Chapter 2 Calorimetry PDF

Chapter 2 Calorimetry 2.1 Heat and Internal Energy At the outset, it is important that we make a major distinction between internal energy and heat. Internal energy is all the energy of a system that is associated with its microscopic components—atoms and molecules—wh...

Chapter 2 Calorimetry 2.1 Heat and Internal Energy At the outset, it is important that we make a major distinction between internal energy and heat. Internal energy is all the energy of a system that is associated with its microscopic components—atoms and molecules—when viewed from a reference frame at rest with respect to the center of mass of the system. The last part of this sentence ensures that any bulk kinetic energy of the system due to its motion through space is not included in internal energy. Internal energy includes kinetic energy of random translational, rotational, and vibrational motion of molecules, potential energy within molecules, and potential energy between molecules. It is useful to relate internal energy to the temperature of an object, but this relationship is limited—we show in Section 3.3 that internal energy changes can also occur in the absence of temperature changes. Heat is defined as the transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings. When you heat a substance, you are transferring energy into it by placing it in contact with surroundings that have a higher temperature. This is the case, for example, when you place a pan of cold water on a stove burner—the burner is at a higher temperature than the water, and so the water gains energy. We shall also use the term heat to represent the amount of energy transferred by this method. Scientists used to think of heat as a fluid called caloric, which they believed was transferred between objects; thus, they defined heat in terms of the temperature changes produced in an object during heating. Today we recognize the distinct difference between internal energy and heat. Nevertheless, we refer to quantities using names that do not quite correctly define the quantities but which have become 40 entrenched in physics tradition based on these early ideas. Examples of such quantities are heat capacity and latent heat. As an analogy to the distinction between heat and internal energy, consider the distinction between work and mechanical energy. The work done on a system is a measure of the amount of energy transferred to the system from its surroundings, whereas the mechanical energy of the system (kinetic plus potential) is a consequence of the motion and configuration of the system. Thus, when a person does work on a system, energy is transferred from the person to the system. It makes no sense to talk about the work of a system—one can refer only to the work done on or by a system when some process has occurred in which energy has been transferred to or from the system. Likewise, it makes no sense to talk about the heat of a system—one can refer to heat only when energy has been transferred as a result of a temperature difference. Both heat and work are ways of changing the energy of a system. It is also important to recognize that the internal energy of a system can be changed even when no energy is transferred by heat. For example, when a gas in an insulated container is compressed by a piston, the temperature of the gas and its internal energy increase, but no transfer of energy by heat from the surroundings to the gas has occurred. If the gas then expands rapidly, it cools and its internal energy decreases, but no transfer of energy by heat from it to the surroundings has taken place. The temperature changes in the gas are due not to a difference in temperature between the gas and its surroundings but rather to the compression and the expansion. In each case, energy is transferred to or from the gas by work. The changes in internal energy in these examples are evidenced by corresponding changes in the temperature of the gas. Units of Heat As we have mentioned, early studies of heat focused on the resultant increase in temperature of a substance, which was often water. The early notions of heat based on caloric suggested that the 41 flow of this fluid from one substance to another caused changes in temperature. From the name of this mythical fluid, we have an energy unit related to thermal processes, the calorie (cal), which is defined as the amount of energy transfer necessary to raise the temperature of 1 g of water from 14.5°C to 15.5°C. (Note that the “Calorie,” written with a capital “C” and used in describing the energy content of foods, is actually a kilocalorie.) The unit of energy in the U.S. customary system is the British thermal unit (Btu), which is defined as the amount of