Chapter 10 Temperature And Heat - Doane...

Physics Including Human Applications

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Chapter 10 Temperature And Heat GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: temperature mechanical equivalent of heat thermometer heat capacity heat specific heat linear expansion latent heat of fusion volumetric expansion latent heat of vaporization calorie heat of combustion Calorimetry Solve problems in calorimetry. Gas Laws Solve problems using the gas laws involving the pressure, volume, and temperature of a confined gas. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 5, Energy, and Chapter 9, Transport Phenomena.


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Chapter 10 Temperature And Heat

10.1 Introduction Your interactions with your environment provide a variety of experiences that are related to the ideas of temperature and heat. Some of your first autonomous decisions may have been your choices of clothing, choices at least partly influenced by the answer to the question, "What is the temperature going to be today?" You have had many opportunities to influence the temperature of your food. Have you not burned your mouth on something too hot? What methods have you learned to use to cool your food down rapidly so that it is at the proper eating temperature? Your experiences may have included a number of injunctions that contradict one another. For example, you may think that all water boils at the same temperature, and that water never gets hotter than its boiling temperature, yet the cookbook says to cook the rice in water that is at a "rolling boil." Why? In addition to contradictions, our cultural lore is full of interesting, and perhaps false, statements about temperature and heat: "Hot water placed outside on a very cold winter day will freeze faster than cold water." "A steam burn is worse than a hot water burn." "White clothes are cooler than dark clothes." Furthermore, your first-hand experiences with heat and temperature are not without puzzles. On a hot summer day why would a piece of metal lying on the sidewalk seem hotter to your bare feet than the concrete? Such puzzles helped lead to the present quantitative ideas about temperature and heat. Count Rumford, one of the forefathers of our modern concepts of heat, noted in his personal diary that each time he saw a person burn his mouth on hot apple pie, he wondered what it was about apples that made them seem to retain their heat much longer than other foods. The point of these observations is that you already have had many encounters with the concepts of temperature and heat. The purpose of this chapter is to discuss these concepts in a way that will give you some quantitative understanding of them and will correct any misunderstandings that you have. 10.2 Temperature One of the first discriminations you learned was the relative hotness or coldness of an object using the heat sensing organs of your body. However, you also learned that this discrimination was not very reliable. Bath water that feels only slightly warm to your hand may feel rather hot when you sit down in it. The cookbook says, "Bake in a hot oven for two hours." The novice cook is puzzled by the term "hot" oven, just what is a hot oven anyway? How hot does an oven have to be to be a hot oven? To answer this question we need to define some common measures of hotness that can be used by different people. We use the term temperature to speak about the relative hotness or coldness of an object. We then set out to develop some temperature scales that can have universal use. The human heat sensors are transducers that change the heat energy input into electrical signals that are sent to the brain (see Chapter 1). But we know how unreliable those transducers are for use in defining any universal temperatures. Therefore, we must find some other kinds of transducers that


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have reproducible, quantitative responses to changes in temperature. Such transducers are called thermometers. In this context temperature can be operationally defined as the property of a system measured with a thermometer. A system that has a reproducible linear relationship between temperature and a change in a physical property of the system seems to be a logical choice for a thermometer. A list of some of the kinds of thermometers that are presently in use is given in Table 10.1. None of these transducers is completely satisfactory for all uses and for all ranges of temperature. The best thermometer for a particular use must be selected by taking into account such factors as cost, accuracy, temperature range, and durability. Once you have selected a transducer system to use for your thermometer, you must establish some way of calibrating it. That is, you need to determine a numbering system that you can use to measure various temperatures. If you wish other people to be able to use your temperature system, you will need to find some events in nature that always occur at the same temperature. You can use these events to serve as reference, or fixed, points on your thermometer.

Based on these ideas, the first accurate thermometer was made in 1641 using alcohol in glass. Then in 1714 the German physicist G. D. Fahrenheit built a mercury-in-glass type thermometer and used three fixed points to calibrate it: the freezing temperature of a mixture of water, ice, and salt which he labeled 0ø, the freezing point of pure water labeled 32ø, and the temperature of the human body labeled 96ø. Please note that this temperature does not agree with current normal body temperature. Although Fahrenheit did not use it to calibrate his thermometer, it was found that water boils at 212ø on the Fahrenheit temperature scale at sea level. In 1742 the so-called centigrade scale was developed, which uses the freezing point of pure water (0ø) and the boiling point of pure water (100ø) as its fixed points with 100 scale divisions between these two points. This scale is attributed to Anders Celsius and is now known as the Celsius temperature scale. Another international temperature scale has also been developed. This scale is based upon a principle of thermodynamics of Lord Kelvin which implies that there exists an absolute minimum temperature. This international scale, called the Kelvin scale, assigns the number zero to this lowest possible temperature. Additional


