368 11 Thermal Properties of Matter Mass of one Cu atom = M(Cu) 63.546 kg/kmol = NA 6.022 × 1026 atoms/kmol = 1.059 × 10−25 kg/atom Example 11.5 The main constituents of air are nitrogen molecules of molar mass M(N2 ) = 28 kg/kmol and oxygen molecules of molar mass M(O2 ) = 32 kg/kmol with approximate proportions of 80% and 20%, respectively Using the ideal gas law, find the mass of air in a volume of 50 cm3 at a pressure of 700 torr and temperature of 20 ◦ C Solution: The molar mass of air can be obtained from the ratios of the two gases as follows: M(air) = 0.8 M(N2 ) + 0.2 M(O2 ) = 0.8 (28 kg/kmol) + 0.2 (32 kg/kmol) = 28.8 kg/kmol The volume, pressure, and temperature values can be written as: V = 50 cm3 = × 10−5 m3 1.01 × 105 Pa atm × = 9.3 × 104 Pa 760 torr atm T = 20 ◦ C = 20 + 273 = 293 K (From now on, we ignore the 0.15 K) P = 700 torr = 700 torr × We can use the ideal-gas equation PV = nRT with n = m/M(air), where m is the mass of air under consideration, to find m as follows: m= PVM(air) (9.3 × 104 Pa)(5 × 10−5 m3 )(28.8 kg/kmol) = = 5.5 × 10−5 kg RT (8.314 × 103 J/kmol.K)(293 K) Example 11.6 (a) How many molecules are there in cm3 of air at room temperature (27 ◦ C)? (b) How many kilomoles of air are in that volume? (c) The best vacuum that can 11.3 The Ideal Gas 369 be produced corresponds to a pressure of about 10−16 atm How many molecules remain in cm3 ? Solution: The number of molecules in cm3 can be calculated from the ideal gas equation PV = NkB T (a) Rewriting the quantities given, we have: V = cm3 = 10−6 m3 P = atm 105 Pa T = 27 ◦ C = 27 + 273 = 300 K Thus : N = PV (105 Pa)(10−6 m3 ) = = 2.4 × 1019 molecules kB T (1.38 × 10−23 J/K)(300 K) (b) We use the ideal gas equation PV = nRT to calculate the number of kilomoles as follows: n= PV (105 Pa)(10−6 m3 ) = = × 10−8 kmol RT (8.314 × 103 J/kmol.K)(300 K) Also, we can use the relation N = n NA to get n as follows: n= N 2.4 × 1019 molecules = × 10−8 kmol = NA 6.022 × 1026 molecules/kmol (c) Rewriting the quantities given, we have: V = cm3 = 10−6 m3 P = 10−16 atm 10−11 Pa T = 27 ◦ C = 27 + 273 = 300 K Thus : N= (10−11 Pa)(10−6 m3 ) PV = = 2,415 molecules kB T (1.38 × 10−23 J/K)(300 K) There are still a large number of molecules left in this cm3 vacuum Example 11.7 A metal barrel is filled with air and is closed firmly when the pressure is atm and the temperature is 20 ◦ C On a hot sunny day, the barrel’s temperature rises 370 11 Thermal Properties of Matter to 60 ◦ C while its volume remains almost the same Find the final pressure inside the barrel Solution: We mark the initial state of air with P1 , V1 , T1 and final state with P2 , V2 , T2 , see Fig 11.7 If no air escapes from the barrel, the number of moles of air n remains constant Therefore, using the ideal gas law PV = nRT in the initial and final states and the fact that V2 = V1 , we get: T2 P1 V1 P2 V2 P1 P2 = ⇒ = ⇒ P2 = P1 T1 T2 T1 T2 T1 ⎧ ⎪ ⎪ ⎨ P1 = atm = 1.01 × 10 Pa The quantities given are: T1 = 20 ◦ C = 20 + 273 = 293 K ⎪ ⎪ ⎩ T = 60 ◦ C = 60 + 273 = 333 K nR = Thus: P2 = (1 atm) 333 K = 1.14 atm = 1.15 × 105 Pa 293 K Fig 11.7 V2 =V1 P1 P2 Air T1 inside Before V2 V1 T2 Air inside After Example 11.8 The initial volume, pressure, and temperature of helium gas trapped in a container with a movable piston are × 10−3 m3 , 150 kPa, and 300 K, respectively; see Fig 11.8 If the volume is decreased to 1.5 × 10−3 m3 and the pressure increases to 300 kPa find the final temperature of the gas, assuming it behaves like an ideal gas Fig 11.8 Pi ,Vi , Ti Pf ,Vf , Tf 11.3 The Ideal Gas 371 Solution: The initial state of helium is Pi , Vi , Ti and final state is Pf , Vf , Tf With the use of the ideal gas law PV = nRT , we get: Pi Vi Pf Vf = Ti Tf ⇒ Tf = Ti Pf Vf Pi Vi = (300 K) 11.4 (300 kPa)(1.5 × 10−3 m3 ) = 450 K (150 kPa)(2 × 10−3 m3 ) Exercises Section 11.1 Temperature (1) Convert the temperatures −30 ◦ C, 10 ◦ C, and 50 ◦ C to Kelvin and Fahrenheit (2) Express the normal human body temperature, 37 ◦ C, and the sun’s