energy transfer required to raise the temperature of 1 lb of water from 63°F to 64°F. Scientists are increasingly using the SI unit of energy, the joule, when describing thermal processes. In this textbook, heat, work, and internal energy are usually measured in joules. (Note that both heat and work are measured in energy units. Do not confuse these two means of energy transfer with energy itself, which is also measured in joules.) The Mechanical Equivalent of Heat Whenever friction is present in a mechanical system, some mechanical energy is lost—in other words, mechanical energy is not conserved in the presence of nonconservative forces. Various experiments show that this lost mechanical energy does not simply disappear but is transformed into internal energy. We can perform such an experiment at home by simply hammering a nail into a scrap piece of wood. What happens to all the kinetic energy of the hammer once we have finished? Some of it is now in the nail as internal energy, as demonstrated by the fact that the nail is measurably warmer. Although this connection between mechanical and internal energy was first suggested by Benjamin Thompson, it was Joule who established the equivalence of these two forms of energy. 42 Fig. 2.1 Joule’s experiment for determining the mechanical equivalent of heat. The falling blocks rotate the paddles, causing the temperature of the water to increase. A schematic diagram of Joule’s most famous experiment is shown in Figure 2.1. The system of interest is the water in a thermally insulated container. Work is done on the water by a rotating paddle wheel, which is driven by heavy blocks falling at a constant speed. The temperature of the stirred water increases due to the friction between it and the paddles. If the energy lost in the bearings and through the walls is neglected, then the loss in potential energy associated with the blocks equals the work done by the paddle wheel on the water. If the two blocks fall through a distance h, the loss in potential energy is 2mgh, where m is the mass of one block; this energy causes the temperature of the water to increase. By varying the conditions of the experiment, Joule found that the loss in mechanical energy 2mgh is proportional to the increase in water temperature T. The proportionality constant was found to be approximately 4.18 J/g.°C. Hence, 4.18 J of mechanical energy raises the temperature of 1 g of water by 1°C. More precise measurements taken later 43 demonstrated the proportionality to be 4.186 J/g.°C when the temperature of the water was raised from 14.5°C to 15.5°C. We adopt this “15-degree calorie” value: 1 cal  4.186 J (2.1) This equality is known, for purely historical reasons, as the mechanical equivalent of heat. Example.1 A student eats a dinner rated at 2 000 Calories. He wishes to do an equivalent amount of work in the gymnasium by lifting a 50.0-kg barbell. How many times must he raise the barbell to expend this much energy? Assume that he raises the barbell 2.00 m each time he lifts it and that he regains no energy when he lowers the barbell. Solution Because 1 Calorie = 1.00103 cal, the total amount of work required to be done on the barbell–Earth system is 2.00106 cal. Converting this value to joules, we have   W  2.00  106 cal 4.186 J/cal   8.37  106 J The work done in lifting the barbell a distance h is equal to mgh, and the work done in lifting it n times is nmgh. We equate this to the total work required: W  nmgh  8.37 106 J If the student is in good shape and lifts the barbell once every 5 s, it will take him about 12 h to perform this feat. Clearly, it is much easier for this student to lose weight by dieting. 44  In reality, the human body is not 100% efficient. Thus, not all of the energy transformed within the body from the dinner transfers out of the body by work done on the barbell. Some of this energy is used to pump blood and perform other functions within the body. Thus, the 2 000 Calories can be worked off in less time than 12 h when these other energy requirements are included. 