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details of the Kelvin scale will be given later in this chapter. The Celsius, Kelvin, Fahrenheit and Rankine scale relationships are shown in the following table. Absolute Freezing point Boiling point Scale zero of water of water ------------------------------------------------------------------------------------------------------------ Kelvin(øK) 0ø 273.15ø 373.15ø Celsius(øC) -273.15 ø 0ø 100ø Fahrenheit(øF) -459.67 ø 32ø 212ø

Rankine(oR) 0 ø 491.67o 671.64o Using this table, derive the equations required to convert any temperature from one of these scales to the other three. You will find that either the Kelvin scale (øK) or the Celsius scale (øC) is used in this book. 10.3 Changes in Size with Changes in Temperature You already know that the size of an object changes as its temperature changes. Joints (or cracks) are put in highways, bridges, buildings, and sidewalks to allow for changes in size that occur with the seasonal changes in temperature. As you know, most objects increase in size as their temperature increases. Can you think of any exceptions? Do you know any systems in nature that get larger as their temperatures decrease? The simplest assumption that we can make about the quantitative changes in the size of an object is to assume that the changes are linearly related to changes in temperature, that is, the changes in length of an object are directly proportional to its temperature changes.

ΔL ∝ ΔT (10.1) This relationship can be stated as an equality by the use of a proportionality constant a called the coefficient of linear expansion. The coefficient of linear expansion is defined as the increase in length per unit length per degree change in temperature, hence α = ΔL/(L0ΔT) (10.2)


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where ΔL is the change in length, L0 is the original length, and ΔT is the change in temperature. The values for the coefficient of linear expansion for various materials are given in Table 10.2. The length of an object at the temperature T is given by LT = L0 + ΔL = αL0ΔT (10.3) where ΔT is the difference between the original temperature and the temperature T. It is customary to choose the standard length L0 to be the length as measured at 0øC. Then Equation 10.3 can be written as LT = L0 (1 +αT) (10.4) where T is the temperature in degrees Celsius. EXAMPLE If a steel railroad rail of length 10.0 m is laid on a day when the temperature is 0.00øC, how much space should be left between the ends of the adjoining rails for a maximum summer temperature of 45.00øC? Lo = 10.0 m; ΔT = (45.00 - 0.00ø) = 45.00øC ΔL = (10.0)(11.0 x 10-6/øC)(45.00øC) =4.95 x 10-3 m = 0.495 cm So about one-half centimeter should be left between the ends of the 10-m rails. 10.4 Volumetric Expansion When a solid or liquid is heated, it expands in all directions. In a manner similar to the definition of the linear expansion coefficient, the coefficient of volumetric expansion can be defined: β =ΔV/VoΔT or ΔV = VoβΔT (10.5) where ΔV is the change in volume Vo for a temperature change of ΔT. Can you show that the coefficient for the volumetric expansion of a solid is approximately equal to three times the coefficient of linear expansion? Typical values for the volumetric coefficient of expansion of materials are given in the second half of Table 10.2 Water has many peculiar properties, and one of these is its behavior upon heating. The volume of pure water changes in an unusual way in the range from 0ø to 10øC. Starting with a given volume of water at 0ø and warming it, you will find that at first the water contracts until a temperature of 4øC is reached. Above this temperature it expands. Water has maximum density at 4øC and about the same density at 0øC and 8øC. This unusual behavior is shown graphically in Figure 10.1.


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This behavior of water is of great importance in nature. If water did not change in density this way, large bodies of water would more easily freeze at the bottom. (You will be able to explain this after you have studied convection.) All of the animal life in the lakes or ponds would be destroyed by complete freezing of the water. In Lake Superior the temperature of the water at depth greater than about 74 m is about 4øC during the entire year. You should be able to explain why a liquid in a glass thermometer using water is useless below about 10øC. Water is not a good liquid for thermometers above 10øC as the rate of expansion is not constant up to 100øC. EXAMPLES 1. A 100-cm3 Pyrex glass flask is filled with mercury at a temperature of 10øC. The flask

is stored in a room that reaches a summer temperature of 40øC. Will the mercury overflow? If so, how much will be lost?

The approximate coefficient of volumetric expansion for Pyrex is three times its coefficient of linear expansion, 3 x (3.2 x 10-6) or 9.6 x 10-6. This is much smaller than the volumetric expansion for mercury so the mercury will expand and overflow the flask.

