius to Fahrenheit: Celsius to Kelvin: These conversion relaDon can be obtained from their respecDve values for melDng point of ice and boiling point of water: TC T − 32 TK − 273.15 =F = 100 180 100 Temperature Boling water Human body Pleasant day Freezing water Boiling point of liquid nitrogen Absolute zero oC oF K 373.15 310.15 293.15 273.15 77.36 0 100 37.0 20 0  ­195.79  ­273.15 212 98.6 68 32  ­320.42  ­459.67 Thermal Expansion In general most materials expand when they are heated. If a metal rod is heated it’s expansion (change in length here) is proporDonal to its iniDal length and the rise in temperature: ΔL = α Li ΔT The proportionality constant α is called the linear expansion coefficient and depends on the type of material being heated/expanded. It has the unit of inverse temperature, oC ­1 or K ­1. Example: A 2 m long rod of aluminum is heated from 0oC to 70oC. By how much does it expand ? For aluminum α = 23x10-6/oC Example: What is the temperature required to extend a length of steel railroad track iniGally of length 103m = L, with a = 12E ­6/oC, by 60cm = ΔL? This is why train tracks have gaps along their length, so that they don’t buckle from expansion due to the heat increase. ApplicaDons Bimetallic strips in thermostats Expansion joints in a bridge WINTER SUMMER Volume expansion Of course an object will not only expand in one dimension, but in all three, i.e. its volume will change. Assume we have a cube of edge length L of some material. Then: If the cube expands we have: Since ΔL is small, we neglect terms terms with ΔL2 and ΔL3. Now use the equaDon for linear expansion: ΔV = βV ΔT with Example By how much do I have to increase the temperature of a 20 cm3 piece of Aluminum in order to increase its volume to 21 cm3? ΔV = βV ΔT with Unusual expansion expansion of water If we heat water from 0 C to 4 C, it does not expand like other materials – it contracts. This anomalous behavior of water can be seen from the following density vs temperature diagram of water. Density increases from 0 oC to 4 oC and then increase again from there because volumes decreases from 0 oC to 4 oC and and increases again from 4 oC. Density of water is therefore maximum at 4 oC. This behavior of water explains why a pond begins freezing at the surface rather than at the boCom. Even when the surface totally freezes, there will be warmer water (e.g. at 4 oC) at the boCom. This is very important for the survival of fish and other forms of marine life in cold regions. Macroscopic descripDon of an ideal gas Consider a gas of mass m confined in a container of volume V at pressure P and temperature T (kelvin scale). An equaDon that interrelates these quanDDes is called the equa4on of state of the gas. EquaDons of state of a gas may be very complicated. However, for a gas at very low pressure or very low density, the equaDon of state turns out to be a simple and can be found experimentally. Such low pressure or low density gas is called an ideal gas. Amount of substance in a volume is usually expressed in number of moles. A mole of a substance, is the amount that contains NA = 6.022× 1023 consGtuent parGcles (molecules in most gases). The number NA is called the Avogadro’s number. For a gas, for example, of mass m and molar mass M, the number of moles n is given by: m n= M where M = molar mass of the substance, usually expressed in gram/mole. For example, for oxygen M = 32.0 g. Example 1 What is the number of moles in 60 g of O2? How many oxygen molecules are there? Here m = 60 g, M = 32g/mol n = m/M = 60g/(32g/mole) = 1.875 mole = 1.9 moles Number of oxygen molecules = nNA = 1.875×6.022× 1023 = 1.1×1024 Example 2 How many SiO2 units are in a grain of sand, having a mass of 10 ­7 kg ? Here m = 10 ­7 kg, molar mass of SiO2 , M = 28 + 2 × 16 = 60 g /mol Number of moles, n = m/M = 10 ­7×1000g/(60g/mol) = 1.7 x 10 ­6 mol Number of SiO2 units = nNA = 1.7 x 10 ­6 mol × 6.022× 1023/mol = 1018 Example 3 How large are copper atoms ? Here M = 63.54 g /mol, density of copper = 8.96 g /cm3. Thus for 1 cm3 of copper, n = m/M = (8.96 g /cm3)/(63.54 g /mol) = 0.141 mol Number of copper atoms in 1 cm3 of copper = nNA = 0.141× 6.02 x 1023 = 8.5 x 1022 Therefore volume occupied by a copper atom = 1 cm3/8.5 x 1022 = 1.2 x 10 ­23 cm3 If we assume that the atoms are spherical, volume = 4πR3/3. The radius of copper atom is EquaDon of state of an ideal gas There are two important observaDons on ideal gas: 1.  Pressure of an ideal gas P, at constant temperature T, is inversely proporDonal to its volume. (Boyle’s law) 2.  At constant pressure, volume of an ideal gas is directly proporDonal to its temperature. (Charles’ law). These two empirical ‘laws’ can be combined into a single equaDon: pV = nRT Here n is the number of moles of the gas and R a constant called universal gas constant. As the name indicates it is the same for all ideal gases. In SI, its value is R = 8.32 J/(mol.K) (R = 0.082 L.atm/(mol.K), if pressure is expressed in atmosphere and volume in liters.) We see that for an ideal gas: pV = const nT , or for the same amount n: pV = const T We can write the equaDon of state in terms of total numbers of atoms/molecules. For this, note that n = N/NA and have pV = N R RT = N T ⇒ pV = NkBT NA NA Here kB is called the Boltzmann constant. kB = 1.38 10 ­23 J/K. Example A hair ­spray can contains a gas pressure of twice atmospheric pressure (=202kPa) with a volume of 125cm3 at temperature Ti = 22oC. It is then heated to a Tf= 195oC, what is the Pf inside the can? (assume volume changes are negligible) pV p V p f Vf = const ⇒ i i = T Ti Tf ⇒ pf = Vi T f ⋅ ⋅ pi V f Ti Here Vi = Vf (volume does not change). Ti = 22oC = 273.15 + 22 = 295.15K, Tf = 195oC = 273.15 + 195 = 468.15K , Pi = 202 kPa. Therefore, ⇒ pf = 1⋅ 468.15 K ⋅ 202 kPa = 320 kPa 295.15 K ...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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