2.2 Specific Heat and Calorimetry When energy is added to a system and there is no change in the kinetic or potential energy of the system, the temperature of the system usually rises. (An exception to this statement is the case in which a system undergoes a change of state—also called a phase transition—as discussed in the next section.) If the system consists of a sample of a substance, we find that the quantity of energy required to raise the temperature of a given mass of the substance by some amount varies from one substance to another. For example, the quantity of energy required to raise the temperature of 1 kg of water by 1°C is 4 186 J, but the quantity of energy required to raise the temperature of 1 kg of copper by 1°C is only 387 J. In the discussion that follows, we shall use heat as our example of energy transfer, but keep in mind that we could change the temperature of our system by means of any method of energy transfer. The heat capacity C of a particular sample of a substance is defined as the amount of energy needed to raise the temperature of that sample by 1°C. From this definition, we see that if energy Q produces a change T in the temperature of a sample, then Q  CT (2.2) The specific heat c of a substance is the heat capacity per unit mass. Thus, if energy Q transfers to a sample of a substance with mass m and the temperature of the sample changes by T, then the specific heat of the substance is 45  Q c (2.3) mT Specific heat is essentially a measure of how thermally insensitive a substance is to the addition of energy. The greater a material’s specific heat, the more energy must be added to a given mass of the material to cause a particular temperature change. Table 2.1 lists representative specific heats. From this definition, we can relate the energy Q transferred between a sample of mass m of a material and its surroundings to a temperature change T as Q  mcT (2.4) For example, the energy required to raise the temperature of 0.500 kg of water by 3.00°C is (0.500 kg)(4 186 J/kg.°C) (3.00°C) = 6.28103 J. Note that when the temperature increases, Q and T are taken to be positive, and energy transfers into the system. When the temperature decreases, Q and T are negative, and energy transfers out of the system. Specific heat varies with temperature. However, if temperature intervals are not too great, the temperature variation can be ignored and c can be treated as a constant. For example, the specific heat of water varies by only about 1% from 0°C to 100°C at atmospheric pressure. Unless stated otherwise, we shall neglect such variations. Measured values of specific heats are found to depend on the conditions of the experiment. In general, measurements made in a constant-pressure process are different from those made in a constant- volume process. For solids and liquids, the difference between the two values is usually no greater than a few percent and is often neglected. Most of the values given in Table 2.1 were measured at atmospheric pressure and room temperature. The specific heats for gases measured at constant pressure are quite different from values measured at constant volume. 46 Table 2.1 Specific Heats of Some Substances at 25C and Atmospheric Pressure Substance J/kg°C cal/g °C Elemental solids Aluminum 900 0.215 Beryllium 1 830 0.436 Cadmium 230 0.055 Copper 387 0.092 4 Germanium 322 0.077 Gold 129 0.030 8 Iron 448 0.107 Lead 1 28 0.030 5 Silicon 703 0. 1 68 Silver 234 0.056 Other solids Brass 380 0.092 Glass 837 0.200 Ice(-5C) 2090 0.50 Marble 860 0.21 Wood I 700 0.41 Liquids Alcohol (ethyl) 2400 0.58 Mercury' 140 0.033 Water (15°C) 4186 1.00 Gas Steam (100°C) 2010 0.48 Quick Quiz 2.1 Imagine you have 1 kg each of iron, glass, and water, and that all three samples are at 10°C. Rank the samples from lowest to highest temperature after 100 J of energy is added to each sample. 47 Quick Quiz 2.2 Considering the same samples as in Quick Quiz 2.1, rank them from least to greatest amount of energy transferred by heat if each sample increases in temperature by 20°C. It is interesting to note from Table 2.1 that water has the highest specific heat of common materials. This high specific heat is responsible, in part, for the moderate temperatures found near large bodies of water. As the temperature of a body of water decreases during the winter, energy is transferred from the cooling water to the air by heat, increasing the internal energy of the air. Because of the high specific heat of water, a relatively large amount of energy is transferred to the air for even modest temperature changes of the water. The air carries this internal energy landward when prevailing winds are favorable. For example, the prevailing winds on the West Coast of the United States are toward the land (eastward). Hence, the energy liberated by the Pacific Ocean as it cools keeps coastal areas much warmer than they would otherwise be. This explains why the western coastal states generally have more favorable winter weather than the eastern coastal states, where the prevailing winds do not tend to carry the energy toward land. Conservation of Energy: Calorimetry One technique for measuring specific heat involves heating a sample to some known temperature Tx, placing it in a vessel containing water of known mass and temperature Tw < Tx, and measuring the temperature of the water after equilibrium has been reached. This technique is called