knight ch 12 Thermal Properties of Matter ocr.pdf

404 CHAPTER 12 Thermal Properties of Matter r=IGURE 12. 7 A tire gauge measures the difference between the tire's pressure and atmospheric pressure. This fixed disk holds the end of the spring. The scale can slide The difference·········· between p,;,e and Pmmo, pushes the piston forward agai...

404 CHAPTER 12 Thermal Properties of Matter r=IGURE 12. 7 A tire gauge measures the difference between the tire's pressure and atmospheric pressure. This fixed disk holds the end of the spring. The scale can slide The difference·········· between p,;,e and Pmmo, pushes the piston forward against the spring. moving the scale outward. tire gauge, the difference between an absolute pressure and atmospheric pressure, so we can write Pg = P - Patmos P =Pg+ Patmos If you are at sea level, you can assume that you '11 need to use the local pressure. EXAMPLE 12.4 Patmos = (12.14) 1.0 atm; at other elevations, Finding the force due to a pressure difference Patients uffering from decompre sion icknes may be treated in a hyperbaric oxygen chamber filled with oxygen at greater than atmospheric pre ure. A cylindrical cham­ ber with flat end plat.e of ctiameter 0.75 mis filled with oxygen to a gauge pre sure of 27 kPa. What is the resulting force on the end plate of the cylinder? STRATEGIZE The net force F0e, = Al:!..p on the end plate depends on 6.p, the difference between the pres ure in ide and the pre sure outside the chamber. Thi is the gauge pre ure, 27 kPa. PREPARE SOLVE The end plates are circular, with area A = ,,,-,- 2 = r,(0.375 m) 2 = 0.442 m 2 . The pres ure difference re ults in a net force Fnet = A 6.p = (0.442 m2 )(27,000 Pa)= 12 kN The area of the end plate i large so we expect a large force, which i exactly what we get-the force is equivalent to the weight of about 15 people. If you've seen chambers like this, you know that the end plate is fastened in place with stouc bolts. It i remarkable to think that chi large force results from the colli ion of tiny air mol­ ecule with the plate! ASSESS The Ideal-Gas Law We can use the fact that the pressure in a gas is due to the collisions of particles with the walls to make some qualitative predictions. FIGURE 12.8 presents a few such predictions. FIGURE 12.8 Relating gas pressure to other variables. .. Increasing the temperalllre of the gas means the pmticles move at higher speeds. They hit the walls more often and with more force, so there is more pressure. Decreasing the volume of the container means more frequent collisions with the walls of the container, and thus more pressure. Increasing the number of particles i ;1 the container means more frequent collisions with the walls of the container. and thus more pressure. Based on the reasoning in Figure 12.8, we expect the following proportionalities: ■ Pressure should be proportional to the temperature of the gas: p et:. T. ■ Pressure should be inversely proportional to the volume of the container: p et:. 1/V. ■ Pressure should be proportional to the number of gas particles: p er. N. In fact, careful experiments back up each of these predictions, leading to a single equation that expresses these proportionalities: Pumping up a bicycle tire adds more particles to the fixed volume of the tire, increasing the pressure. NT p=C­ V The proportionality constant C turns out to be none other than Boltzmann's constant k 8 , which allows us to write 12.2 pV=Nk 8 The Atomic Model of an Ideal Gas 405 (12.15) T Ideal-gas law, version I Equation 12.15 is known as the ideal-gas law. Equation 12.15 is written in terms of the number N of particles in the gas, whereas the ideal-gas law is stated in chemistry in terms of the number n of moles. But the change is straightforward to make. The number of particles is N = nN;1, so we can rewrite Equation 12.15 as (12.16) Ideal-gas law, version 2 In this version of the equation, the proportionality constant-known constant-is R = NAk 8 = 8.31 J/rnol as the gas •K The units may seem unusual, but the product of Pa and 1113, the units of pV, is equivalent to J. Let's review the meanings and the units of the vaiious quantities in the ideal-gas law: Number of moles in the sample or container of gas Absolute pressure (Pa)··········· \ ._..•··········· Gas constant, 8.31 J/mol • K .. "pV= nRT,.. Volume of the sample or·········_..container of gas (m 3) EXAMPLE 12.5 ·············Temperature in kelvin (K) Finding the volume of a mole of a gas What volume is occupied by I mole of an ideal gas at a pressure of 1.00 arm and a temperature of 0°C? We recall from earlier in the chapter that 1.00 m3 = I 000 L, so we can write STRATEGfZE We are given the pressure. the temperature. and the number of moles. We will use version 2 of the ideal-gas law to find the volume. \I= 22.4 L The first step in ideal-gas law calculation is to convert all quantities to SI units: PREPARE At this temperature and pressure. we find that the volume of I mole of a gas is 22.4 L. a result you might recall from chemistry. When we do calculations using gases, it will be useful to keep this volume in mind to see if our answers make physical sense. ASSESS p = 1.00 atm = IOI .3 X !03 Pa T = 0 + 273 = 273 K We rearrange version 2 of the ideal-gas law equation to compute SOLVE nRT V = -- P ( 1.00 mo1)(8.3 I J/mol • K)(273 K) = -'----_:_.:.._ ____ 101.3Xl0 3 --'-'---- = 0.0224 m3 Pa STOP TO THINK 12.2 A sample of ideal gas is in a sealed container. The temperature of the gas and the volume of the container ai·e both increased. What other properties of the gas necessarily change? (More than one answer may be correct.) A. The rms speed of the gas atoms C. The pressure of the gas B. The thermal energy of the gas D. The number of molecules of gas 406 c HAP TE A 12 Thermal Properties of Matter 12.3 Ideal-Gas Processes Suppose you measure the pressure in the tires on your car on a cold morning. How much will the tire pressure increase after the tires warm up in the sun and the temperature of the air in the tires has increased? We will solve this problem later in the chapter, but, for now, note these properties of this process: FIGURE 12.9 The state of the gas and ideal-gas processes can be shown on a ■ pV diagram. ■ (a) Each state of an ideal gas is represented as a point on a p V diagram. ~O(kPa\ = I mol ···--•••........ ,; Processes with these properties are called ideal-gas processes. For gases in sealed containers, the number of moles (and the number of molecules) does not change. In that case, the ideal-gas law can be written as T2= 3600 K T1 ■ The quantity of gas is fixed. No air is added to or removed from the tire. There is a well-defined initial state. The initial values of pressure, volume, and temperature will be designated Pi, Vi, and T;. There is a well-defined final state in which the pressure, volume, and temperature have values /Jr, Vr, and Tr. • = 900 K pV 5 - T = nR = constant The values of the variables in the initial and final states are then related by 3 0 +------------- V (m ) 0 2 3 (b) A process that changes the gas from one state to another is represented /J (k.Pa) by a trajectory on a pV diagram. 2 Pr Vr P; Vi Tr T; (12.17) Initial and final states for an ideal gas in a sealed container 10 This before-and-after relationship between the two states, reminiscent of a conservation law, will be valuable for solving many problems. EmDI ► Because pressure and volume appear on both sides of the equation, 5 \I (m 3) 0 +---~---------.