Chapter Notes: Temperature, Energy and Thermal Properties of Materials Mr. Kiledjian

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1 Chapter Notes: Temperature, Energy and Thermal Properties of Materials Mr. Kiledjian 1) Temperature 2) Expansion of Matter 3) Ideal Gas Law 4) Kinetic Theory of Gases 5) Energy, Heat transfer and Specific Heat 6) Heat of Fusion and Vaporization and Phase Change 7) Radiation 1) Temperature: Temperature is measured either in the Celsius, Fahrenheit or Kelvin scales. The Celsius scale is used all over the world and is the typical metric unit for temperature. It is set up so that 0 Celsius corresponds to the melting point of ice and 100 Celsius to the boiling point of water. The coldest possible temperature is Celsius which is called Absolute Zero. The Kelvin scale is set up so that Celsius corresponds to 0 Kelvin so the equation is K = C Therefore, ice melts at Kelvin and water boils at Kelvin; typical room temperature of about 27 Celsius corresponds to 300 Kelvin. The Kelvin scale is used in most scientific applications, particularly in equations such as the ideal gas law. [Notice that 1 unit of Celsius scale is the same as 1 unit of Kelvin scale.] The Fahrenheit scale is used in America and is set up so that 0 Celsius corresponds to 32 F and 100 Celsius to 212 Fahrenheit. The equation is F = 9C/ or C = 5(F 32)/9. Ex 1: The typical human temperature is 98.6 Fahrenheit. What is the corresponding Celsius and Kelvin measurement? C = 5( )/9 = 37 Celsius, K = = Ex 2: What temperature is there where the Fahrenheit and the Celsius measurements are the same? F = 9C/ = C ----> 32 = -4C/ > C = -40 Celsius = -40 Fahrenheit Ex 3: What is the equivalent Fahrenheit measure for absolute zero? F = 9C/ = 9( )/ = Fahrenheit 2) Expansion of Matter: When a substance is heated, it expands an amount L = L final - L initial. This expansion amount is proportional to its original length, L initial, and the change in temperature of the material, T = T final T initial, and a property of the material itself known as the Coefficient of Linear Expansion,. This formula can be written as L = L i T and is known as the formula for linear expansion. Usually L and L i are measured in meters, T is measured in Celsius, and therefore, the units of are 1/Celsius. The coefficients of different materials are: Alum = 24 x 10-6, Copper = 17 x 10-6, Lead = 29 x 10-6, Concrete = 12 x 10-6, Steel = 11 x 10-6 For objects that are two or three dimensional, then their area and volume expand as given by these equations: A = 2 A i T V = 3 V i T and for liquids and gases, V = V i T EthylAlcohol = 1.12 x 10-4, Mercury = 1.82 x 10-4, Gasoline = 9.6 x 10-4, Air = 3.67 x 10-3

2 All of the coefficients are 10-6 or 10-4 or 10-3, so that the expansion of materials is not that noticeable to the human eye, but its effects are felt especially if the object is large to begin with. Ex 1. Let s say L initial = 92.3 m, T initial = 24.5 Celsius, T final = 83.4 Celsius and the metal is Copper, then L = (17 x 10-6 )(92.3)( ) = meters = 9.24 cm, which is definitely noticeable. Ex 2. Let s say there is a square concrete block with sides of 2 m when T initial = 24.5 Celsius, what will be the new side dimensions when T final = 50 Celsius? A = 2(12 x 10-6 )(2) 2 ( ) = m 2, the original area A = 2 2 = 4 m 2 ; Therefore, the final area, A f = = m 2 ; setting this equal to s 2, where s is the new sides of the block gives us, s f = ( ).5 = m, and the difference in the sides gives us s f s i = m =.61 mm, which is very small, but still if you had a bunch of blocks lying side by side on the street and they each grew by that much, then that would have a big effect! Ex 3; If a certain car takes 16 gallons of gasoline, and the gas costs $3.00/gallon, how much more is a person paying in the summer for a full tank of gasoline than in the winter? Assume the summer temperature is 40 Celsius and the winter is 20 Celsius. In the winter, if you fill 16 gallons at 3 dollars per gallon, you will pay $48 In the summer, the 16 gallons of gas will expand V = V i T = (9.6 x 10-4 )(16)(20) =.3072 g, so the new volume it occupies is gallons, but you are still getting the same mass of gasoline. So, you will be charged $48.92 for the same amount of gas. You are paying 92 cents extra and percentage wise, this amounts to.92/48 x 100% = 1.92% higher that you are paying in summers!! 