calorimetry, and devices in which this energy transfer occurs are called calorimeters. If the system of the sample and the water is isolated, the law of the conservation of energy requires that the amount of energy that leaves the sample (of unknown specific heat) equal the amount of energy that enters the water. Conservation of energy allows us to write the mathematical representation of this energy statement as 48  Qcold  Qhot (3.5) The negative sign in the equation is necessary to maintain consistency with our sign convention for heat. Suppose mx is the mass of a sample of some substance whose specific heat we wish to determine. Let us call its specific heat cx and its initial temperature Tx. Likewise, let mw, cw, and Tw represent corresponding values for the water. If Tf is the final equilibrium temperature after everything is mixed, then from Equation 20.4, we find that the energy transfer for the water is mwcw(Tf  Tw), which is positive because Tf > Tw , and that the energy transfer for the sample of unknown specific heat is mxcx(Tf  Tx), which is negative. Substituting these expressions into Equation 2.5 gives m w c w Tf  Tw   m x c x Tf  Tx  Solving for cx gives mw c w T f  Tw  cx  m x Tx  T f  Example 2.2 A 0.050 0-kg ingot of metal is heated to 200.0°C and then dropped into a beaker containing 0.400 kg of water initially at 20.0°C. If the final equilibrium temperature of the mixed system is 22.4°C, find the specific heat of the metal. Solution According to Equation 2.5, we can write m w c w Tf  Tw   m x c x Tf  Tx  0.400 kg4 186 J/kg C22.4  C  20.0  C  49    0.050 0 kg c x  22.4  C  200.0  C  From this we find that cx  453 J/kg C The ingot is most likely iron, as we can see by comparing this result with the data given in Table 2.1. Note that the temperature of the ingot is initially above the steam point. Thus, some of the water may vaporize when we drop the ingot into the water. We assume that we have a sealed system and that this steam cannot escape. Because the final equilibrium temperature is lower than the steam point, any steam that does result recondenses back into water. What If? Suppose you are performing an experiment in the laboratory that uses this technique to determine the specific heat of a sample and you wish to decrease the overall uncertainty in your final result for cx. Of the data given in the text of this example, changing which value would be most effective in decreasing the uncertainty? Answer The largest experimental uncertainty is associated with the small temperature difference of 2.4°C for Tf  Tw. For example, an uncertainty of 0.1°C in each of these two temperature readings leads to an 8% uncertainty in their difference. In order for this temperature difference to be larger experimentally, the most effective change is to decrease the amount of water. Example 2.3 A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume that all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet? Solution The kinetic energy of the bullet is 50  1 K mv 2 2 Because nothing in the environment is hotter than the bullet, the bullet gains no energy by heat. Its temperature increases because the kinetic energy is transformed to extra internal energy when the bullet is stopped by the wall. The temperature change is the same as that which would take place if energy Q = K were transferred by heat from a stove to the bullet. If we imagine this latter process taking place, we can calculate T from Equation 20.4. Using 234 J/kg.°C as the specific heat of silver (see Table 2.1), we obtain K 1 2 m200 m/s  2 Q T     85.5 C  mc mc m 234 J/kg C   (1) Note that the result does not depend on the mass of the bullet. What If? Suppose that the cowboy runs out of silver bullets and fires a lead bullet at the same speed into the wall. Will the temperature change of the bullet be larger or smaller? Answer Consulting Table 2.1, we find that the specific heat of lead is 128 J/kg.°C, which is smaller than that for silver. Thus, a given amount of energy input will raise lead to a higher temperature than silver and the final temperature of the lead bullet will be larger. In Equation (1), we substitute the new value for the specific heat: K 1 2 m200 m/s  2 Q T     156 C  mc mc m 128 J/kg C   Note that there is no requirement that the silver and lead bullets have the same mass to determine this temperature. The only requirement is that they have the same speed. 