The overflow will be given by the difference between the increase in the volume of the mercury and the increase in the volume of the flask. ΔVflask = 100 cm3 x 9.6 x 10-6 x (40 - 10) øC = 2.9 x 10-2 cm3 ΔVHg = 100 cm3( 0.18 x 10-3) x (40 - 10) øC = 5.4 x 10-1cm3 overflow = ΔVHg - ΔVflask = 0.54 cm3 - 0.029 cm3 = 0.51 cm3


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2. On a hot summer day gasoline from an underground service station tank is pumped into and completely fills a warm automobile tank. The gasoline is at 20.0øC when it is put into the auto. How much will overflow when it reaches the temperature of the 80.0 liter steel auto tank, which is at 40.0øC? Assume that the coefficient of expansion of gasoline is equal to that of petroleum. ΔVgasoline = (80.0)(0.900 x 10-3)(40.0 - 20.0) = 1.44 liters

10.5 Bimetallic Switch A very common use of the thermal expansion properties of materials is in a temperature sensing switch. An on-off switch that responds to changes in temperature has a wide variety of applications, from safety switches in factories to control knobs for electric frying pans. Many of these temperature switches consist of two thin pieces of different metals welded together to form a bimetallic strip Figure 10.2. Since the two metals have different coefficients of linear expansion, as the temperature changes the two metals will have different lengths, and the bimetallic strip will bend to accommodate the length difference. Describe the change in curvature of the strip as the temperature is increased and as it is decreased.

For small temperature changes, the bimetallic strip will bend only slightly. List some variables that influence the amount of deflection of the strip, x. 10.6 Heat In Chapters 1, 2, and 9 you have been introduced to the concepts of energy and heat. The amount of heat energy of an object depends upon the type of material of the object, the quantity of material, and its temperature. It is possible for a given object at low temperature to have more heat energy than another object at a high temperature. Also, an object has thermal inertia toward a change in temperature. You know that if you bring together two objects of different temperatures, they will reach a common temperature that is intermediate between the two initial temperatures. For example, if you have a cup of coffee too hot to drink, you can lower the temperature of the coffee by adding cool water, cream, or milk. You may think of this as a system of mixtures. Qualitatively, you can explain the phenomenon by saying the hot coffee gave up some heat energy, and the cool liquid gained some heat energy. To set up a quantitative equation we need to be able to calculate the transfer of energy involved in a temperature change for a system. In order to do this we must introduce the concept of


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the amount of energy required to change the temperature of a system. Once again we will fall back upon the simplicity of the linear relationship and assume that the amount of energy DQ needed to change the temperature of a system is directly proportional to the temperature change: ΔQ ∝ ΔT (10.6) We can relate these two quantities by the use of a proportionality constant, which we call the heat capacity C of the system. The heat capacity is the amount of heat energy absorbed or liberated from the object for a change in temperature of one degree. ΔQ = CΔT (10.7) where ΔQ is the change in heat energy of the system for a temperature change of ΔT. Since energy has the units of joules in the SI units, C will have the units of J/øC. You know that it takes longer to heat a large volume of water on a stove than it does to heat a small volume of water. So you already know that somehow the heat capacity is related to the amount of matter in the system. In fact, the heat capacity of an object is directly proportional to the amount of matter of the object. Therefore, to compare the heat capacity of different systems, we must take into account their different quantities of matter, or masses. It is useful to define the change in heat energy of a system per degree temperature per unit of mass. This property of a system is called its specific heat, c, where c = C/m = ΔQ/mΔT (10.8) Table 10.3 gives the value of the specific heat for various materials.

In other words, the quantity of heat energy required for a given temperature change of a system is the product of the mass of the matter in the system, the specific heat of the system, and the change in temperature: ΔQ = mc ΔT (10.9) Unfortunately, the early studies of the properties of heat developed apart from the studies of mechanical energy; so an independent unit for measuring the quantity of heat energy was defined. This quantity of energy is called the calorie and is the amount of energy required to raise the temperature of one gram of water one degree Celsius. This, it turns out, is the same amount of energy as 4.186 joules. 1 calorie = 4.186 joules This is called the mechanical equivalent of heat.


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EXAMPLES 1. An 800-g silver gravy bowl is filled with hot gravy at 100øC. If the silver bowl is

originally at room temperature (20øC), how much heat energy is required to warm the bowl up to the temperature of the gravy? FromTable 10.3, we see that the specific heat of silver is 0.56 x 10-1 cal/g-øC. Q = 800 g x 0.056 cal/g-Cø x (100 - 20)øC Q = (64 x 103)(5.6 x 10-2 cal) = 3.6 x 103 calories

2. We can now make a quantitative calculation as a result of conservation of energy: Heat energy lost by hot system = Heat energy gained by cold system. Assume you have 250 cm3 of coffee in a cup of thermal capacity 20.0 cal/øC and and a silver spoon of mass 80.0 g all at 80.0øC. You desire to have your coffee at 45.0ø for drinking. How much water at 10.0øC will you have to add? Assume coffee has the same specific heat as water.

heat energy lost = heat energy lost by coffee + heat energy lost by cup + heat energy lost by spoon

HELcoffee = 250 cm3 x 1.00 g/cm3 x 1.00 cal/g x (80.0 -45.0)øC = 8750 cal HELcup = 20.0 cal/øC x (80 - 45)øC = 700 cal HELspoon = 80.0 g x 0.056 cal/gm-øC x (80.0 -45.0)øC = 160 cal heat energy gained = mw x 1 x (45.0 -10.0)øC = 35.0mw 35.0mw = 9610 cal mw = 275 g of water The physics and math may be correct, but would you want to drink the coffee? What changes would result?