surface temperature, ∼6000 ◦ C, in Fahrenheit and Kelvin (3) A Celsius thermometer indicates a temperature of −40 ◦ C (a) What Fahrenheit and Kelvin temperatures correspond to this Celsius temperature? (b) If the temperature changes from −40 ◦ C to +10 ◦ C, find the change in temperature on the Fahrenheit scale (4) The normal melting point of gold is 1064.5 ◦ C and its boiling point is 2660 ◦ C (a) Convert these two values to the Fahrenheit and Kelvin scales (b) Find the difference between those two values in Celsius (c) Repeat (b) using the Kelvin scale (5) The height of an alcohol column in an alcohol thermometer has a length 12 cm at ◦ C and a length 22 cm at 100 ◦ C Assume that the temperature and the length of the alcohol thermometer are linearly related What is the temperature that the thermometer will measure if the alcohol column has a length 12.5 cm? Section 11.2 Thermal Expansion of Solids and Liquids (6) The Eiffel tower is built from iron and it is about 324 m high Its coefficient of linear expansion is approximately 12 × 10−6 (C◦ )−1 and assumed constant What is the increase in the tower’s length when the temperature changes from ◦ C in winter to 30 ◦ C? 372 11 Thermal Properties of Matter (7) A copper rod is m long at 20 ◦ C and has a coefficient of linear expansion α = 17 × 10−6 (C◦ )−1 What is the increase in the rod’s length when it is heated to 40 ◦ C? (8) A road is built from concrete slabs, each of 10 m long when formed at 10 ◦ C, see Fig 11.9 How wide should the expansion cracks between the slabs be at 10 ◦ C to prevent road buckling if the range of temperature changes from −5 ◦ C in winter to +40 ◦ C in summer? The coefficient of linear expansion for concrete is α = 12 × 10−6 (C◦ )−1 Concrete slab Concrete slab 10 m 10 m Expansion crack Fig 11.9 See Exercise (8) (9) An iron steam pipe is 100 m long at ◦ C and has a coefficient of linear expansion α = 10 × 10−6 (C◦ )−1 What will be its length when heated to 100 ◦ C? (10) An ordinary glass window has a coefficient of linear expansion α = × 10−6 (C◦ )−1 At 20 ◦ C the sides a and b have the values m and 0.8 m respectively, see Fig 11.10 By how much does the area increase when its temperature rises to 40 ◦ C? Fig 11.10 See Exercise (10) Δa a b Δa Δb Δb (11) A steel tape measure has a coefficient of linear expansion α = 12×10−6 (C◦ )−1 and is calibrated at 20 ◦ C On a cold day when the temperature is −20 ◦ C, what will be the percentage error for a reading made using this tape measure? 11.4 Exercises 373 (12) A bar of length L = m and linear expansion α = 25 × 10−6 (C◦ )−1 has a crack at its center The ends of the bars are fixed as shown in Fig 11.11 As a result of a temperature rise of 40 C◦ , the bar buckles upwards, see Fig 11.11 Find the vertical rise d of the bar’s center Fig 11.11 See Exercise (12) T L T+ΔT d L (13) A composite rod of length L is made from two different rods of lengths L1 and L2 with linear expansion coefficients of α1 and α2 , respectively, see Fig 11.12 (a) Show that the coefficient of linear expansion α for this composite rod is given by α = (α1 L1 + α2 L2 )/L (b) Using the linear expansion coefficients of steel and brass given in Table 11.2, find L1 and L2 in the case where L = 0.8 m and α = 14 × 10−6 (C◦ )−1 Fig 11.12 See Exercise (13) L1 L2 L (14) A homogeneous metal ring of temperature T has inner and outer radii a and b, respectively As the metal ring is heated to a temperature of T + T , its inner and outer radii increase linearly to a + a and b + b respectively, see Fig 11.13 Show that the heating has no effect on the ratio between the inner and the outer radii b+Δb Fig 11.13 See Exercise (14) b a T a+Δa T+ΔT 374 11 Thermal Properties of Matter (15) A spherical brass plug has a diameter d of 10 cm at T = 150 C◦ and has a coefficient of linear expansion α = 19 × 10−6 (C◦ )−1 , see Fig 11.14 At what temperature will its diameter be 9.950 cm? Fig 11.14 See Exercise (15) r−|Δ r| r T −| Δ T | T (16) Two rods of the same diameter, one made