51 Applications Mechanical equivalent of heat ( Joule's experiment) James Prescott Joule (1818 - 1889) performed a series of experiments to prove convincingly that heat is also a form of energy. His findings that there is definite equivalence between the work done and the corresponding amount of heat and amount of heat and any other form of energy put an end to the controversy that existed between the caloric theory and the dynamic theory of heat was firmly established. He developed the argument that if an amount of work W (or any other form of energy ) disappears a definite quantity of heat Q is produced W Q W = JQ J= W/Q The constant J is called the mechanical equivalent of heat. The mechanical equivalent of heat (J) is defined as the amount of mechanical work done in joule to produce a unit quantity (1 calorie ) of heat. The value of J in C.G.S system is 4.18 *107 erg/calorie while in the M.K.S system its value is 4.18 joules / calorie. Joule's method for determining the mechanical equivalent of heat: The apparatus consists of a calorimeter with the fixed blades a, a etc., fixed to the calorimeter radically. S is the spindle which is fitted with the vanes b , b radically and these vanes are called the movable vanes ( next Fig) 52  When the spindle rotates , the vanes a ,a are stationary and the vanes b , b rotate. The water between any two vans get warmed up due to the friction forces between the two vans. W is the wheel with a handle H, which can rotate about a vertical axis. Two strings are wound in the same direction over the drum W and these strings pass over the pulleys and they carry the masses M, M at the other ends. When the masses are moved up and left they move down due to gravity thus rotating the drum and the spindle. When the masses are moved up the pin P is removed and when the masses move down the P is fixed. This is because when the masses are moved up, the work done is not taken into calculation Let the masses move down by a distance of h cm. Loss of potential energy of the two masses = 2 Mgh If the masses fall down n times , The loss of potential energy = n x 2 Mgh Let the velocity of the mass at the end of the fall be v cm / sec Kinetic energy of the two masses = 2½ M v2 = M v2. For n times the kinetic energy = n M v2 Energy used up in heating the water and the calorimeter , W = n (2Mgh -Mv2) 53  Let the total water equivalent of the calorimeter , water , fixed vanes , movable vanes etc. be w in W   m i , Ci i 1 where m is the mass and c is the specific heat of ( water, fixed vanes, movable vanes etc. ) Initial temperature = I Final temperature = 2 Heat produced(Q) = w (2 - 1) J , the mechanical equivalent of heat J  W n 2 Mgh  MV1   Q w 2  1  The value of J is taken as a mean of a number of observations, the value of J is 4.18 Joule/calorie.. Determination of the Electrical Equivalent of Heat (Using Joule's Calorimeter) The apparatus consists of a copper calorimeter containing water to about 2/3 of its volume. A resistance coil R is immersed in water and two ends of R are connected to two binding terminals fixed to the lid A stirrer and a thermometer are inserted through holes in the lid. The calorimeter is closed in a wooden box to minimize the loss of heat ( next Fig ) 54  (Fig2) The electrical connections are shown in the Fig. A current of I ampere is passed through the resistance coil R at a potential difference of V volts for a time t seconds , Then the amount of energy liberated = V. I.t Joule (1) Knowing the total water equivalent w, and the initial and finial temperatures of the calorimeter 1 and 2 the produced Q can be calculated Q = w(2-1) (2) mechanical equivalent of heat J = work done / heat produced J = W/Q using equations (1) and (2) VI t VI t J  w  2  1  mc Cc  mwCw  2  1  55 where mc is the mass of the calorimeter ,mw is the mass of the water, cc and CW are the specific heats of the calorimeter and the water respectively. In case, the\ resistance of the voltmeter is not high, the actual value of current passing through R is calculated, knowing the values of R and the resistance of the voltmeter. The drawback of this experiment is the small rate of rise of temperature due to low currents that can be drawn from accumulators used as the source of supply. This experiment can also be used to determine the specific heat of a liquid when J is known. Calendar and Barnes' continues flow method (determination of.J): The Calendar and Barnes' apparatus consists of a resistance coil R enclosed inside a narrow glass tube (Fig. 3 ). The two ends of the wire are connected to two thick copper connectors. Continues flow of water can be maintained through the tube and the temperature of inlet and outlet water are measured with thermometer in and out. The central tube containing a resistance R is surrounded by a vacuum jacket which minimizes the loss of heat by conduction and convection. The coil R is connected in series with a battery, a rheostat ,an ammeter and a key.A voltmeter is connected across the terminals of (R). For more accurate work the current through the