10.7 Changes in the State of Matter From past experience you know that there are three states of matter: solid, liquid, and gaseous (vapor). A transition from one state of matter to another, liquid to solid, for example, is called a change of state. Take the substance H2O as an example. Its solid state is ice, its liquid state is water, and its gaseous state is steam. Transition from one state to another is accompanied by absorption or liberation of heat energy and usually by a change in volume. Let us now consider a simple experiment in which we take some ice cubes from a refrigerator at a temperature of -20øC. We have a thermometer and a heat source, which operates at a constant rate. We will read the temperature at regular intervals of time, plot the temperature as a function of time on a graph, and we will draw a curve through the data points. The curve in Figure 10.3 is a typical curve for an experiment of this kind. !


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The temperature of the ice will increase steadily from -20øC to 0øC (portion ab of the curve). When the temperature of 0øC is reached there will be no increase in temperature until all ice is melted (bc). We now have water, and the temperature will again increase steadily (but at different rate than for ice - why?) until the temperature reaches the boiling point of water (cd). The temperature will remain constant at 100øC until all of the water is converted into steam (de). We must have a closed vessel to finish the transition to 120øC. The temperature will again increase steadily as heat is added (ef). The above curve is reversible; that is, if we remove heat from an equivalent system at the same rate as we heated the system in the original experiment, we will duplicate the above results in the reverse order. Although water was used in this example, the same type curve would be obtained for many other crystalline substances that have a definite melting point. Amorphous substances such as glass and butter, which soften gradually as the temperature is raised, would not show the constant temperature melting point. A body has a thermal inertia for a change in its state of matter. Heat energy is required to change an object from one state of matter to another. In other words, heat energy must be added to an object to change it from the solid state to the liquid state, even though the temperature remains constant at the melting point temperature. If there is a change from the liquid state back to the solid state at the melting point, heat energy is liberated. For example, you must add heat energy to convert the ice at 0øC to water at 0øC. And the reverse process occurs when water at 0øC is frozen to ice at 0øC; heat is liberated. This process serves as a source of heat energy. Can you think of an application? The amount of heat required to change a unit mass of material from the solid state to the liquid state at the melting point is called the latent heat of fusion Lf. For ice this is 334 joules per gram (J/g) or 79.7 cal/g. Similarly, for the change from the liquid state to vapor state or vice versa at the boiling point, there is a change of heat energy. In going from the liquid state to the vapor state, heat energy is absorbed, and in the reverse direction, therefore, heat energy is liberated. The process of changing a substance to the vapor state is called vaporization. The term vaporization is a general one and covers the process of evaporation, boiling, and sublimation. Evaporation is the process by which the change from a liquid to a gas occurs at the liquid-gas interface. Evaporation lowers the internal energy of the liquid, making evaporation a cooling process. This process is important in maintaining a constant body temperature. If one exercises vigorously, one perspires. The skin is cooled by evaporation. A fever can be reduced by bathing the skin in alcohol. The alcohol evaporates and produces a cooling effect. Boiling is the change from liquid to gas that may take place anywhere in the entire liquid. Bubbles of vapor are formed throughout the liquid and increase in size as they rise to the surface. This is a process you have surely observed. Sublimation is the process by which a solid changes directly to the vapor state without going through the liquid state. This process also occurs at the surface. Some examples of sublimation at ordinary room temperatures are the change of iodine crystals, moth balls (naphthalene), and dry ice (solid carbon dioxide) to their respective vapors. Under the proper conditions ice will sublime, as it sometimes does in a freezer. The heat required to convert one unit of mass of a material from the liquid state to the vapor state at the boiling point temperature is called the latent heat of vaporization, Lv.


