
CHAPTER 11 ENERGY IN THERMAL PROCESSES Goals: 1. INTRODUCTION
We prepare for the first and second
law of thermodynamics which will be introduced in chapter 12. The first law of
thermodynamics is a statement of conservation of energy which this time, (unlike
the conservation of ME in chapter 5), it is also dealing with the
changes of a system's internal energy. The changes in the internal energy are
caused by heat Q flowing into or out of a system, (Q is examined in this
chapter) and work W done on or by a system (W will be revisited in chapter 12). 2. INTERNAL ENERGY 1) General U = KE + PE = Internal energy U is the grand total energy of all the particles of a system 2) Random nature of the internal KE For example, hot or cold tea do not leap out of the glass, because, even though the individual particles of tea are moving, their constant random change in direction produces no net motion of the bulk of tea. Pouring hot tea into another glass makes its particles obtained an additional preferred direction of motion (the bulk KE). That additional, bulk KE is not considered to be heat energy, rather the KE of the system. In another example on the randomness is the equally distributed air in a room. The notion of randomness will be important for the second law of thermodynamics, at lesson 10. 3. HEAT 1) General: Heat is energy in transit between objects of different temperatures. Objects of same temperature also exchange heat but the net amount exchange is zero because of the random nature of internal KE. 2) Consequences of heat transfer are (a) a temperature change (b) a phase transformation (see latent heat) but not change in temperature, or (c) neither temperature change nor phase transformation (isothermal processes, more in lesson 10)
4. SPECIFIC HEAT 1) Specific heat = c Consider those cases that Q causes a temperature change. Then, we can ask the question: how much heat is required to flow into or out of an object of mass m, in order that its temperature will change by an amount DT? The answer is given by c, the specific heat: c = Q/(mDT) => Q=mcDT 2) Example: c_{w} = 4186 j/(kg Celsius) 3) Then, assuming that the system does not experience a phase transition, we have: a) Q > 0 when heat flows into a system because DT > 0 b) Q < 0 when heat flows out of a system because DT < 0. As a result c is always a positive scalar. 4) Sea breezes 5. PHASE TRANSFORMATIONS 1) What happens during a phase transformation? Q has been converted only into PE, causing physical alteration (changing the atomic bonds, thus creating a phase change) and no portion of this Q has been converted into internal KE, thus the temperature remains the same. 2) Latent heat: Consider a phase transition for which the exchanged heat Q does not cause a temperature change. Then, ask this: how much heat is required to flow into or out of an object of mass m, in order that it will experience a phase transformation? The answer is given by L, the latent heat: L = Q/m => Q = mL Q may be + when heat flows into a system or  when heat flows out of a system L is the amount of heat required to change the phase of an object of mass m. 3) Example: For water, the melting point is 0^{o}C and L_{f} = 3.33 x 10^{5} j/kg. It means that positive Q = 3.33 x 10^{5} j must be supplied to 1 kg of ice initially at 0^{o}C, to melt it and keep it at temperature 0^{o}C. Or, equivalently, negative Q = 3.33 x 10^{5} j must be taken away from 1 kg of water initially at 0^{o}C, to solidify it, thus make it ice and keep it at temperature 0^{o}C. 4) terminology: L_{f} = latent heat of fusion (solid <> liquid) L_{v} = latent heat of vaporization (liquid <> gas) 6. CONSERVATION OF HEAT ENERGY In an isolated system the net flow of heat energy between the objects of the system is zero. Thus, Heat gain = Heat loss In this equation use ΔT
7. EXAMPLES 1) A 0.05Kg metal at 200 Celsius is dropped into a beaker containing 0.4kg of water at 20 degrees Celsius. If the final equilibrium temperature is 22.4 degrees Celsius, find the specific heat of the metal. Ignore any heat absorbed by the beaker. Use the specific heat of water c = 4186 j/(kg Celsius) 2) What mass of steam initially at 130 ^{o}C is needed to warm 0.2 kg of water in a 0.1 kg glass container from 20 ^{ o}C to 50 ^{o}C? Use c(water) = 4186 j/(kg ^{o}C) c(glass) = 837 j/(kg ^{o}C) c(steam) = 2010 j/(kg ^{o}C) Lv (water) = 2.26x10^{6} j/kg 3) An 0.078 kg piece of copper, initially at 290 ^{0}C, is dropped into a 0.245 kg of water contained in a 0.250 kg aluminum calorimeter. The water and calorimeter are initially at 12 ^{0}C. What is the final temperature of the system? The specific heats of water, copper and aluminum are 4186 j/(kg ^{0}C), 387 j/(kg ^{0}C) and 900 j/(kg ^{0}C) respectively. 8. SIMULATION Program 2 The three phases of water
1. INTRODUCTION
2. INTERNAL ENERGY 1) General: Internal energy U = KE + PE = it is the energy related with the microscopic components of a system such as atoms and molecules or in general the particles it is composed of. It includes the KE (translational, rotational and vibrational  note temperature is a measure of this KE) and PE of those particles (i.e., PE of the electrons in an atom) plus interparticle PE (i.e., PE between atoms). PE is stored energy which can be reclaimed to potentially do work. The internal energy does not include the bulk KE of the system (i.e motion of the center of mass of the system or about the center of mass) or the bulk PE of the system due to its interaction with an external force. 2) Random nature of internal KE: The KE portion of the internal energy, a measure of which is obtained through temperature, is a special kind of KE in which many particles move randomly in all directions instead of together in the same direction. The KE portion of the internal energy does not cause an overall motion of the group of particles, that is, of the bulk of the system (of the center of mass of the system) because of its random nature. For example, hot or cold tea do not leap out of the glass, because, even though the individual particles of tea are moving, their constant random change in direction produces no net motion of the bulk of tea. Pouring hot tea into another glass makes its particles obtained an additional preferred direction of motion (the bulk KE). That additional, bulk KE is not considered to be heat energy, rather the KE of the system. The notion of randomness will be important for the second law of thermodynamics, at lesson 10.