coil and the 56 potential difference across its ends are measured using a potentiometer initially calibrated with a standard cell. When the rate of flow of water and the current passing through (R) are maintained constant, a steady state will be reached and the thermometers TI and T2 will show constant readings. At this stage ,the temperatures of every part of the apparatus will remain steady. When thermometers in and out show constant readings , say in and out ,the mass of water (m) flowing in time (t) seconds is measured.If Cw is the mean value of » specific heat of water between the temperatures in and out and Qrad is the quantity of heat lost by radiation, then: The amount of heat produced by passage of electric current through the coil= mcw (1- 2) + Qrad (1) If (V) and (I) are the voltmeter and ammeter readings then the amount of energy liberated in time (t) sec. - V I t joules. The JVt corresponding quantity of heat produced = cals (2) J from (1),(2) IVt J    m Cw  out   in  Qrad (3) To eliminate Qrad a second set of observations is taken by altering the values of current of water such that the difference of temperature (out - in) remains the same after the steady state is reached. If V' , I' and m' are the corresponding values for the same time t, then I V t  mCw  out   in   Qrad (4) J Note ; As the temperature difference (out - in) is maintained the same in the two cases, the temperatures of the various parts of the apparatus will be the same and hence the loss of heat by radiation etc. will also be the same. Subtracting (4) from (3) 57  IV  IVt  m  mc     w out in J J IV  IVt joule / cal m  mc w out  in  All the quantities in the above expression can be measured very accurately. With calibrated platinum resistance thermometers , the temperature difference can be measured with great precision. Similarly, the values of the current and the potential difference can also be measured very accurately. Thus this method ensures an accurate determination of the value of J, This method can also used to determine the specific heat of a liquid if J is known. 2.3 Latent Heat A substance often undergoes a change in temperature when energy is transferred between it and its surroundings. There are situations, however, in which the transfer of energy does not result in a change in temperature. This is the case whenever the physical characteristics of the substance change from one form to another; such a change is commonly referred to as a phase change. Two common phase changes are from solid to liquid (melting) and from liquid to gas (boiling); another is a change in the crystalline structure of a solid. All such phase changes involve a change in internal energy but no change in temperature. The increase in internal energy in boiling, for example, is represented by the breaking of bonds between molecules in the liquid state; this bond breaking allows the molecules to move farther apart in the gaseous state, with a corresponding increase in intermolecular potential energy. As you might expect, different substances respond differently to the addition or removal of energy as they change phase because their internal molecular arrangements vary. Also, the amount of energy transferred during a phase change depends on the amount of 58 substance involved. (It takes less energy to melt an ice cube than it does to thaw a frozen lake.) If a quantity Q of energy transfer is required to change the phase of a mass m of a substance, the ratio L = Q/m characterizes an important thermal property of that substance. Because this added or removed energy does not result in a temperature change, the quantity L is called the latent heat (literally, the “hidden” heat) of the substance. The value of L for a substance depends on the nature of the phase change, as well as on the properties of the substance. From the definition of latent heat, and again choosing heat as our energy transfer mechanism, we find that the energy required to change the phase of a given mass m of a pure substance is Q  mL (2.6) Latent heat of fusion Lf is the term used when the phase change is from solid to liquid (to fuse means “to combine by melting”), and latent heat of vaporization Lv is the term used when the phase change is from liquid to gas (the liquid “vaporizes”). The latent heats of various substances vary considerably, as data in Table 2.2 show. The positive sign in Equation 2.6 is used when energy enters a system, causing melting or vaporization. The negative sign corresponds to energy leaving a system, such that the system freezes or condenses. 