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The conversion of 1g of water at the boiling point of 100øC to steam at the same temperature requires 2260 J, or 540 cal. The freezing point of a liquid is generally lowered by dissolving some other substance in it. Some examples are antifreeze in water for automobile radiators, salt dissolved in water, sugar dissolved in water. The boiling point is also affected by dissolved substances. It may be either increased or decreased depending on the substance. As an example, a salt solution has a higher boiling point than water and a water-alcohol solution has a lower boiling point than water. Both freezing and boiling points are affected by the external pressure. In general if the volume increases during the change of state, then an increase in the external pressure will raise the temperature of the change of state. EXAMPLE Assume you have a 100-g aluminum cup containing 200 g of water and 20.0 g of ice at equilibrium temperature. Dry steam is passed into the mixture and the final temperature is found to be 50.0øC. How much steam is added, and how much water is in the cup at 50.0øC? The cup and its contents gain an amount of energy calculated as follows: heat energy gained = heat energy needed to melt ice + heat energy needed to raise

temperature of water + heat energy needed to raise temperature of cup = 20 x 80 + (mw + mI)x 1(50 - 0) + mc x cc x (50 - 0) = 20 x 80 + (200 + 20) (50 - 0) + 100 (.22)(50 - 0) = 1,600 + 11,000 + 1,100 = 13,700 cal This must equal the total heat loss in the system.

heat energy lost = heat energy given up on condensing steam + heat energy given up in

cooling water = ms x Lv + ms x 1.0(100 -50) = ms(540 +50) = 590ms

energy lost by steam and hot water = energy gained by cup, ice, and cold water 590ms = 13,700 cal; ms = 23.2 g total water in cup at 50øC = mi + m w +ms = 20 + 200 + 23.2 = 243 g

10.8 Heat of Combustion What does it mean for you to speak of a daily diet of 1000 Calories? Let us analyze the physics of the situation. First, the Calorie, which is used in diet discussions, is the kilocalorie or the "large calorie." It is equivalent to 1000 of the gram calories we have been using. In speaking of Calories in food, we are talking about the heat energy of the food in terms of complete oxidation of the food. This involves the concept of heat of combustion. The heat of combustion of a material is the amount of energy produced by complete oxidation per unit measure of the material. This may be expressed in calories per gram, Calorie per liter, or any other units of energy per quantity. When the complete oxidation of the food you eat in one day release 1000 kcal of heat energy, you


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have a daily diet of 1000 Calories. In Table 10.4 are given the approximate values for the heat of combustion of some common foods

On the basis of the principle of energy conservation, in order to lose weight you must consume less food energy per day than your daily energy use. Also in order to gain weight, you must consume more food energy per day than you use daily. From energy considerations only, determine how much daily intake of some common foods you would need for a daily diet of 1000 kcal. The concept of heat of combustion applies to other fuels as well as to foods. Gasoline has a heat of combustion of about 1150 kcal/100 gm. From the standpoint of economics, we are all interested in the maximum number of calories per dollar. Compare the economics of using butter vs. gasoline as a source of energy.

10.9 Gas Laws As was noted in the discussion of temperature transducers in Section 10.2 gases can be used to build thermometers. In this section we will describe the experiments that can be done to find the quantitative relationships between the pressure, volume, and temperature of a confined gas. BOYLE'S LAW In the seventeenth century Robert Boyle performed a number of key experiments on gases. You can replicate his results in a simple laboratory experiment. Take a vertical cylinder with a movable piston that fits snugly into it. Trap some air in the cylinder see Figure 10.4.


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As the weight applied to the piston is increased, the distance of the piston from the bottom of the cylinder will decrease. In other words, as you increase the force on the piston, the volume of the enclosed air decreases. If you double the total force on the piston, the volume of the confined air will become one-half its original volume. What happens when you change the size of the container? Suppose you double the diameter of your cylinder and piston. What increase in force will be necessary to reduce the volume by a factor of two? Boyle performed such experiments and found that the ratio of the total force applied to the piston divided by the area of the piston was the essential physical variable. This ratio is the pressure on the piston (see also Chapter 8). pressure = force/area (8.2) If the diameter of the piston is doubled, the area is increased by a factor of four since the area of the piston is proportional to the square of the diameter. Hence, for a piston of twice the diameter, the total force must be increased by a factor of four to keep the pressure on the confined gas constant see Figure 10.4b. Robert Boyle's results are known as Boyle's law, which can be stated as follows: The volume of a confined gas whose temperature is held constant will vary in inverse proportion to the pressure on the gas. V ∝ 1/P where V is the volume of the confined gas (in m3), andP is the pressure (N/m2). See Figure 10.5.

This inverse proportion is equivalent to stating that the product of the pressure times the volume of a confined gas at constant temperature is a constant, PV = constant (10.10) EXAMPLE A constant mass of gas is confined in a cylinder by a movable piston. The original volume of the gas is 500 cm3 and the pressure is 1.02 x 105 N/m2. The gas is compressed to 100 cm3 at constant temperature. What is the new pressure? P1V1 =P2V2


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(1.02 x 105 N/m2) x 500 x 10 -6 m3 = P2(100 x 10-6 m3) P2 = 5.10 x 105 N/m2. CHARLES' LAW Almost one hundred years after the work of Robert Boyle, Jacques Charles and Joseph Gay-Lussac were independently studying the properties of gases at constant pressure. They found that the volume of a confined gas changed when the temperature changed. You can also duplicate their experiments with simple laboratory apparatus. Take a small- diameter glass tube closed at one end. Enclose a quantity of gas in the tube by putting a drop of mercury in the tube. Measure the length of the air column in the tube at room temperature. Then place the tube in an ice bath. You will notice that the air column shortens. Next, place the tube in a steam bath. You now will find that the air column lengthens. (see Figure 10.6.)