of brass of length L1 = 25 cm, and the other rod made of steel of length L2 = 50 cm, are placed end-to-end and pinned to two rigid supports, see Fig 11.15 The Young’s modulus for the brass and steel rods are Y1 = 100 × 109 N/m2 and Y2 = 200 × 109 N/m2 respectively, and their respective coefficients of linear expansion are α1 = 18 × 10−6 (C◦ )−1 and α2 = 12 × 10−6 (C◦ )−1 The two rods are heated until the rise in temperature becomes T = 40 C◦ What is the stress in each rod? Fig 11.15 See Exercise (16) T L L2 30.00 1mm L' 27.35 mm F F 59.63 mm T + ΔT L'2 62.27 mm F F (17) Two parallel metal bars with the same length L and negligible width, but different linear expansion coefficients α1 and α2 , are fixed at a distance d apart, see Fig 11.16 When their temperature changes by T , they will bend into two circular arcs intercepting at an angle θ as shown in Fig 11.16 Find their mean radius of curvature r (18) Find the change in volume of an aluminum sphere that has a radius of cm when it is heated from ◦ C to 300 ◦ C Assume that the coefficient of volume expansion is β = 7.2 × 10−5 (C◦ )−1 (19) A glass flask holds 50 cm3 at a temperature of 20 ◦ C What is its capacity at 30 ◦ C? Assume the coefficient of volume expansion of this glass flask is 2.7 × 10−5 (C◦ )−1 11.4 Exercises 375 Fig 11.16 See Exercise (17) d 14.82 14 d16 45.00 mm L2 L1 L 2 r1.30 53 r 61.0 q r2 69.51 mm T + ΔT T (20) A flask is completely filled with mercury at 20 ◦ C and is sealed off, see Fig 11.17 Ignore the expansion of the glass and assume that the bulk modulus of mercury is B = 2.5 × 109 N/m2 and its coefficient of volume expansion is β = 1.82 × 10−4 (C◦ )−1 Find the change in pressure inside the flask when it is heated to 100 ◦ C Fig 11.17 See Exercise (20) Sealed flask T=20 °C Mercury (21) A glass flask of volume 200 cm3 is filled with mercury when the temperature is T = 20 ◦ C, see Fig 11.18 The coefficient of volume expansion of the glass and mercury are β = 1.2 × 10−5 (C◦ )−1 and β = 18 × 10−5 (C◦ )−1 respectively How much mercury will overflow when the temperature of the flask is raised to 100 ◦ C? T + ΔT T Mercury Fig 11.18 See Exercise (21) (Take atm Mercury 105 Pa unless specified) 376 11 Thermal Properties of Matter Section 11.3 The Ideal Gas (22) Find the density of nitrogen (N2 ) and oxygen (O2 ) at STP assuming they behave like an ideal gas (23) A tank contains 0.5 m3 of nitrogen at a pressure of 1.5 × 105 N/m2 and a temperature of 27 ◦ C (a) What will be the pressure if the volume is increased to 5.0 m3 and the temperature is raised to 327 ◦ C? (b) Answer part (a) if the volume remains constant (24) A tank contains nitrogen N2 at an absolute pressure of 2.5 atm What will be the pressure of an equal mass of CO2 that replaces the nitrogen at the same temperature? (25) A tire is filled with air at 27 ◦ C in a normal day to a gauge pressure of atm Then its temperature reaches 40 ◦ C in a hot day What fraction of the original air must be removed if the original pressure is to be restored? (26) A 1,000 L container holds 50 kg of argon gas at 27 ◦ C The molar mass of argon is M = 40 kg/kmol What is the pressure of the gas? (27) A bubble of air rises from the bottom of a lake, where the pressure is atm and the temperature is ◦ C, to the surface, where the pressure is atm and the temperature is 27 ◦ C, see Fig 11.19 What is the ratio of the volume of the bubble just as it reaches the surface to its volume at the bottom? Fig 11.19 See Exercise (27) atm 27 °C bubble atm °C (28) (a) How many molecules are there in L of air at a temperature of 27 ◦ C? (b) How many kilomoles of air are in that volume? (c) The best vacuum that can be produced corresponds to a pressure of about 10−16 atm How many molecules remain in L? (29) A cylindrical metallic container is filled with air and is closed firmly when the pressure is Pi = atm and the temperature is Ti = 27 ◦ C In a very hot sunny day, the container’s temperature rises to Tf = 70 ◦ C while its volume remains almost the