59 Table 2.2 Latent Heat of Fusion and. Vaporization Latent Heat Latent Heat of Melting Boiling Substance of Fusion Vaporization Point (°C) Point (°C) (J/kg) (J/kg) Helium - 269.65 5.23  103 - 268.93 2.09  104 Nitrogen - 209.97 * 2.55  10 - 195.81 5 2.01  105 Oxygen -218.79 1.38  104 - 182.97 2.13  105 Ethyl alcohol - 114 1.04  104 78 8.54  105 Water 0.00 3.33  10 5 100.00 2.26  106 Sulfur 119 3.81  104 444.60 3.26  106 Lead 327.3 2.45  104 1 750 8.70 , 107 Aluminum 660 3.97  104 2 450 1.14  107 Silver 960.80 8.82  104 2 193 2.33  105 Gold 1 063.00 6.44  104 2660 1.58 106 Copper 1 083 1.34  105 1 187 5.06  106 Fig. 2.2 A plot of temperature versus energy added when 1.00 g of ice initially at 30.0°C is converted to steam at 120.0°C. 60  To understand the role of latent heat in phase changes, consider the energy required to convert a 1.00-g cube of ice at 30.0°C to steam at 120.0°C. Figure 2.2 indicates the experimental results obtained when energy is gradually added to the ice. Let us examine each portion of the red curve. Part A. On this portion of the curve, the temperature of the ice changes from 30.0°C to 0.0°C. Because the specific heat of ice is 2 090 J/kg.°C, we can calculate the amount of energy added by using Equation 2.4:     Q  mi ci T  1.00  103 kg 2 090 J/kg C 30.0 C  62.7 J Part B. When the temperature of the ice reaches 0.0°C, the ice–water mixture remains at this temperature—even though energy is being added—until all the ice melts. The energy required to melt 1.00 g of ice at 0.0°C is, from Equation 20.6,   Q  mi Li  1.00  103 kg 3.33  105 J/kg  333 J  Thus, we have moved to the 396 J (= 62.7 J + 333 J) mark on the energy axis in Figure 20.2. Part C. Between 0.0°C and 100.0°C, nothing surprising happens. No phase change occurs, and so all energy added to the water is used to increase its temperature. The amount of energy necessary to increase the temperature from 0.0°C to 100.0°C is     Q  m w c w T  1.00  103 kg 4.19  103 J/kg C 100.0 C  419 J Part D. At 100.0°C, another phase change occurs as the water changes from water at 100.0°C to steam at 100.0°C. Similar to the ice–water mixture in part B, the water–steam mixture remains at 100.0°C—even though energy is being added—until all of the liquid has been converted to steam. The energy required to convert 1.00 g of water to steam at 100.0°C is 61     Q  m w L w  1.00  103 kg 2.26  106 J/kg  2.26  103 J Part E. On this portion of the curve, as in parts A and C, no phase change occurs; thus, all energy added is used to increase the temperature of the steam. The energy that must be added to raise the temperature of the steam from 100.0°C to 120.0°C is     Q  ms cs T  1.00  103 kg 2.01  103 J/kg C 20.0 C  40.2 J The total amount of energy that must be added to change 1 g of ice at 30.0°C to steam at 120.0°C is the sum of the results from all five parts of the curve, which is 3.11103 J. Conversely, to cool 1 g of steam at 120.0°C to ice at 30.0°C, we must remove 3.11103 J of energy. Note in Figure 2.2 the relatively large amount of energy that is transferred into the water to vaporize it to steam. Imagine reversing this process—there is a large amount of energy transferred out of steam to condense it into water. This is why a burn to your skin from steam at 100°C is much more damaging than exposure of your skin to water at 100°C. A very large amount of energy enters your skin from the steam and the steam remains at 100°C for a long time while it condenses. Conversely, when your skin makes contact with water at 100°C, the water immediately begins to drop in temperature as energy transfers from the water to your skin. We can describe phase changes in terms of a rearrangement of molecules when energy is added to or removed from a substance. (For elemental substances in which the atoms do not combine to form molecules, the following discussion should be interpreted in terms of atoms. We use the general term molecules to refer to both chemical compounds and elemental substances.) Consider first the liquid-to-gas phase change. The molecules in a liquid are close together, and the forces between them are stronger than those between the more widely separated molecules of a gas. Therefore, work must be done on the liquid against these attractive molecular forces if the molecules are to 62 separate. The latent heat of vaporization is the amount of energy per unit mass that must be added to the liquid to accomplish this separation. Similarly, for a solid, we imagine that the addition of energy causes the amplitude of vibration of the molecules about their equilibrium positions to become greater as the temperature increases. At the melting point of the solid, the amplitude is great enough to break the bonds between molecules and to allow molecules to move to new positions. The molecules in the liquid also are bound to each other, but less strongly than those in the solid phase. The latent