The results of this experiment for the volume of a confined gas at constant pressure show that the volume is a linear function of temperature as shown in Figure 10.7.

The curve of volume against temperature has a slope of Vo divided by 273.16 where Vo is the volume of the confined gas at 0øC. Thus, the change in volume of the gas at constant pressure is directly proportional to the change in temperature. This can be expressed using the idea of a coefficient of volumetric expansion we discussed earlier, ΔV = Vt -Vo = Voβt (10.5)


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or Vt = Vo (1 + βt) (10.11) where β is the coefficient of volumetric expansion of the gas 1/273.16, t is the temperature in degrees Celsius, and Vt is the volume at the temperature t. You may realize the implication of Equation 10.11: If the temperature becomes -273.16øC, the volume Vt becomes zero. The law of Charles and Gay-Lussac can be reformulated using the Kelvin (or absolute) temperature scale by replacing the temperature t(in øC) by its equivalent Kelvin reading T(øK)=t(øC) + 273.15 VT = Vo [1 +β (T - 273.16)] where β = 1/273.16. Thus the volume of a confined gas at constant pressure is directly proportional to its absolute temperature in degrees Kelvin VT = (Vo/273.16) T (10.12) This is equivalent to stating that V/T = constant, for constant P. This equation provides the basis for a gas thermometer with a linear scale, except for temperatures near 0øK. Such a thermometer can be constructed using mercury and a glass tube as described in the experiment above. EXAMPLE If 2 liters of a gas at 0øC at constant pressure are cooled to - 136.58øC, what is the volume of the gas? original temperature (øK) = 0øC + 273.15 = 273.16øK final temperature (øK) =-136.58 + 273.15 = 136.58øK So, 2 liters/273.15øK = Vfinal/136.58øK Vfinal = 1 liter CONSTANT VOLUME To keep a mass of gas at constant volume one needs a tight vessel made of material with negligible expansion as represented in Figure 10.8a. The application of heat causes the pressure and temperature to increase. The pressures are taken at a number of different temperatures. A pressure- temperature curve is then plotted. See Figure 10.8b. The equation of the curve is given by Pt = Po(1 +B't) Where Pt is pressure at temperature t, P0 is the pressure at 0øC and B' is the pressure coefficient, which has a numerical value of 1/273.15. The significance of this curve is that we can extend the curve to find the temperature at which Pt = 0. The temperature is 0øK, or absolute zero. So we can now say that absolute zero is the temperature at which the pressure of a confined gas is zero. It follows that


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the pressure of a confined gas (at constant volume) is directly proportional to its absolute temperature in degrees Kelvin, PT = Po/273.16 T (10.13)

EXAMPLE If the pressure on a constant volume of a gas is doubled, what is the change in the temperature of the gas? The answer is that the absolute temperature in degrees Kelvin will be doubled. The constant volume gas thermometer is an application of the change of pressure produced by heating or cooling a gas. Boyle's law and the law of Charles and Gay-Lussac can be combined into a single gas law. Since Boyle's law shows that the product of the pressure and volume of a gas is constant for a constant temperature and the law of Charles and Gay-Lussac shows that the ratio of the volume to the absolute temperature is a constant for constant pressure, these two laws can be combined to show that the product of the absolute pressure times the volume of the confined gas divided by the absolute temperature is a constant, PV/T = constant (10.14) This expression is equivalent to the following equation: PiVi /Ti =PfVf /Tf (10.15) where the subscripts i and f refer to the initial and to the final states of the system, and where P and T are the absolute pressure and the temperature, respectively, and where V is the volume of the confined gas. EXAMPLE A weather balloon carrying 3 m3 of helium gas at 1 atmosphere of pressure and 27øC is sent aloft to an elevation where the pressure is one- twelfth of an atmosphere and the temperature is a -73øC. What is the volume of the balloon at that elevation? What is the change in the radius of the balloon. Let us begin by rounding off the Kelvin temperature scale to T (øK) =t(øC) + 273. Ti = initial temperature = 27øC + 273 = 300øK Tf = final temperature = -73øC + 273 = 200øK Pi = 1 atmosphere and Pf = 1/12 atmosphere (1 atm)(Vi m3) / 300øK = (1/2 atm) Vf/200øK


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8 Vi m3 =Vf The final volume is eight times the initial volume, and since the volume increases as the cube of the radius (V = 3/4πr3 for a sphere), the radius will increase by 3√8 or by a factor of 2. SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the section number where you can find the related content material. Definitions Fill in the blanks with correct words.