3) Ideal Gases: ideal gases are those gases which obey the Ideal Gas law pv = nrt which can also be written as pv = NkT. P = pressure of the gas measured in Pascals (in Chemistry, this is also measured in atm) V = volume of the gas measured in m 3 (also measured in Liters in Chemistry) n = number of moles N = number of molecules of the gas = nn A where N A is Avogrado s number = 6.02 x R = Universal gas constant = 8.31 J/mol.K (in chemistry, they use.0821 L.atm/mol.K) k = Boltzmann s constant = 1.38 x J/K Ex 1: An ideal gas in a chamber starts at a pressure of 1.5 atm, volume 5 liters and temperature 40 Celsius. Then, a piston pushes down on it reducing its volume to.8 liters, while a burner applied to the chamber increases the temperature of the gas to 200 Celsius. What is its new pressure? P i V i = nrt i, P f V f = nrt f, therefore, P i V i /T i = P f V f /T f so the equation becomes, > (1.5)(5)/( ) = P f (.8)/( ) ----> P f = atm [Always use Kelvin for T] % increase = 100% x /1.5 = % This is sort of what happens in a car engine!! Ex 2: A helium gas balloon occupying a volume of 2 liters starts out at the surface of the Earth at 20 Celsius and begins to rise until the temperature drops down -50 Celsius. What is the initial and final pressure in the balloon expressed in atmospheres and Pascals? We have to first find the mass of helium in the balloon from its density. The density of helium is.179 kg/m 3. Since, 1m 3 = 1000 liters, therefore, >

3 2 Liters x (1m 3 /1000 liters) x (.179 kg/m 3 ) x (1000 g/1 kg) =.358 grams, then we find the number of moles because we know 1 mole of helium has a mass of 4 grams from its atomic mass > n =.358 grams x (1 mole/4 grams) =.0895 moles. Therefore, PV = nrt becomes, P i (2) =.0895(.0821)( ) -----> P i = atm x (1.013 x 10 5 Pa/1 atm) = 109,100.1 Pa So, for the final pressure, we assume that the volume of the balloon stays at 2 liters, so P f (2) =.0895(.0821)( ) -----> P i =.82 atm x (1.013 x 10 5 Pa/1 atm) = 83,066 Pa 4) Kinetic Theory of Gases: The average kinetic energy of molecules is what determines the temperature of the gas. The equation can be written as KE avg = mv 2 avg/2 = 3kT/2 = 3nRT/(2N), so that this could also be written as Mv 2 avg/2 = 3RT/2, since N = the number of particles = nn A and the n cancels, and N A m = M = mass of 1 mole of the gas, m = mass of 1 particle or molecule of the gas, k is Boltzmann s constant, and R is the gas constant. From this we see that if the temperature doubles, then the average velocity goes up by a factor of radical 2 and if the temperature quadruples, then the velocity doubles, but in order for this to work, the temperature has to be measured in Kelvin. Ex 1: Find the average kinetic energy and velocity of a Helium atom at a temperature of 25 Celsius. KE avg = 3kT/2 = 3(1.38 x )( )/2 = 6.17 x Joules, v = (3RT/M).5 = (3(8.31)( )/.004).5 = 1,363.2 m/sec, which is fast. Notice that I used.004 for M since 1 mole of Helium has an atomic mass of 4 grams =.004 kg. Ex 2: If the gas in the above example doubles its average velocity, then what is its new temperature? If v doubles, then the temperature quadruples so T f = 4( ) = 1,192.6 Kelvin, so the new temperature in Celsius, T f = = Celsius. Notice that the temperature in the Celsius scale more than quadruples. 5) Energy, Heat transfer and Specific Heat When two substances are mixed together, then heat flows from the hotter object to the colder object, so heat is the transfer of energy between two objects, not a property of a single object whereas energy is a property of a single object and not the transfer of it. Energy and Heat are both measured by the same units, either Joules or calories or Calories, which are food calories. 