heat of fusion is equal to the energy required per unit mass to transform the bonds among all molecules from the solid-type bond to the liquid-type bond. As you can see from Table 2.2, the latent heat of vaporization for a given substance is usually somewhat higher than the latent heat of fusion. This is not surprising if we consider that the average distance between molecules in the gas phase is much greater than that in either the liquid or the solid phase. In the solid-to-liquid phase change, we transform solid-type bonds between molecules into liquid- type bonds between molecules, which are only slightly less strong. In the liquid-to-gas phase change, however, we break liquid-type bonds and create a situation in which the molecules of the gas essentially are not bonded to each other. Therefore, it is not surprising that more energy is required to vaporize a given mass of substance than is required to melt it. Quick Quiz 2.3 Suppose the same process of adding energy to the ice cube is performed as discussed above, but we graph the internal energy of the system as a function of energy input. What would this graph look like? 63 Quick Quiz 2.4 Calculate the slopes for the A, C, and E portions of Figure 2.2. Rank the slopes from least to greatest and explain what this ordering means. Calorimetry Problems If you have difficulty in solving calorimetry problems, be sure to consider the following points: Units of measure must be consistent. For instance, if you are using specific heats measured in J/kg.°C, be sure that masses are in kilograms and temperatures are in Celsius degrees.  Transfers of energy are given by the equation Q = mcT only for those processes in which no phase changes occur. Use the equations Q = mLf and Q = mLv only when phase changes are taking place; be sure to select the proper sign for these equations depending on the direction of energy transfer.  Often, errors in sign are made when the equation Qcold = Qhot is used. Make sure that you use the negative sign in the equation, and remember that T is always the final temperature minus the initial temperature. Example 2.4 What mass of steam initially at 130°C is needed to warm 200 g of water in a 100-g glass container from 20.0°C to 50.0°C? Solution The steam loses energy in three stages. In the first stage, the steam is cooled to 100°C. The energy transfer in the process is 64     Q1  ms cs T  ms 2.01  103 J/kg C  30.0 C  ms 6.03  104 J/kg   where ms is the unknown mass of the steam. In the second stage, the steam is converted to water. To find the energy transfer during this phase change, we use Q = mLv , where the negative sign indicates that energy is leaving the steam:  Q2  ms 2.26  106 J/kg  In the third stage, the temperature of the water created from the steam is reduced to 50.0°C. This change requires an energy transfer of    Q3  ms cw T  ms 4.19  103 J/kg C  50.0 C  ms 2.09  105 J/kg   Adding the energy transfers in these three stages, we obtain Q hot  Q1  Q 2  Q3     m s 6.03  10 4 J/kg  2.26  106 J/kg  2.09  105 J/kg  m s 2.53  106 J/kg  Now, we turn our attention to the temperature increase of the water and the glass. Using Equation 2.4, we find that    Qcold  0.200 kg  4.19  103 J/kg C 30.0 C  0.100 kg  837 J/kg C30.0 C  2.77 10   4 J Using Equation 20.5, we can solve for the unknown mass: Q cold  Q hot   2.77  10 4 J    m s 2.53  10 6 J/kg  ms  1.09  102 kg  10.9 g 65  What If? What if the final state of the system is water at 100°C? Would we need more or less steam? How would the analysis above change? Answer More steam would be needed to raise the temperature of the water and glass to 100°C instead of 50.0°C. There would be two major changes in the analysis. First, we would not have a term Q3 for the steam because the water that condenses from the steam does not cool below 100°C. Second, in Qcold, the temperature change would be 80.0°C instead of 30.0°C. Thus, Qhot becomes Q hot  Q1  Q 2     m s 6.03  10 4 J/kg  2.26  106 J/kg  m s 2.32  106 J/kg  and Qcold becomes    Qcold  0.200 kg  4.19 103 J/kg C 80.0 C  0.100 kg  837 J/kg C80.0 C  7.37 10   4 J leading to ms  3.18  102 kg  31.8 g S Example 3.5 Liquid helium has a very low boiling point, 4.2 K, and a very low latent heat of vaporization, 2.09 % 104 J/kg. If energy is transferred to a container of boiling liquid helium from an immersed electric heater at a rate of 10.0 W, how long does it take to boil away 1.00 kg of the liquid? Solution Because Lv = 2.09104 J/kg, we must supply 2.09104 J of energy to boil away 1.00 kg. Because 10.0 W = 10.0 J/s, 10.0 J of energy is transferred to the helium each second. From  = E/t, the time interval required to transfer 2.09104 J of energy is 66  E 2.09  10 4 J t     2.09  103 s  35 min Q 10.0 J/s 67