same, see Fig 11.20 Find the final pressure inside the container 11.4 Exercises 377 Fig 11.20 See Exercise (29) V2 =V1 P2 P1 V1 Air inside T1 Before V2 T2 Air inside After (30) The main constituents of air are nitrogen molecules of molar mass M(N2 ) = 28 kg/kmol and oxygen molecules of molar mass M(O2 ) = 32 kg/kmol with approximate ratios of 80 and 20%, respectively Using the ideal gas law, find the mass of air in a volume of L at atmospheric pressure and temperature of 27 ◦ C (31) The initial volume, pressure, and temperature of helium gas trapped in a container with a movable piston are Vi = L, Pi = 150 kPa, and Ti = 300 K, respectively, see Fig 11.21 If the volume is decreased to Vf = 2.5 L and the pressure increases to Pf = 300 kPa, find the final temperature of the gas assuming that it behaves like an ideal gas Fig 11.21 See Exercise (31) Pi ,Vi , Ti Initial Pf ,Vf , Tf Final (32) The volume of an oxygen tank is 50 L As oxygen is withdrawn from the tank, the pressure of the remaining gas in the tank drops from 20 atm to atm, and the temperature also drops from 30 to 10 ◦ C (a) How many kilograms of oxygen were originally in the tank? (b) How many kilograms of oxygen were withdrawn from the tank? (c) What volume would be occupied by the oxygen that withdrawn from the tank at a pressure of atm and a temperature of 27 ◦ C? (33) A balloon filled with helium is left free on the surface of the ground when the temperature is 27 ◦ C When the balloon reaches an altitude of 3,000 m, where the temperature is ◦ C and the pressure is 0.65 atm, how will its volume compare to the original volume on the ground? 378 11 Thermal Properties of Matter (34) The density of water vapor at exactly 100 ◦ C and atm = 1.013 × 105 Pa is ρ = 0.598 kg/m3 Calculate the density of water vapor, with a molecular mass M = 18 kg/kmol, from the ideal gas law Why would you expect a difference? (35) An empty room of volume V contains air having a molar mass M At atmospheric pressure Pa , the mass and temperature of the room are initially mi and Ti , respectively Assuming that the room is maintained at atmospheric pressure while its temperature is increased to Tf , show that the final mass of air left in the room, mf , will be given by: mf = mi − Pa V M R 1 − Ti Tf Heat and the First Law of Thermodynamics 12 Our focus in this chapter will be on the concept of internal energy, energy transfer, the first law of thermodynamics, and some applications of this law The first law of thermodynamics expresses the general principle of conservation of energy According to this law, an energy transfer to or from a system by either heat or work can change the internal energy of the system 12.1 Heat and Thermal Energy It is important to make a major distinction between heat and internal energy (thermal energy) Internal energy is all the energy of a system that is associated with its microscopic constituents Internal energy includes kinetic energy of random translational, rotational, and vibrational motion of molecules, potential energy of molecules and between molecules Heat is defined as the transfer of energy from one system to another due to a temperature difference between them 12.1.1 Units of Heat, The Mechanical Equivalent of Heat Previously, heat was measured in terms of its ability to raise the temperature of water Thus, the calorie (cal), in cgs units, was defined as the amount of heat required to raise the temperature of g of water from 14.5 to 15.5 ◦ C The H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_12, © Springer-Verlag Berlin Heidelberg 2013 379 380 12 Heat and the First Law of Thermodynamics ‘Calorie’ with a capital C, used by nutritionists, is a kilocalorie (1 Cal = kcal = 103 cal).The British Thermal Unit (Btu) was also defined as the amount of heat required to raise the temperature of lb of water from 63 to 64 ◦ F Since heat is now known as transferred energy, the SI unit for it is the joule (J) In a famous experiment, see Fig 12.1, Joule measured the