1 Calorie = 1000 calories = 4,186 Joules. When heat flows into an object, its temperature rises according to a property of the object known as its Specific Heat Capacity, c. Water has the highest specific heat. Therefore, its temperature does not rise much if a certain amount of heat flows into it. On the other hand, most metals have low specific heats and their temperatures rise a lot for a given amount of heat flow. The equation for the phenomenon is written as H or Q = mc T where H or Q is the amount of heat (measured in Joules or calories), m is the mass of the substance (measured in kg or grams), T is the change in temperature (measured in Celsius), and c is the specific heat (measured in Joules/Kg.Celsius or calories/gram.celsius). The specific heats of some substances are: c water = 1 cal/g.c = J/g.C = 1 Cal/Kg.C = 4,186 J/Kg.C, c ice =.5 cal/g.c = J/g.C, c Alum =.215 cal/g.c =.90 J/g.C, c gold =.030 cal/g.c =.126 J/g.C, c wood =.41 cal/g.c = J/g.C. Notice that water has the highest specific heat. This means that for a given amount of heat transfer, its temperature does not rise or drop a lot. Wood is about half that of water and the metals are much lower.

4 If we pour a hot substance into a cup of water at room temperature and stir the cup, the temperature of the water will rise, and the temperature of the substance will drop. It will finally reach an equilibrium temperature, T final. If the cup is well insulated so that all of the heat flowing from the substance goes into the water, then the heat lost by the substance will equal to the heat gained by the water. This can be written by the equation: H lost by substance = m substance c substance T substance = H gained by the water = m water c water T water Since they reach the same final temperature, the above equation can also be written as: M substance c substance Initialsubstance T final ) = m water c water final - T Initial Water ) This equation can be used to find the experimental value of the specific heat of metals or any other substance. Ex 1. M substance = 75.8 grams, T InitialSubstance = 97.6 Celsius, T final = 53.6 Celsius, M water = 64.9 grams, T initialwater = 24.5 Celsius. Find the specific heat of this substance. The equation becomes 75.8c substance ( ) = 64.9(1)( )----> c substance = > c substance =.566 Ex 2. M substance = 75.8 grams, T InitialSubstance = 23.8 Celsius, T final =? Celsius, M water = 64.9 grams, T initialwater = 74.5 Celsius, Find the final temperature of the system; assume that the substance is gold. The equation becomes 75.8(.030)(T f 23.8) = 64.9(1)(74.5 T f )----> 2.274(T f 23.8) = 64.9(1)(74.5 T f ) ----> 2.274T f = T f -----> T f = > T f = C Notice that the water was not effected by the gold too much!! 6) Heat of Fusion and Vaporization and Phase Change The heat of fusion of a substance is the amount of heat necessary to melt it, and if we want to freeze it, then we need to extract an amount of heat equal to the heat of fusion. The heat of vaporization is the amount of heat necessary to vaporize a substance and change it to its gas form; the opposite of this would be to condense the substance from its gas form back to liquid. The amount of heat needed to melt or vaporize a substance is the product of its mass and these constants. The equations are written as, H f = ml f and H v = ml v where L f and L v are the heats of fusion and vaporization in units of J/g or cal/g or J/kg or Cal/kg. Here are the heats of fusion and vaporization of different substances: L fwater = 333 J/g = 79.7 cal/g, L fgold = 64.4 J/g = 15.4 cal/g, L falum = 397 J/g = 94.8 cal/g L vwater = 2260 J/g = 540 cal/g, L fgold = 1580 J/g = 377 cal/g, L falum = J/g = 2720 cal/g Notice that it is much tougher to vaporize an object that to melt it because you are trying to break the bonding of the atoms and change the substance to its gas form. If