1. The thermal energy of a system is characterized by its _________ which you can measure using a ________.

2. Many thermal transducers have a positive ________ or increase in length with increase in temperature.

3. Water is unusual in the fact that its ______ is negative from 0øC to 4øC. 4. The study of heat energy developed separately from the study of mechanical

energy; so the unit for thermal energy is called a ______ and is equivalent to the amount of energy required to ______ .

5. James Prescott Joule carried out a number of experiments which gave a ______ of 1 calorie = 4.186 joules.

6. The amount of energy required to raise the temperature of a system 1øC is called the ______ of the system.

7. The amount of energy required to raise the temperature of 1 kg of material 1øC is the ______ of the material.

8. If a gram of a substance is completely burned in oxygen and gives off 80 calories of energy, then we say its ______ is 80 cal/g.

9. When one gram of ice is melted to water, it absorbs heat energy from its environment in an amount equal to its ______.

10. When one gram of water vapor condenses on the outside of a glass of iced tea, the water has ______ heat energy to its environment in an amount equal to its ______.

Calorimetry

11. Given: average specific heat of ice = 0.55 cal/g-øC; average specific heat of steam = 0.48 cal/g-øC; heat of fusion of ice = 80 cal/g; heat of vaporization of water = 540 cal/g. Calculate the amount of heat needed to heat 20 g of ice from -25øC to steam at 126øC.


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Gas Laws 12. As the human diaphragm contracts, it moves down-ward and expands the

volume of the lungs. As a result, the air pressure in the lungs drops from 76.0 cm Hg to 75.8 cm Hg. In a deep breath your lungs will hold 3.5 liters of air at 37øC. What is the volume of cold fresh air at 7øC that is necessary to fill your lungs?

Answers

1. temperature, thermometer (Sections 10.6 and 10.2) 2. linear coefficient of expansion (Section 10.3) 3. volumetric expansion (Section 10.4) 4. calorie, raise the temperature of 1 g of water 1øC (Section 10.6) 5. mechanical equivalent of heat (Section 10.6) 6. heat capacity (Section 10.6) 7. specific heat (Section 10.6) 8. heat of combustion (Section 10.8) 9. latent heat of fusion (Section 10.7) 10. liberated latent heat of vaporization (Section 10.7) 11. 14,900 calories (Section 10.7) 12. 3.15 liters (Section 10.9)

ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single concept problems. Equations

øC = 5/9 (øF - 32) LT = L0 (1 +αT) (10.4) ΔV = VoβΔT (10.5) ΔQ = mc ΔT (10.9) PiVi /Ti =PfVf /Tf (10.15)

Problems 1. The normal temperature of the human body is said to be 98.6øF. What temperature is

this on the Celsius scale? 2. If a doctor tells you that you have 6 degrees of fever, what is the equivalent number

of Celsius degrees? 3. What is the change in length of a 30-m steel tape in a change of temperature from 0øC

to 40øC? 4. What is the percentage change in volume of the mercury in a thermometer for a

temperature change from 0øC to 100øC? 5. If you do 1000 J of work to produce heat, how many calories of heat do you produce?

6. How much heat will be required to warm 200 g of ice from - 10øC to 0øC.


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7. What is the calorie value of a 300-g apple? 8. Suppose you want to consume only milk to get a 1000 Calories/day diet. How much

milk would you have to drink per day? 9. An automobile tire, initially with a volume of 0.36 m3, a pressure of 3 atmospheres,

and a temperature of 7øC, will increase in volume by 5 percent and in pressure by 15 percent at highway speeds. What is the temperature of the high-speed tire?

Answers

1. 37øC 2. 3.3øC 3. 1.32 cm 4. 1.8 percent 5. 240 cal 6. 1000 cal 7. 174Kcal 8. 1540 g (slightly more than 1.5 liters) 9. 65.1øC

EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. When appropriate the numerical answer is given at the end of each exercise. Section 10.2 1. A human rectal temperature is 102øF. What is this reading on the Celsius scale? A

patient is said to have 4 Fahrenheit degrees of fever. What is the equivalent in Celsius degrees? [38.9øC, 2.2øC]

2. At what temperature is the reading on the Celsius and Fahrenheit scales the same? [-40ø]

Section 10.3 3. A distance is measured by a steel tape graduated in millimeters. The tape is correct at

15øC, but a measurement is made on a hot day of 35øC. The length read on the tape was 300.250 m. What was the true distance? [300.316 m]