--0 2 3 FIGURE 12.10 A constant-volume (a) process. As the temperature increases, so does the pressure ...... '::, (b) p f /Jr ... •····"A constant-volume •····· process appears on a p V diagram i as a vertical line. f ~-------v V pressure and volume can be in any units; we don't necessarily need to convert to SI units. However, temperature must be in K. Unit-conversion factors for pressure and volume are multipl.icative factors, and the same factor on both sides of the equation cancels. But the conversion from K to °C is an additive factor, and additive factors in the denominator don't cancel. ◄ pVDiagrams It's useful to represent ideal-gas processes on a graph called a pV diagram. The important idea behind a pV diagram is that each point on the graph represents a single, unique state of the gas. This may seem surprising because a point on a graph specifies only the values of pressure and volume. But knowing p and V, and assuming that n is known for a sealed container, we can find the temperature from the ideal-gas law. Thus each point on a p V diagram actually represents a triplet of values (p, V, T) specifying the state of the gas. For example, FIGURE 12.sa is a p V diagram showing three states of a system consisting of 1 mol of gas. The values of p and V can be read from the axes, then the temperature at that point calculated from the ideal-gas law. An ideal-gas process-a process that changes the state of the gas by, for example, heating it or compressing it-can be represented as a "trajectory" in the pV diagram. FIGURE 12.sb shows one possible process by which the gas of Figure 12.9a is changed from state 1 to state 3. Constant-Volume Processes Suppose you have a gas in the closed, rigid container shown in FIGURE 12.10a. Warming the gas will raise its pressure without changing its volume. This is an example of a constant-volume process. Vr = Vi for a constant-volume process. Because the value of V doesn't change, this process is shown as the vertical line i - f on the pV diagram of FIGURE 12.,ob. A constant-volume process appears on a p V diagram as a vertical line. 12.3 EXAMPLE 12.6 the ratio of the two volumes. which is equal volume process: STRATEGIZEA tire is (to a good approximation) a sealed container with constant volume. so this is a con tant-volume process. pr= 44.7 p Tr = 30°C + 273 = 303 K SOLVE We can use Equation 12.17 10 solve for the final pressure. In thi equation. we divide both ides by Vr.and then cancel Processes Many gas processes take place at a constant, unchanging pressure. A constant-pressure process is also called an isobaric process. For a constant-pressure process Pr= Pi· One way to produce a constant-pressure process is shown in FIGURE 12.11a, where a gas is sealed in a cylinder by a Jjghtweight, tight-fitting cap-a piston-that is free to slide up and down. Tn fact, the piston will slide up or down, compressing or expanding the gas inside, until it reaches the position at which Pgas = /Jcxt· That's the equilibrium position for the piston, the position at which the upward force fgas = /JgasA, where A is the area of the face of the piston, exactly balances the downward force F0 , 1 = /Jcx,A due to the external pressure Pext· The gas pressure in thjs situation is thus equal to the external pressure. As long as the external pressure doesn't change, neither can the gas pressure inside the cylinder. (a) process. (b) In equilibrium, pressure must be the same on both ~ Pex1 /Jp., p t- Initial A constant-pressure process appears on a p V diagram as a horizontal line. .;. p Fexl /J\:.t ···........ . F,._, Final V; ~ V1 Movable piston IX 303 K K = 49.6 psi 273 ASSESS The temperature has changed by 30 K, which is a bit more than 10% of the initial temperature. so we expect a large change in pressure. Our re ult seems reasonable. and it has practical implications: If you check the pressure in your tires when they are at a panicular temperature, don't expect the pressure to be the same when conditions change! o·c+ 213 = 213 K FIGURE 12.11 A constant-pressure . (pg)r = Pr- 1.00 atm = 49.6 psi - 14.7 psi= 34.9 p i Temperatures must be in kelvin, o we conven: Constant-Pressure I for this constant- This is an ab olute pressure. but the problem asks for the measured pressure in the tire-a gauge pressure. Convening to gauge pressure gives 1.00 atm = 30.0 psi+ 1-U psi= 44.7 psi T; = 10 The units for Pr will be the same as tho e for p;, so we can keep the initial pressure in psi. The pressure at the higher temperature is PREPAREWe weren't told anything about elevation, so we assume that the atmospheric pressure is 1.0 atm. When we solve this problem. we don't need to conven units, but we do need to be sure that we are using the correct form of the pressure. The measured tire pressure is a gauge pressure, but the ideal-gas law requires an absolute pre sure. We can use Equation 12.14 10 conven to absolute pre ure. The initial pressure i = (pg);+ 407 Computing tire pressure on a hot day The pressure in a car tire measures 30.0 psi on a cool morning when the air temperature is 0°C. After the day warms up and bright sun shines on the black tire, the temperature of the air inside the tire reaches 30°C. What is the tire pressure at this temperature? p, Ideal-Gas Processes L---,----~-- V • ••••Because the external pressure doesn't change, the gas pressure remains (,~ constant as the gas expands. Suppose we heat the gas in the cylinder. The gas pressure doesn't change because the pressure is controlled by the unchanging external pressure, not by the temperature. But as the temperature rises, the faster-moving atoms cause the gas to expand, pushing the piston outward. Because the pressure is always the same, this is a constant-pressure process, with a trajectory as shown in FIGURE 12.11b. A constantpressure process appears on a pV diagram as a horizontal line. 408 CHAPTER EXAMPLE 12.7 12 Thermal Properties of Matter Finding the final volume of a compressed gas A gas in a cylinder with a movable piston occupies 50.0 cm3 at 50°C. The gas is cooled at constant pressure until the temperature is 10°C. What is the final volume? SOLVE STRATEGIZE The pressure of the gas doesn't change, so this is a constant-pressure process, with p/JJr = l. Our answer makes sense. The gas cools, so we expect a smaller volume. The cooling is-in absolute temperature termsrelatively modest, so we expect a modest change in volume. In this example and in Example 12.6, we have not converted pressure and volume units because these multipUcative factors cancel. But we did convert temperature to kelvin because this additive factor does not cancel. The temperatures must be in kelvin, so we convert: PREPARE Ti = 50°C + 273 = 323 K Tr= 10°C +273 = 283 K We can use Equation 12.17 to solve for Vr: Pi Tr Vr = Vi - 0 ~ A constant-temperature process. (a) Push - ,_ -~ V; Vr , Final Initial I ConstantT ConstantT (b) JJ .. /Jr _..-A constanHemperature process appears on a p V diagram as a hyperbola. 