we mix two substances together and one of them is going to experience a phase change, then we have to include this additional heat in the equation to find the final temperature. Ex 1. Imagine there is some ice at a temperature of -15 Celsius and you drop it in a beaker of water at 80 Celsius, what is the final temperature of the mixture? m ice cubes = 25.6 grams, T iice = -15 Celsius, T iwater = 80 Celsius, T f =? Celsius, m water = grams. The ice has to heat up to 0 Celsius, then melt, and then heat up to T f, whereas the water just has to cool down to T f. We will need to use the specific heat of ice which is.5, but when it turns to water, it is (.5)(15) (79.7) (1)T f = 184.6(1)(80 T f ) ---> T f = T f > 210.2T f = > T f = Celsius

5 Ex 2: A block of gold with a mass of 20 kg at a temperature of -15 Celsius is dropped into a container of steam at 100 Celsius with a mass of 80 grams. What is the final temperature of the mixture? How much heat was extracted from the steam and given to the gold? The steam has to condense and cool down to a final temperature T f but the gold block just needs to heat up to T f ; it experiences no phase change since its melting point is 1063 Celsius. 20(.030)(T f -15) =.08(540) +.08(1)(100 T f ) ---->.6T f + 9 = T f >.68T f = > T f = Celsius H extractedfromsteam = 80grams(540cal/gram) + 80grams(1cal/g.C)( ) = calories = Calories = x 10 5 Joules = kjoules H transferredtogold = (20,000 grams)(.03)( ) = calories (it should be the same answer; the only reason for the difference is rounding error!) 7) Radiation The power radiated by an object is given by the formula P = AeT 4 where T is its temperature in Kelvins, A = surface area of the object in meter 2 = 4 R 2 for a spherical object, is the radiation constant = x 10-8, and e = emissivity of the object which is between 0 and 1. The emissivity of black objects is 1 and their albedo, which is the ratio of their reflected energy to their absorbed energy, is 0. Therefore, black objects absorb the most but also radiate the most. So, on hot summer days, it is not good to wear black shirt because the shirt will absorb a lot of energy even though it will also radiate a lot, but since the temperature of the outside is hotter than the shirt, its absorption will be more than its radiation. On cold winter days, it is also not good to wear black because this time it will radiate more than absorb because a person s body temperature will be higher than the outside and they will lose heat at a fast rate. Similarly, a black coffee without creamer cools down faster because it radiated more energy than the white coffee even though it also absorbs from its surroundings. Ex 1: Find the power radiated by the sun, also called its luminosity. How much energy does it radiate in a day? What would happen if its temperature doubles and becomes 11,600 Kelvin? The emissivity of the sun is about 1 because it is considered a black body object which radiates at all wavelengths. We will use R sun = 6.96 x 10 5 km = 6.96 x 10 8 meters and T sun = 5,800 Kelvin > L sun = P sun = ( x 10-8 )4 (6.96 x 10 8 ) 2 (1)(5800) 4 = 3.91 x watts Since power is energy divided by time, E = Pt = 3.91 x (1 day) x(24hrs/1day) x(3600sec/1hr) = 3.37 x Joules. That s a lot of energy!! If the temperature doubles, since power is proportional to the fourth power of temperature, the power is multiplied by a factor of 16, so P f = 16P i = 16(3.91 x ) = x watts Ex 2: Imagine there is a cup of black coffee at 75 Celsius and another cup of coffee with creamer which is usually colder to begin with, let s say 50 Celsius and take its emissivity to be.6 What is the ratio of their emitted power? Assume that both cups are the same size. P blackcoffee = AeT 4 = A(1)( ) 4 = x A P whitecoffee = AeT 4 = A(.6)( ) 4 = x 10 9 A P blackcoffee /P whitecoffee = 1.469/.6543 = 2.25 So, the black coffee cools about 2 and ¼ times faster initially until their temperatures are about the same and then it cools 1/.6 = 1.67 times faster!