Section 10.4 4. Mercury is to be stored in iron spheres which have a capacity of one liter at 10øC. The

temperature of the storage room may reach a high of 40øC during the summer. How much mercury can be put in the iron flask at 10øC if no mercury is to overflow at the peak temperature? α = 12 x 10-60C. [1 - 0.0044 liters]

5. The density of mercury is 13.6 g/cm3 at 0øC. What is its density at 200øC? [13.1 g/cm3] 6. A vertical Pyrex glass tube has an internal area of 10 mm2 at 0øC, and it is filled to a

height of 80 cm with Hg at this temperature. How long would the tube have to be so


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that no mercury would overflow if the system is heated to 100øC? Could this system be used as a thermometer? If so, how high would the mercury column be for a temperature of 50øC? Is it a linear system? [81.4 cm, 80.7 cm, yes]

Section 10.6 7. The British thermal unit (BTU) is a unit of heat energy defined as the amount of heat

required to raise the temperature of one pound of water one degree Fahrenheit. How many calories are equivalent to one BTU? [252]

8. The following data are obtained in an experiment to measure the specific heat of a specimen. An aluminum container of mass 100 g contains 180 g of water at 10øC. A 200-g specimen is taken from a furnace at 200øC and put into the container. The final temperature of the mixture is 30øC. What is the specific heat of the specimen? [0.12 cal/g-øC]

Section 10.7 9. If you know that the specific heat of a specimen is 0.98 J/g-øC, what is it in cal/g-øC?

What is the heat of fusion of ice in joules/g? What is the heat of vaporization of water in J/g? [0.234 cal/g-øC. 335 J/g, 2260 J/g]

10. A 200-g copper calorimeter cup contains 400 g of water at 45øC. In the cup 200 g of ice are melted, giving a final temperature of 5øC. Compute the heat of fusion of the ice. [78.7 cal.g]

11. Calculate the heat required to convert 10 g of ice, originally at temperature of -10øC, to steam at a temperature of 120øC. Use 0.5 cal/g-øC as the specific heat of the steam. [7355 cal]

Section 10.8 12. A given individual is said to eat 1500 kcal per day. Considering energy only, design

a day's diet made up of six of the foods listed in Table 10.4. There are many possible answers depending upon food choices.

13. A continuous-flow calorimeter is used to measure the heat of combustion of a gaseous fuel. For a steady condition water flows at 5000 g/min, and the temperature is raised 10øC. Gas flows at the rate of 4 liter/min. What is the heat of combustion of the gas? [12,500 cal/liter]

14. A tank contains 2 m3 of nitrogen at an absolute pressure of 150 cm Hg and a temperature of 7øC. What will the pressure be, if the volume is increased to 10 m3 and the temperature is raised 227øC? [53.6 cm Hg]

PROBLEMS Each problem may involve more than one physical concept. The numerical answer is given in brackets at the end of the problem. 15. Eight grams of steam at 100øC are delivered by a rubber hose into a 100-g aluminum

cup containing 80 g of ice and 200 g of water at 0øC. What is the final state and


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temperature of the contents of the cup? [All at 0øC, 64 gm of ice melted, 16 gm still ice.]

16. Steam burns are often severe. Compare the energy liberated if one gram of steam at 100øC is transformed to water at 37øC, and one gram of water is cooled from 100øC to 37øC. If your body absorbed all the heat released, how many times greater is the heat absorbed from the steam than from the water? [603 cal compared to 63 cal]

17. One hundred square meters of a lake is covered with ice at 0øC. If ice absorbs 0.25 cal/cm2-min, how much ice will melt in one hour? [188 x 103 g/hr]

18. A bubble of air rises from the bottom of the lake where the pressure is 2.8 atmospheres to the surface where the pressure is 1 atmosphere. The temperature at the surface is 27øC and 7øC at the bottom. Compare the size of the bubble at the surface with its size at the bottom of the lake? [Vt/Vb = 3]

19. Suppose your coffee maker is rated at 300 watts. How long does it take to boil water for one cup of instant coffee? (Assume one cup contains 250 ml of water at 15øC.) [nearly 297 sec]

20. If the average loss of heat due to evaporation of perspiration is about 10 kcal/hr, find the rate of water loss in g/hr. Assume the heat of evaporation 575 cal/g. [17.4 g/hr].

21. Some insects that dive beneath the water surface carry an air bubble with them. Compare the volume of the bubble at a depth of 1 m to the surface volume if the temperature at the surface is 27øC, and the temperature at 1 m is 22øC. [0.90 = V1m/Vs]

22. If the human body loses 4 kg of water at 300øK each day due to evaporation, find the body heat loss this represents. [10.1 x 106 J]

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Chapter 10 Temperature And Heat - Doane...

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