283 K 323K X 1 X -- = 43.8 cm 3 ASSESS Constant-Temperature FIGURE 12.12 = 50.0 cm3 Processes A constant-temperature process is also called an isothermal process. For a constant-temperature process Tr= Ti. One possible constant-temperature process is illustrated in FIGURE 12.12a. A piston is being pushed clown to compress a gas, but the gas cylinder is submerged in a large container of liquid that is held at a constant temperature. If the piston is pushed slowly, then heat-energy transfer through the walls of the cylinder will keep the gas at the same temperature as the surrounding liquid. This is an isothermal compression. The reverse process, with the piston slowly pulled out, is an isothermal expansion. Representing a constant-temperature process on the pV diagram is a little more complicated than the two preceding processes because both p and V change. As long as T remains fixed, we have the relationship p.116 nRT constant (12.18) P --- i V V ~\I INVERSE .: /Ji V V; Vr (c) JJ Different temperatures correspond to different isotherms. Because there is an inverse relationship between p and V, the graph of a constant-temperature process is a hyperbola. Figure l 2. l 2a shows an isothermal compression, which is represented graphically as the hyperbola i ~fin FIGURE 12.12b. An isothermal expansion would move in the opposite direction along the hyperbola. The graph of an isothermal process is known as an isotherm. The location of the hyperbola depends on the value of T. If we use a higher constant temperature for the process in Figure 12.12a, the isotherm will move farther from the origin of the pV diagram. FIGURE 12.12c shows three isotherms for this process at tlu·ee different temperatures. A gas undergoing a constant-temperature process will move along the isotherm for the appropriate temperature. EXAMPLE 12.8 .____________ v Why does the bag expand? Residents of Colorado, where it's easy to drive rapidly to different elevations, are familiar with this scenario: You buy a sealed bag of chips at a low elevation and then drive your car up into the mountains. As you climb, the atmospheric pressure drops, the air inside the bag expands, and the bag gets "puffy." Explain what is happening. The chip bag is sealed-no gas gets in or out-so we can treat this as an ideal-gas process. Folks in the car will use the heater or the air conditioner to keep the temperature in the car steady, so this is a constant-temperature process. Chip bags aren't very sturdy, so the pressure inside the bag is the same as the pressure outside the bag. As the pressure outside the bag drops, so does the pressure inside the bag. The volume of air in the bag is inversely proportional to the pressure, so as the pressure drops, the volume of the air increases and the bag expands. REASON The result makes sense. A change in pressure also occurs on an airplane, so you may have seen something similar during a fUght, or you may have had bottles with caps that pop off and contents pushed out as the gas inside expands. ASSESS 12.3 Ideal-Gas Processes EXAMPLE 12.9 409 Finding the volume of air in the lungs (iJ A snorkeler takes a deep breath at the surface, filling his lungs with 4.0 L of air. He then descends to a depth of 5.0 m, where the pressure is 0.50 atm hjgber than at the surface. At thjs depth, what is the volume of air in the snorkeler's lungs? Most of the snorkeler's air is nitrogen, whlch his body doesn't use. Oxygen is used by the body, but for every oxygen molecule taken from tbe lungs, a molecule of carbon dioxide is returned. The ideal-gas law doesn't care what the molecules are, just the total number. The total number of gas molecules stays approximately constant, and we can treat this as an ideal-gas process. The air stays at body temperature, so this is a constanttemperature process with Tr= T;, so Tr/Ti = 1. As the snorkeler descends, the pressure inside his lungs rises to match the pressure of the surrounding water because the body can't sustain large pressure differences between inside and out. STRATEGIZE Thermodynamics We need absolute pressures for our calculation. At the surface, the pressure of tbe air inside the snorkeler's lungs is 1.0 atm-it's the pressure at sea level. The additional pressure at the noted depth is 0.50 atm, so the absolute pressure at thjs depth is 1.5 atm. PREPARE SOLVE The ideal-gas law for a sealed container (the lungs) gives Vr= vi Pi Tr Pr Ti J.5 atm 1 =2.7 L Notice that we didn't need to convert pressure to SI uruts. As long as the units are the same in the numerator and the denorrunator, they cancel. ASSESS The air has a smaller volume at the higher pressure, as we would expect. The air inside your lungs does compresssignificantly 1-when you dive below the surface. of Ideal-Gas Processes Chapter 11 introduced the first law of thermodynamics, and we saw that heat and work are just two different ways to add energy to a system. We've been considering the changes when we heat gases, but now we want to consider the other form of energy transfer-work. When gases expand, they can do work by pushing against a piston. This is how the engine under the hood of your car works: When the spark plug foes in a cylinder in the engine, it ignites the gaseous fuel-air mixture inside. The hot gas expands, pushing the piston out and, through various mechanical linkages, turning the wheels of your car. Energy is transferred out of the gas as work; we say that the gas does work on the piston. Similarly, the gas in Figure 12.11 a does work as it pushes on and moves the piston. You learned in ~SECTION 10.2 that the work done by a constant force Fin pushing an object a distanced is W = Fcl. Let's apply this idea to a gas. FIGURE 12.13a shows a gas cylinder sealed at one end by a movabl~piston. Force Fgas is due to the gas pressure and has magnitude Fgas = pA. Force Fext, perhaps a force applied by a piston rod, is equal in magnitude and opposite in direction to Fgas. The gas pressure would blow the piston out if the external force weren't there! Suppose the gas expands at constant pressure, pushing the piston outward from x; to xr, a distanced= xr - x;, as shown in FIGURE 12.13b. As it does, the force due to the gas pressure does work Wgas 1.0 atrn --=4.0LX--X The expanding gas does work on the piston. FIGURE 12.13 The gas pu~hes on the piston with force Fgas· To keep the piston in place, (a) an external force must be equal an/ opposite to F'gns· Pressure p \ \ - " Fg.,---- = Fgasd = (pA)(xr - x;) = p(xrA - X;A) But x;A is the cylinder's initial volume V; (recall that the volume of a cylinder is the length times the area of the base) and xrA is the final volume Vr. Thus the work done is (b) (12.19) As the piston moves a distance d, the ga__ ..?loes work F'ga,d, Piston areaA Work done by a gas in a constant-pressure process where LlV is the change in volume. Equation 12.19 has a particularly simple interpretation on a pV diagram. As FIGURE 12.14a on the next page shows, p fl V is the "area under the p V graph" between V; and Vr, Although we've shown this result for only a constant-pressure process, it turns out to be true for all ideal-gas processes. That is, as FIGURE 12.14b shows, W gas = area under the p V graph between V; and Vr 0 / X; Xf The gas volume changes by ~V = dA. 410 c HAP TE R 12 Thermal Properties of Matter FIGURE 12.14 Calculating the work done There are a few things to clarify: in an ideal-gas process. ■ (a) In order for the gas to do work, its volume must change. No work is done in a constant-volume process. ■ The simple relationship of Equation 12.19 applies only to constant-pressure processes. For any other ideal-gas process, you must use the geometry of the p V diagram to calculate the area under the graph. ■ To calculate work, pressure must be in Pa and volume in m3 . The product of Pa (which is N/m 2) and m3 is N • m. But l N ·mis J J-the unit of work and energy . ■ Wgas is positive if the gas expands ( I:::. V > 0). The gas does work by pushing against the piston. In this case, the work done is energy transferred out of the system, and the energy of the gas decrea~s. Wgas is negative if the piston compresses the gas ( I:::. V < 0) because the force Fgas is opposite the displacement of the piston. Energy is transferred into the system as work, and the energy of the gas increases. We often say "work is done on the gas," but this just means that Wgas is negative. For a constant-pressureprocess, the area under the graph p!).V is the work done by the gas as it e7ands from V~to Vr, p p ... • I .w V c_-1------~ V, Vr (b) Tngeneral,the work done by a /J i In the first law of thermodynamics, l:::.E 111= Q + W, W is the work done by the environment-that is, by force F'ext acting on the system. But F'ext and F'gas are equal and opposite forces, as we noted previously, so the work done by the environment is the negative of the work done by the gas: W = -Wgas· Consequently, the first law of thermodynamics can be written as gas is th~t=~du~~-erthe graph f lVg.,, area (12.20) V c_-+-----l-- V; The thermal energy of an ideal gas depends only on its temperature as £ 1h = ~ Nk8 T. Comparing Equations l2.15 and 12.16 shows that Nk8 is equal to nR, so we can write the change in thermal energy of an ideal gas as 3 3 !:::.E = -nR l:::.T 1h = -NkB l:::.T 2 (12.2]) 2 Finding the heat for an ideal-gas process EXAMPLE 12.10 A cylinder with a movable piston contains 0.016 mol of helium. A researcher expands the gas via the process illustrated in FIGURE12.15. To achieve this, does she need to heat the gas? If so, bow much heat energy must be added or removed? FIGURE 12.15 pVdiagram PREPARE The graph tells us the pressure and the volume, so we can use the ideal-gas law to compute the temperature at the initial and final poinrs. To do this we need the volumes in SI units. Reading the initial and final volumes on the graph and convening. we find for Example 12.10. l m3 V; = 100 cm 3 X --- p (kPa) 106 cm 3 400 I m3 I06cm3 Vr = 300 cm 3 X --200 = 1.0 X 10-4 m3 = 3.0 X 10-4 m3 SOLVE The initial and final temperatures are found using the V (cm3 ) 0-+--~--~-~0 100 200 ideal-gas law: 300 T= As the gas expands, it does work on the piston. Its temperature may change as weU. implying a change in them1al energy. We will use die version of the first law of thennodynamics in Equation 12.20 to describe the energy changes in the gas. We'll first find the cbange in thermal energy by computing the temperature change, then calculate how much work is done by looking at the area under the graph. Once we know Wgasand tlE 111 , we will use the first law to detemune the sign and the magnitude of the heat-telling us whether heat energy goes in or out, and how mucb. ' P; V; (4.0 X 1()5 Pa)( 1.0 X 10-4 m3 ) = --------= 300 K 11R (0.016 mol)(8.31 J/mol • K) STRATEGIZE PrVr (2.0 X 105 Pa)(3.0 X 10-4 m 3 ) Tr= = -'---------11R (0.016 mol)(8.31 J/mol • K) = 450 K The temperature increases, and so must the thermal energy. We can use Equation 12.21 to compute this change: .lE111= 3 (0.016 2 - mol)(S.31 J/mol • K)(4:>0 K - 300 K) = 30J The other piece of the puzzle is to compute the work done. We do this by finding the area under the graph for the process. FIGURE 12.16 shows that we can do this calculation by viewing the area as a triangle on top of a rectangle. Notice that the areas are in joules because they are the product of Pa and m3. The total work is 411 Ideal-Gas Processes 12.3 FIGURE 12.16 The work done by the expanding gas is the total area under the graph. Area of triangle: .t()() X ½(2.00 X 105 Pa)(2.00 X 10-4 m 3) p (Pa) 105 = 20.0J Area of rectangle: / (2.00 X 105 Pa)(2.00 X 10--.1m 3) W835 = area of triangle + area of rectangle = 20 J + 40 J = 60 J =40.0J 2.00 Now we can use the first law as wrillen in Equation 12.20 to find the heat: X 105 f 0 -I----+-----!--0 1.00 X 10-4 3.00 X 10-4 V (m3) Q = 6.Elh+ Wg:is= 30 J + 60 J = 90 J This is a positive number, so--using the conventions introduced in Chapter I I-we see that 90 J of heat energy must be added to the gas. ASSESS The gas does work-a loss of energy-but its temperature increases, so it makes sense that heat energy must be added. Adiabatic Processes You may have noticed that when you pump up a bicycle tire with a hand pump, the pump gets warm. We noted the reason for this in Chapter l 1: When you press down on the handle of the pump, a piston in the pump's chamber compresses the gas, doing work on it. According to the first law of thermodynamics, doing work on the gas increases its thermal energy. So the gas temperature goes up, and heat is then transferred through the walls of the pump to your hand. Now suppose you compress a gas in an insulated container, so that no heat is exchanged with the environment, or you compress a gas so quickly that there is no time for heat to be transferred. In either case, Q = 0. If a gas process has Q = 0, for either a compression or an expansion, we call this an adiabatic process. An expanding gas does work, so Wgas > 0. If the expansion is adiabatic, meaning Q = 0, then the first law of thermodynamics as written in Equation 12.20 tells us that LiE111< 0. Temperature is proportional to thermal energy, so the temperature will decrease as well. An adiabatic expansion lowers the temperature of a gas. Jf the gas is compressed, work is done on the gas ( Wgas < 0). If the compression is adiabatic, the first law of thermodynamics implies that AE 111> 0 and thus that the temperature increases. An adiabatic compression raises the temperature of a gas. Adiabatic processes allow you to use work, rather than heat, to change the temperature of a gas. ·coNCEPTUAL EXAMPLE 12. ,, Warm mountain winds This image shows surface temperatures (in °F) in N01thAmerica on a winter day. Notice the b1ightgreen area of unseasonablywrum temperaniresextending n01thand west from the center of the continent. On this day, a strong westerly wind, known as a Chinook wu1d,was blowing down off the Rocky Mountains, rapidly moving from high elevations (ru1dlow pressures)to low elevations (and higher pressures).The air was rapidly compressed as it descended.1l1e compression was so rapid that no heat was exchru1gedwith the environment, so this was an adiabatic process that significantly increasedthe air temperantre. What is the shape of the curve? shows the pV diagram of a gas undergoing an isothermal compression from point I to point 2. Sketch how the p V diagram would look if the gas were compressed from point I to the same final pressure by a rapid adiabatic compression. FIGURE 12.11 FIGURE 12.17 p\l p Pi Path of the isothennal _.·compression REASON An adiabatic compression increases the temperature of the gas as the work done on the gas is transformed into thermal energy. Consequently, as seen in FIGURE 12.18 on the next page. the curve of the adiabatic compression cuts across the isotherms to end on a higher-temperature isotherm when the gas pressure reaches p 2 . diagram for an isothermal compression. ,,:· Pt ~-....----------....