calorie (cal) by converting mechanical energy into heat energy, expressed as an increase in water temperature Fig 12.1 Joule’s experiment for measuring the mechanical Thermometer equivalent of heat from the m temperature rise in water h Thermal insulator Water Joule found that the loss in mechanical energy is proportional to the increase in temperature of the water The proportionality constant was found to be equal to 4,180 J/kg.C◦ Hence, 4,180 J of mechanical energy will raise the temperature of kg of water from 14.5 to 15.5 ◦ C One kilocalorie (1 kcal) is now defined to be exactly 4,186 J without reference to the heating of substance Thus: kcal = 4,186 J (12.1) The relations among the various heat units are as follows: J = 2.389 × 10−4 kcal = 9.478 × 10−4 Btu or (12.2) kcal = 4,186 J = 3.968 Btu 12.1.2 Heat Capacity and Specific Heat The quantity of heat energy Q required to raise the temperature of an object by some amount T varies from one substance to another 12.1 Heat and Thermal Energy 381 The heat capacity C of an object is defined as: The Heat Capacity C: The heat capacity C of an object of a particular material is defined as the amount of heat energy needed to raise the object’s temperature by one degree Celsius Accordingly, if Q units of heat energy are required to change the temperature by T = Tf − Ti , where Ti and Tf are the initial and final temperatures of the object, then: Q=C T , where T = Tf − Ti (12.3) Heat capacity C has the unit J/C◦ (≡J/K) or kcal/C◦ (≡kcal/K) The heat capacity for any object is proportional to its mass m For this reason, we define the “heat capacity per unit mass” or the specific heat c which refers to a unit mass of the material of which the object is made Thus, with C = m c, Eq 12.3 becomes: Q = m c T , where T = Tf − Ti (12.4) Specific heat c has the unit: J/kg.C◦ ≡ J/kg.K or kcal/kg.C◦ ≡ kcal/kg.K The specific heat of water at 15 ◦ C and atmospheric pressure is: cwater = 4,186 J/kg.K = kcal/kg.C◦ Note that, when heat energy is added to objects, Q and T are both positive, i.e the temperature increases Likewise, when heat is removed from objects, Q and T are both negative, i.e the temperature decreases In general, specific heat c varies with temperature However, if temperature intervals are not too big, the temperature variation can be ignored, and c can be treated as a constant For example, the specific heat of water varies by about 1% from to 100 ◦ C at atmospheric pressure Table 12.1 presents some specific heat values for various substances, measured at room temperature and atmospheric pressure 382 12 Heat and the First Law of Thermodynamics Table 12.1 Specific heat c of some substances at atmospheric pressure and room temperature (20 ◦ C) with few exceptions Substance Specific heat J/kg.C◦ kcal/kg.C◦ Silver 230 0.0564 Copper 390 0.0923 Iron or steel 450 0.107 Aluminum 900 0.215 Brass 380 0.092 Granite 790 0.19 Glass 840 0.20 Ice (−5 ◦ C) 2,100 0.50 2,220 0.530 Mercury 140 0.033 Alcohol (Ethyl) 2,400 0.58 Seawater 3,900 0.93 Water (15 ◦ C) 4,186 2,010 0.48 Ice (−10 ◦ C) Steam (100 ◦ C) Measuring Specific Heat Figure 12.2 shows an example of a calorimeter, which is a device used to determine the specific heat of a solid or liquid substance The substance (represented by a circular object, having a specific heat cx and mass mx ) is heated up to some known initial temperature Tx , and then placed in a perfectly insulated vessel containing water of specific heat cw , mass mw , and initial temperature Tw If Tf is the final temperature after reaching equilibrium, then Tw < Tf < Tx Using Eq 12.4, we calculate the heat gained by the water to be Q = mw cw (Tf − Tw ), and calculate the heat energy lost by the object to be −Q = mx cx (Tf − Tx ) Assuming that the entire system does not lose or gain any heat from its surrounding, then the heat gained by the water must equal the heat lost by the object That is: Q = mw cw (Tf − Tw ) = −mx cx (Tf − Tx ) (12.5) 12.1 Heat and Thermal Energy 383 Solving for cx