-- V Con1ir111ed 414 c HAP T ER 12 Thermal Properties of Matter 12.5 Specific Heat and Heat of Transformation If you hold a glass of cold water in your hand, the heat from your hand will raise the temperature of the water. The heat from your hand will also melt an ice cube; melting is an example of what we will call a phase change, a change of state from one phase to another. In this section, we'll consider these two types of changes. - Specific Heat TABLE 12.4 Specific heats of solids and liquids Substance c (J/kg • K) Solids Lead 128 Gold 129 Copper 385 Iron 449 Aluminum 900 Water ice 2090 Mammalian body 3400 Adding 4190 J of heat energy to l kg of water raises its temperature by 1 K. If you are fortunate enough to have 1 kg of gold, you need only 129 J of heat to raise its temperature by 1 K. The amount of heat that raises the temperature of l kg of a substance by l K is called the specific heat of that substance. The symbol for specific heat is c. Water has specific heat Cwaier = 4190 J/kg • K, and the specific heat of gold is cgold = 129 J/kg • K. Specific heat depends only on the material from which an object is made. TABLE 12.4 lists the specific heats of some common liquids and solids. If heat c is required to raise the temperature of l kg of a substance by 1 K, then heat Mc is needed to raise the temperature of mass M by I Kand Mc !::,Tis needed to raise the temperature of mass M by t:.T. In general, the heat needed to bring about a .temperature change t:.T is Liquids Mercury Ethyl alcohol 2400 Water 4190 II Video Heating Water and Aluminum (12.24) Q=Mct:.T 140 Heat needed to produce a temperature change J.T for mass M with specific heat c Q can be either positive (temperature goes up) or negative (temperature goes down). It takes more heat energy to change the temperature of a substance with a large specific heat than to change the temperature of a substance with a small specific heat. Water, with a very large specific heat, is slow to warm up and slow to cool down. This large "thermal inertia" of water is essential for the biological processes of life. EXAMPLE 12. 14 How much energy is needed to run a fever? A 70 kg student catches the flu, and his body temperature increases from 37.0°C (98.6°F) to 39.0°C ( 102.2°F). How much energy is required to raise his body's temperature? STRATEGIZEThe increase in temperature requires the addition of energy. PREPAREThe change in temperature 6.T is 2.0°C. or 2.0 K. SOLVE Raising the temperature of the body uses energy supplied i111emally from the chemical reactions of the body's metabolism, which transfer heat to the body. The specific heat of the body is given in Table 12.4 as 3400 J/k:g • K. We can use Equation 12.24 io find the necessary heat energy: Temperate lakes At night, the large specific heat of water preventsthe temperan1reof a body of water from dropping nearly as much as that of the surrounding air. Early in the morning, water vapor evaporatingfrom a warm lake quickly condenses in the colderair above, forming mist. During the day, the opposite happens: The air becomes much wanner than the water. Q = Mc 6.T = (70 kg)(3400 J/kg • K)(2.0 K) = 4.8 X 105 J ASSESS The body is mostly water, with a large specific heat, and the mass of the body is large. so we·d expect a large amount of energy to be necessary. Looking back to Chapter 11, we see tbat this is approximately the energy in a large apple. or the amount of energy required to walk I mile. 12.5 415 Specific Heat and Heat of Transformation Phase Changes Suppose you remove a few ice cubes from the freezer and place them in a sealed container with a thermometer. Then, as in FIGURE 12.21a, you put a steady flame under the container. We'll assume that the heating is done slowly so that the inside of the container always has a single, uniform temperature. FIGURE 12.21b shows a graph of the temperature as a function of time. In stage l, the ice steadily warms without melting until it reaches 0°C. During stage 2, the temperature remains fixed at 0°C for an extended period of time during whjch the ice melts. As the ice is melting, both the ice temperature and the Liquid water temperature remain at 0°C. Even though the system is being heated, the temperature doesn't begin to rise again until all the ice has melted. I KEYCONCEPT FIGURE 12.21 The temperature as a function of time as water is transformed from solid to liquid to gas. (b) T(oC) Temperature is constant as Steam is the system changes from warming. 10<?%solid to 100% liquid. \ 150 (a) Ice Ice Water Steam I Water 0 Ice is warming. into water. warming. LOO Steam 50 0 -50 8 Water is turning into steam. 0 Steam is warming. ( 0 .;. 6 \ 0 Temperature is constant as the system changes from 100% liquid to 100% gas. lee is Liquid water warming. is warm.ing. In Figure 12.21, by comparing the slope of the graph dming the time the liquid water is wanning to the slope as steam is wanning, we can say that STOP TO THINK 12.5 A. B. C. D. The The The The specific heat of water is larger than that of steam. specific heat of water is smaller than that of steam. specific heat of water is equal to that of steam. slope of the graph is not related to the specific heat. At the end of stage 2, the ice has completely melted into liquid water. During stage 3, the flame slowly warms this Jjquid water, raising its temperature until it reaches 100°C. During stage 4, the liquid water remains at l00°C as it turns into water vapor-steam-which also remains at 100°C. Even though the system is being heated, the temperature doesn't begin to rise again until all the water has been converted to steam. Finally, stage 5 begins, in which the system is pure steam whose temperature rises continuously from 100°C. lmiEI ► In everyday language, the three phases of water are called ice, ware,; and steam. The term "water" implies the Liquid phase. Scientifically, these are the solid, liquid, and gas phases of the compound called wate1: When we are working with different phases of water, we'll use the term "water" in the scientific sense of a collection of H2 0 molecules. We'll say either liquid or liquid water to denote the liquid phase. ◄ CONCEPTUAL EXAMPLE 12.15 Strategy for cooling a drink If you have a warm soda that you wish to cool. is it more effective to add 25 g of liquid water at 0°C or 25 g of water ice at 0°C? 1f you add liquid water at 0°C, heat will be transferred from the soda to the water, raising the temperature of the water and lowering that of the soda. If you add water ice at 0°C, heat will first be transferTedfrom the soda to the ice to melt it, REASON transforming the 0°C ice to 0°C liquid water, then will be transferred to the liquid water to raise its temperature. Thus more thermal energy will be removed from the soda, giving it a lower final temperature. if ice is used rather than liquid water. This makes sense because you know that this is what you do in practice. To cool a drink, you drop in an ice cube. ASSESS 416 c HAP TE R 12 Thermal Properties {I Frozen frogs it seems impossible,but common wood frogssurvivethe winter with much of their bodiesfrozen. When you dissolve substancesin water,the freezing point lowers. Althoughthe liquid water between cells in the frogs' bodiesfreezes,the water inside their cells remainsliquidbecause of high concentrationsof dissolvedglucose. This preventsthe cell damage that accompanies the freezingand subsequentthawing of tissues. When springan-ives,the frogs thaw and appear no worsefor their winter freeze. ' CONCEPTUAL EXAMPLE 12.16 of Matter The temperature at which-if the thermal energy is increased-a solid becomes a liquid is called the melting point; if the thermal energy is instead decreased, a liquid becomes a solid at the freezing point. Melting and freezing are phase changes. A system at the melting point is in phase equilibrium, meaning that any amount of solid can coexist with any amount of liquid. Raise the temperature ever so slightly and the entire system soon becomes liquid. Lower it slightly and it all becomes solid. You can see another region of phase equilibrium of water in Figure 12.21b at 100°C. This is a phase equilibrium between the liquid phase and the gas phase, and any amount of liquid can coexist with any amount of gas at this temperature. As heat is added to the system, the temperature stays the same. The added energy is used to break bonds between the ljquid molecules, allowing them to move into the gas phase. The temperature at wruch a gas becomes a liquid is called the condensation point; the temperature at which a liquid becomes a gas is the boiling point. £mill ► Liquid water becomes solid ice at 0°C, but that doesn't mean the temperature of ice is always 0°C. Ice reaches the temperature of its surroundings. If the air temperature in a freezer is -20°C, then the ice temperature is -20°C. Likewise, steam can be heated to temperanires above 100°C. That doesn't happen when you boil water on the stove because the steam escapes, but steam can be heated far above 100°C in a sealed container. ◄ Fast or slow boil? You are cooking pasta on the stove: the water is at a slow boil. Will the pasta cook more quickly if you tum up the burner on the stove so that the water is at a fast boil? Adding beat at a faster rate will make the water boil away more rapidly but will not change the temperature-and will not alter the cooking time. REASON Water boils at l00°C: no matter how vigorously Lbe water is boiling, the temperature is the same. It is the temperature of the water that determines how fast the cooking takes place. ASSESS This result may seem counterintuitive, but you can try the experiment next time you cook pasta! Heat of Transformation Lava-molten rock-undergoes a phase change from liquid to solid when it contacts liquid water; the transfer of heat to the water causes the water to undergo a phase change from liquid to gas. In Figure 12.21b, the phase changes appeared as horizontal line segments on the graph. During these segments, heat is being transferred to the system but the temperature isn't changing. The thermal energy continues to increase during a phase change, but, as noted, the additional energy goes into breaking molecular bonds rather than speeding up the molecules. A phase change is characterized by a change in thermal energy without a change in temperature. The amount of heat energy that causes 1 kg of a substance to undergo a phase change is called the heat of transformation of that substance. For example, laboratory experiments show that 333,000 J of heat are needed to melt l kg of ice at 0°C. The symbol for heat of transformation is L. The heat required for the entire system of mass M to undergo a phase change is Q=ML ( 12.25) "Heat of transformation" is a generic term that refers to any phase change. Two particular heats of transformation are the heat of fusion Lr, the heat of transformation between a solid and a liquid, and the heat of vaporization Lv, the heat of transformation between a liquid and a gas. The heat needed for these phase changes is Q= ± MLr { ± MLv Heat needed to melt/freeze mass M Heat needed to boil/condense mass M (12.26) 12.5 Specific Heat and Heat of Transformation 417 The ± indicates that heat must be added to the system during melting or boiling but removed from the system during freezing or condensing. You must explicitly include the minus sign when it is needed. Carefully consider the nature of the phase change and whether heat is added or removed. When an ice cube melts in your hand, heat goes from your hand to the ice cube. In your freezer, where the temperature is below the freezing point of water, liquid water freezes as heat goes from the liquid water to the surrounding air. If you put a pan of water on the stove, you must add heat to make the water boil. When you breathe on a mirror, water vapor from your lungs condenses on the glass, and heat goes from the water vapor into the cooler glass. TABLE 12.5 lists some heats of transformation. Notice that the heat of vaporization is always much higher than the heat of fusion. TABLE 12.5 Melting and boiling temperatures and heats of transformation at standard atmospheric pressure Liquid to Gas Solid to Liquid Heat of Fusion Lr (J/kg) Substance Melting Point T111 (°C) Nitrogen (N 2) -210 0.26 X 105 Ethyl alcohol -114 l.09 Mercury Water Lead EXAMPLE 12.17 Boiling Point Tb(°C) Heat of Vaporization Lv (J/kg) -196 l.99 X 105 105 78 8.79 X 105 0.11 X 105 357 2.96 X 105 0 3.33 X 105 JOO 22.6 X 105 328 0.25 X 105 -39 X 1750 8.58 X 105 How much energy is needed to melt a popsicle? A girl eats a 45 g frozen popsicle that was taken out of a -10°C freezer. How much energy does her body use to bring the popsicle up to body temperature? STRATEGIZEWe will assume that the popsicle is pure water. There are three parts to the problem, corresponding to stages 1-3 in Figure 12.21: The popsicle must be warmed to 0°C, the popsicle must melt. and then the resulting water must be warmed to body temperature. We will figure out how much heat is needed for each stage; the sum is the total energy required. PREPARENormal body temperature is 37°C. The specific heats of ice and liquid water are given in Table 12.4: the heat of fusion of water is given in Table 12.5. SOLVEThe heat needed to warm the frozen water by ti T = 10°C = IO K to the melting point is Q 1 = Mcicc ti T = (0.045 kg)(2090 J/kg • K)( IO K) = 940 J Note that we use the specific heat of water ice, not liquid water, in this equation. Melting 45 g of ice requires beat Q2 = MLr = (0.045 kg)(3.33 X 105 J/kg) = 15,000 J The liquid water must now be warmed to body temperature; this requires heat Q3 = Mcwa,er tiT= (0.045 kg)(4190 J/kg • K)(37 K) = 7000] The total energy is the sum of these three values: Q,0 ta1 = 23,000 J. ASSESS More energy is needed to melt the ice than to warm the water, as we would expect. A commercial popsicle has 40 Calories. which is about 170 kJ. Roughly 15% of the chemical energy in this frozen treat is used to bring it up to body temperature! Evaporation Water boils at 100°C. But individual molecules of water can move from the liquid phase to the gas phase at lower temperatures. This process is known as evaporation. Water evaporates as sweat from your skin at a temperature well below l00°C. Different particles in a liquid move at different speeds, as they do in a gas. At any temperature, some molecules will be moving fast enough to go into the gas phase. And they will do so, carrying away thermal energy as they go. The molecules that leave the liquid are the ones that have the highest kinetic energy, so evaporation reduces the average kinetic energy (and thus the temperature) of the liquid left behind. 