gives: cx = cw mw (Tf − Tw ) , mx (Tx − Tf ) (Tw < Tf < Tx ) (12.6) When calculating cx , we neglected heat exchange with the vessel, which is acceptable when the mass of the water is considerably larger than that of the vessel, and when the vessel has a negligible specific heat Tw cx , mx Thermal insulator Object Tx + Tw cw , mw Tf = Tf Water Initial Initial Tf Final Fig 12.2 In the method of mixtures, a calorimeter filled with water is used to find the specific heat of unknown heated objects Example 12.1 The specific heat of zinc is 352 J/kg.C◦ for temperatures near 25 ◦ C Determine the amount of heat required to raise the temperature of 0.5 kg zinc from 20 to 30 ◦ C Take the specific heat to be constant in that temperature range Solution: The given values are c = 352 J/kg.C◦ , m = 0.5 kg, Ti = 20 ◦ C, and Tf = 30 ◦ C The temperature change has the following magnitude: T = Tf − Ti = 30 ◦ C − 20 ◦ C = 10 C◦ Using Eq 12.4 we find the amount of heat required as follows: Q = m c T = (0.5 kg)(352 J/kg.C◦ )(10 C◦ ) = 1,760 J 384 12 Heat and the First Law of Thermodynamics Example 12.2 A steel metal object of mass 0.05 kg is heated to 225 ◦ C and then dropped into a vessel containing 0.55 kg of water initially at 18 ◦ C When equilibrium is reached, the temperature of the mixture is 20 ◦ C Find the specific heat of the metal Solution: For the steel metal object, we are given mx = 0.05 kg and Tx = 225 ◦ C, but its specific heat cx is unknown For water, the known values are mw = 0.55 kg, Tw = 18 ◦ C, and cw = 4,186 J/kg.C◦ (Table 12.1) For the mixture, the equilibrium temperature occurs at Tf = 20 ◦ C Since the heat gained by the water is equal in magnitude to the heat lost by the steel, see Eq 12.6 and Fig 12.2, then we must have: mw cw (Tf − Tw ) = −mx cx (Tf − Tx ), (Tw < Tf < Tx ) Solving for cx we get: mw (Tf − Tw ) mx (Tx − Tf ) (0.55 kg) (20 ◦ C − 18 ◦ C) = (4,186 J/kg.C◦ ) (0.05 kg) (225 ◦ C − 20 ◦ C) cx = cw = 449 J/kg.C◦ 12.1.3 Latent Heat When heat energy is transferred from one substance to another, the temperature of the substance often changes However, there are situations in which the transfer of energy does not change the temperature Instead, the substance may change from one form to another Such a change is commonly referred to as a phase change or phase transition, see Sect 13.4 and especially Fig 13.10 We consider the following two main common phase changes: A phase change from solid to liquid (as ice melting) and from liquid to gas (as water boiling), where heat energy is absorbed while the temperature remains constant A phase change from gas to liquid (as steam condensing) and from liquid to solid (as water freezing), where heat energy is released while the temperature remains constant The amount of heat energy per unit mass, L, that must be transferred when a substance completely undergoes a phase change without changing temperature is 12.1 Heat and Thermal Energy 385 called the latent heat (literally, the “hidden” heat) If a quantity Q of heat energy transfer is required to change the phase of a pure substance of a mass m, then L = Q/m characterizes an important thermal property of that substance That is: Q = ±m L (12.7) A positive sign is used in this equation when energy enters the system, causing melting or vaporization of the substance, while a negative sign corresponds to energy leaving the system such that the substance condenses or solidifies When a substance experiences a phase change from solid to liquid by absorbing heat, the heat of transformation is called the latent heat of fusion LF , see Fig 12.3 When the substance releases heat and experiences a phase change from liquid back to solid, the heat of transformation is called the latent heat of solidification and is numerically equal to the latent heat of fusion, see Fig 12.3 In the case of water at its normal melting or freezing temperature, we have: LF = 3.33 × 105 J/kg = 79.5 kcal/kg = 6.01 × 106 J/kmol (12.8) When a substance experiences a phase change