418 c HAP TE R 12 Thermal Properties of Matter The evaporation of water, both from sweat and in moisture you exhale, is one of the body's methods of exhausting the excess heat of metabolism to the environment, allowing you to maintain a steady body temperature. The heat to evaporate a mass M of water is Q = MLv, and this amount of heat is removed from your body. However, the heat of vaporization Lv is a little larger than the value for boiling. At a skin temperature of 30°C, the heat of vaporization of water is (iJ Keeping your cool Humans (and cattle and horses) have sweat glands, so we can perspire to moisten our skin, allowing evaporation to cool our bodies. Animals that do not perspire can also use evaporation lo keep cool. Dogs, goats, rabbits, and even birds pant, evaporating water from their respiratory passages. Elephants spray water on their skin; other animals may lick their fur. Lv( at 30°) = 24 X 105 J /kg This is 6% higher than the value at l00°C in Table 12.5. You should use this value for Lv when you consider the heat transfer due to evaporation of perspiration from the skin and similar situations. Computing heat loss by perspiration EXAMPLE 12.18 4iJ The human body can produce approximately 30 g of perspiration per minute. At what rate is it possible to exhaust heat by the evaporation of perspiration? STRATEGIZE We are looking for a rate of exhausting hear, so we are looking for a power in watts, an energy divided by a time. we·11 assume that perspiration is pure water, which is a good approximation. The energy is the heat required to evaporate 30 g of water from the skin; the time is I minute. PREPARE The water is evaporated from the skin, which we assume is at 30°C. The value for the heat of transformation is therefore L,, = 24 X 105 J/kg. SOLVE The evaporation of 30 g of perspiration at normal body temperature requires heat energy The opposite of sweating You are sitting outside, enjoying a cold beverage and you notice that your glass "sweats"-drops of water collect on the outside.Yourglass isn't really sweating, of course. The water doesn't come from inside the glass but from outside, as water vapor from the air condenses onto the sides of the glass. Heat leaves the water vapor when it condenses-and so heat is added to your glass. Sweatingmay keep you cool due to evaporation,but a "sweating" glass means condensation,and this means a wanning drink. Q = Mlv = (0.030 kg)(24 X 105 J/kg) = 7.2 X 104 J This is the heat losr per minute; the rate of heat loss is Q /11 7.2X 104J 60 s = 1200W Given the metabolic power required for different act1v1t1es.as listed in Chapter l l, this is sufficient to keep the body cool even when exercising in hot weatheras long as the person drinks enough water to keep up this rate of perspiration. ASSESS l{J.iJi,jfoiJiiiJJ l kg of barely molten lead, at 328°C, is poured into a large beaker holding liquid water right at the boiling point, 100°C. What is the mass of the water that will be boiled away as the lead solidifies? II Video What the Physics? Fighting Fire with Evaporation A. 0kg B. < l kg C. l kg D. >l kg 12.6 Calorimetry FIGURE 12.22 Cooling hot coffee with ice. If you've put an ice cube into a hot drink to cool it quickly, you engaged, in a trialand-error way, in a practical aspect of heat transfer known as calorimetry, the quantitative measurement of the heat transferred between systems or evolved in reactions. You know that heat energy will be transfened from the hot drink into the cold ice cube, reducing the temperature of the drink, as shown in FIGURE 12.22. Let's make this qualitative picture more precise. FIGURE 12.23 shows two systems that can exchange heat with each other but that are isolated from everything else. Suppose they start at different temperatures T 1 and T 2 . As you know, heat energy will be transferred from the hotter to the colder system until they reach a common final temperature Tr. In the cooling coffee example of Figure 12.22, the coffee is system l, the ice is system 2, and the insulating barrier is the mug. 12.6 The insulation prevents any heal energy from being transferred to or from the environment, so energy conservation tells us that any energy leaving the hotter system must enter the colder system. The concept is straightforward, but to state the idea mathematically we need to be careful with signs. Let Q 1 be the energy transferred to system I as heat. Q 1 is positive if energy enters system l, negative if energy leaves system 1. Simjlarly, Q2 is the energy transferred to system 2. The fact that the systems are merely exchanging eilergy can be written as IQil = IQ2'- But Q 1 and Q 2 have opposite signs, so Q 1 = -Q 2 . No energy is exchanged with the environment, so it makes more sense to write this relationship as Calorimetry 419 FIGURE 12.23 Heat is being transferred from system 1 to system 2, so Q2 > O and Q1<0. The magnitude IQilof the heat leaving system I equals the magnitude IQ of 21 the heat entering system 2. ~-:::·:. :- ._•. :•:.: . ._.-,--:•:. : . ._•. :-:. :- . :·.:. :·· :i System I System 2 (12.27) EmDI ► The signs are very important in calorimetry problems. /j, T is always Tr - T;, so /j, T and Q are negative for any system whose temperature decreases. The proper sign of Q for any phase change must be supplied by you, depending on the direction of the phase change. ◄ PROBLEM-SOLVING App Ro Ac H 12 . 1 Opposite signs mean that Qnct = Q1 + Q2 = 0 • Calorimetry problems When two systems are brought into thermal contact, we use calorimetry to find the heat transferred between them and their final equilibrium temperature. STRATEGIZE Identify the individual interacting systems. Assume that they are isolated from the environment. PREPARE List the known information and identify what you need to find. Convert all quantities to SI units. SOLVEThe statement of energy conservation is = Mc(Tr - T;). Be sure to have the temperatures T; and Tr in the correct order. ■ For systems that undergo a phase change, Qphase = ± ML. Supply the correct sign by observing whether energy enters or leaves the system during the transition. ■ Some systems may undergo a temperature change and a phase change. Treat the changes separately. The beat energy is Q = Q~r + Qphasc. ■ For systems that undergo a temperature change, Q.!.T ASSESSThe final temperature should be between the initial temperatures. A Tr that is higher or lower than all initial temperatures is an indication that something is wrong, usually a sign error. EXAMPLE 12.19 Using calorimetry to identify a metal 200 g of an unknown metal is heated to 90.0°C, then dropped into 50.0 g of water at 20.0°C in an insulated container. The water temperature rises within a few seconds to 27.7°C, then changes no further. Identify the metal. STRATEGIZE The interacting systems are the metal and the water. We'll assume that the two systems quickly come to a final temperature. so there is no time for heat to be exchanged with the environment. PREPARE We label the temperatures as follows: The initial temperature of the metal is T111 ; the initial temperature of the water is Tw. The common final temperature is Tr. For water, Cw= 4190 J/kg • K is known from Table 12.4. Only the specific beat cm of the metal is unknown. Comi11ued 12.8 423 Heat Transfer Heat-transfer mechanisms When two objects are in direct physical contact, such as the soldering iron and the circuit board, heat is transfe1Ted by conductio11. Energy is transferred by direct contact. This special photograph shows air currents near a flame. Hot gases from the combustion rise, carrying thermal energy in a process known as convectio11. Energy is transferred by the bulk motion of molecules with high thermal energy. The lamp shiJ1es Oil the lambs huddled below, warming them. The energy is transferred by infrared radiation, a form of electromagnetic waves. Energy is transferred by electromagnetic waves. When you blow Oil a cup of cocoa, this increases the rate of evaporation, rapidly cooling it. Energy is transferred by the removal of molecules with high thermal energy. Conduction If you hold a metal spoon in a cup of hot coffee, the handle of the spoon soon gets warm. Thermal energy is transferred along the spoon from the coffee to your hand. The difference in temperature between the two ends drives this heat transfer by a process known as conduction. Conduction is the transfer of thermal energy directly through a physical material. FIGURE12.26 shows a copper rod placed between a hot reservoir (a foe) and a cold reservoir (a block of ice). We can use our atomic model to see how thermal energy is transferred along the rod by the interaction between atoms in the rod; fast-moving atoms at the hot end transfer energy to slower-moving atoms at the cold end. Suppose we set up a series of experiments to measure the heat Q transferred through various rods. We would find the following trends in our data: ■ ■ ■ ■ FIGURE 12.26 Conduction of heat in a solid rod. Q is the heat being trans~~rre: through the rod. Q increases if the temperature difference l!;.T between the hot end and the cold end is increased. Q increases if the cross-section area A of the rod is increased. Q decreases if the length L of the rod is increased. Some materials (such as metals) transfer heat quite readily. Other materials (such as wood) transfer very little heat. The final observation is one that is familiar to you: [f you are stirring a pot of hot soup on the stove, you generally use a wood or plastic spoon rather than a metal one. These experimental observations about heat conduction can be summarized in a single formula. If heat Q is transfeITed in a time interval /!;.t,the rate of heat transfer (joules per second, or watts) is Q//!;.t. For a material of cross-section area A and length L, spanning a temperature difference /!;.T, the rate of heat transfer is Q /!;./= (kA) L l!;.T ( 12.35) Rate of conduction of heat across a temperature difference The quantity k, which characterizes whether the material is a good or a poor conductor of heat, is called the thermal conductivity of the material. Because the heat transfer rate J/s is a powe1; measured in watts, the unjts of k are W /m • K. Values of k Fire Ice Cross-section area A The particles on the left side of the rod are vibrating more vigorously than the pruticles on the right. The particles on the left transfer energy to the particles on the right via the bonds connecting them. 424 c HAP TE R 12 Thermal Properties of Matter for some common materials are listed in material is a better conductor of heat. TABLE 12.1 4iJ Cold feet, warm heart A penguin standing on the coldAntarcticice loses ve1ylittle heat through its feet.Its thick skin with limited thennal conductivityhelps,but more important, the penguin's feet are ve1ycold, which minimizes heat loss by conduction.The penguin can have cold feet so closeto the warmth of the body because of an adaptationcalled counterrnrrellf heat exchange.Aite1iescarrying wa1111 blood to the feet 11111nextto veinscanying cold blood back from the feel.Heat goes from the aiteries to the veins, cooling the blood going to the feet and waiming the bloodgoing to the body-so the feet stay cool whilethe body stays wrum. EXAMPLE 12.22 TABLE 12.1; a larger number for k means a Thermal conductivity values (measured at 20°C) Material k (W!m • K) Material k (W!m • K) Diamond 1000 Skin 0.50 Silver 420 Muscle 0.46 Copper 400 Fat 0.21 Iron 72 Wood 0.2 Stainless steel 14 Carpet 0.04 Ice 1.7 Fur, feathers 0.02-0.06 Concrete 0.8 Air (27°C, 100 kPa) 0.026 Plate glass 0.75 The weak bonds between molecules make most biological materials poor conductors of heat. Fat is a worse conductor than muscle, so sea mammals have thick layers of fat for insulation. The bodies of land mammals are insulated with fur, and those of birds with feathers. Both fur and feathers trap a good deal of air, so their conductivity is similar to that of air, as Table 12.7 shows. Wanning the bench At the stan of the section we noted that you "feel cold" when you sit on a cold concrete bench. "Feeling cold" really means that your body is losing a significant amount of beat. How significant? Suppose you are sitting on a l0°C concrete bench. You are wearing thin clothing that provides negligible insulation. In thjs case, most of the insulation that protects your body's core (temperature 37°C) from the cold of the bench is provided by a 1.0-cm-thick layer of fat on the part of your body that touches the bench. (The thickness varies from person to person, but this is a reasonable average value.) A good estimate of the area of contact with the bench is 0.10 m2 . Given these details, what is the rate of heat loss by conduction? STRATEGIZEHeat is lost to the bench by conduction through the fat layer, so we will compute the rate of heat loss by using Equation 12.35. PREPAREThe thickness of the conducting layer is 0.010 m, the area is O.lO ni2, and the thermal conductivity of fat is given in Table 12.7. The temperature difference is the difference between your body's core temperature (37°C) and the temperature of the bench ( I0°C), a difference of 27°C, or 27 K. SOLVE We have all of the data we need to use Equation 12.35 to compute the rate of heat loss: Q -= 6.1 2 ((0.21 W/m • K)(0.10 m )) ----'-'------'(27K)=57W 0.010 m ASSESS 57 Wis more than half your body's resting power, which we learned in Chapter 11 is approximately 100 W. That's a signilicant loss, so your body will feel cold, a result that seems reasonable if you've ever sat on a cold bench for any leagth of time. Convection A feather coat @liJ A penguin's short, dense feathers serve a different role than the flight feathers of other birds: They trap air to provide thermal insulation. The fluffy feathers of the juvenile penguin makes this trapping of air very clear, but the smooth covering of the adults works in a similar manner. In conduction, faster-moving atoms transfer thermal energy to adjacent atoms. But in fluids such as water or air, there is a more efficient means to move energy: by transferring the faster-moving atoms themselves. When you place a pan of cold water on a burner on the stove, it's heated on the bottom. This heated water expands and becomes less dense than the water above it, so it rises to the surface while cooler, denser water sinks to take its place. Th.is transfer of thermal energy by the motion of a fluid is known as convection. Convection is usually the main mechanism for heat transfer in fluid systems. On a small scale, convection mixes the pan of water that you heat on the stove; on a large scale, convection is responsible for making the wind blow and ocean currents circulate. Air is a very poor thermal conductor, but it is very effective at transferring energy by convection. To use air for thermal insulation, it is necessary to trap the air in small pockets to Jimjt convection. And that's exactly what feathers, fur, doublepaned windows, and fiberglass insulation do. Summary 431 SUMMARY - To use the atomic model of matter to explain many proper