from liquid to gas by absorbing heat, the heat of transformation is called the latent heat of vaporization LV , see Fig 12.3 When the gas releases heat and experiences a phase change from gas back to liquid, the heat of transformation is called the latent heat of condensation and is numerically equal to the latent heat of vaporization, see Fig 12.3 For water at its normal boiling and condensation temperatures, we have: LV = 2.256 × 106 J/kg = 539 kcal/kg = 40.7 × 103 J/kmol Latent heat of fusion LF Latent heat of vaporization LV Vaporization Q (positive) Fusion Q (positive) Solid Gas Liquid Solidification Q (negative) (12.9) Condensation Q (negative) Fig 12.3 A sketch showing heat of fusion/vaporization (positive Q) as well as heat of condensation/ solidification (negative Q) 386 12 Heat and the First Law of Thermodynamics Phase changes can be described in terms of a rearrangement of molecules when heat energy is added or removed from a substance Consider, for example, the solidto-liquid phase change The molecules in the solid are strongly attracted to each other As thermal energy is absorbed, the molecules usually move further apart and their potential energy increases (Water-ice is an exception where there is shrinkage.) This leads to no change in the average kinetic energy of the molecules during the melting process, which involves molecules moving from fixed lattice positions to a random liquid state, the temperature stays constant The latent heat of fusion is equal to the work done in separating the molecules during the melting process and hence breaking their bonds and transforming the substance from the ordered solid phase into the disordered liquid phase Now, we consider the liquid to gas phase change The attractive forces between molecules in liquid form are stronger than in gas form because the average distance between molecules is smaller in the liquid state As described in the solid-to-liquid phase transition, work must be done against these attractive forces The latent heat of vaporization is the amount of energy added to the molecules in liquid form to accomplish this Table 12.2 gives some latent heats of various substances Table 12.2 Latent heats of fusion and vaporization (approximates) Substance Melting Melting point ◦ C Latent heat of fusion J/kg Boiling Boiling point ◦ C Latent heat of vaporization J/kg Helium −270 5.23 × 103 −269 2.09 × 104 Nitrogen −210 2.55 × 104 −196 2.01 × 105 −219 1.38 × 104 −183 2.13 × 105 Water 3.33 × 105 100 2.26 × 106 Sulfur 119 3.81 × 104 445 3.26 × 105 Lead 327 2.45 × 104 1,750 8.70 × 105 Aluminum 660 3.97 × 105 2,450 1.14 × 107 Silver 961 8.82 × 104 2,193 2.33 × 106 Gold 1,063 6.44 × 104 2,660 1.58 × 106 Copper 1,083 1.34 × 105 1,187 5.06 × 106 Silicon 1,410 1.65 × 106 2,447 1.06 × 107 Oxygen 12.1 Heat and Thermal Energy 387 To understand the role of latent heat in phase changes, we calculate the energy required to convert g of ice at −50 ◦ C into steam at 150 ◦ C Figure 12.4 shows the results obtained when energy is added gradually to g of ice The red curve of the figure is divided into the following five stages: 150 C 50 A Ice -50 B Ice + Water 111 Water + steam Water 500 1000 444 863 Thermal energy (J) Steam T (oC) E D 100 3000 3123 3224 Fig 12.4 Temperature as a function of the thermal energy added gradually to convert g of ice at −50 ◦ C into steam at 150 ◦ C Stage A—Changing the temperature of ice from −50 to ◦ C: With a specific heat of ice ci = 2,220 J/kg.C◦ , the amount of heat added QA is: QA = mi ci T = (1 × 10−3 kg)(2,220 J/kg.C◦ )(50 C◦ ) = 111 J Stage B— Ice-water mixture remains at ◦ C (even heat is added): With a latent heat of fusion LF = 3.33 × 105 J/kg, the amount of heatadded QB until all of the ice melts is: QB = m LF = (1 × 10−3 kg)(3.33 × 105 J/kg) = 3.33 × 102 J Stage C— Changing the temperature of water from to 100 ◦ C: With a specific heat of water cw = 4,186 J/kg.C◦ , the amount of heat added QC is: QC = mw cw T = (1 × 10−3 kg)(4,186 J/kg.C◦ )(100 C◦ ) 419 J