3. HEAT 1) General: Heat Q is that portion of the internal energy of a system, that flows from a substance of high temperature to one of lower temperature. Objects of the same temperature are in thermal equilibrium with each other and therefore do not experience a net exchange of energy between them. We emphasize that, even objects in thermal equilibrium are continuously exchanging energy, but the net exchange is zero. Why heat energy is exchanged, and why always flows from the hot to the cold object, will be answered next lesson by the second law of thermodynamics. 2) Q sometimes causes a temperature change and sometimes a phase transformation: Suppose we have a system of two or more objects which can interact thermally (that is, they can exchange energy). Then, heatflow into or out of an object may result in a temperature change in at least one of the objects. If heat flows into or out an object without producing a temperature change, then that object may either experience a phase transformation (see below), or experience an isothermal process (such as that of an ideal gas, see chapter 12). There are cases for which both interacting objects exchange heat isothermally, thus their temperatures remain the same. 3) Why does sometimes temperature change and why does sometimes not? That object that experiences a change in its temperature, as heat Q is exchanged between it and its surroundings, is doing so because either the object's internal KE is increasing (when Q flows into the object), or its internal KE is decreasing (when Q flows out of the object). On the other hand, if an object exchanges heat Q with its surroundings and its internal KE does not change, it means that the exchanged heat Q increases (or decreases) only the internal PE (potential energy) and not at all the internal KE. In such a case, the object is experiencing a phase transformation. Recall that heat can be exchanged between an object and its surroundings via conduction, convection and radiation.
4. SPECIFIC HEAT 1) Specific heat = c Consider those cases that Q causes a temperature change. Then, we can ask the question: how much heat is required to flow into or out of an object of mass m, in order that its temperature will change by an amount DT? Specific heat = c =Q/mDT Þ Q=mcDT: c is a specific amount of heat energy that must be exchanged between a system and its surroundings so that 1 kg of mass experiences 1 degree Celsius temperature change. For example at one atmospheric pressure and at temperature 15 degrees Celsius the specific heat capacity of water is 4186 j/(kg Celsius). Q is the heat exchanged between m and its surroundings resulting in a temperature change DT. Then, assuming that the system does not experience a phase transition, we have: 2a) Q > 0 when heat flows into a system because DT > 0 2b) Q < 0 when heat flows out of a system because DT < 0. As a result c is always a positive scalar. 3) Explain sea breezes based on the different specific heat of land and water.
5. PHASE TRANSFORMATIONS 1) What happens during a phase transformation? During a phase transformation of a system (substance, say water) heat Q is exchanged between the system and its surroundings resulting in physical alteration of the system (phase change) but no temperature change in the system. Since temperature is a measure of the average KE of the particles of a system, during a phase transformation of a system during which its temperature does not change, one can conclude that no portion of the heat Q which has been exchanged between the system and its surroundings has been converted into KE. Rather, all heat Q has been converted into PE causing physical alteration (thus a phase change) in the system. As a result of no temperature change during a phase change we can have coexisting phases at the same temperature. 2) Latent heat Consider a phase transition for which the exchange heat Q does not cause a temperature change. Then, ask this: how much heat is required to flow into or out of an object of mass m, in order that it will experience a phase transformation? L is called the latent heat and is the amount of energy required to change the phase of 1 kg of mass. Then if we have mass m the heat required to change its phase is Q = ± mL. The ± indicates that the energy may flow into, +, or out, , the system. Note that in the formula Q = mcΔT the "direction" of energy flow is determined by the sign of ΔT. Q > 0 when heat flows into a system, and Q < 0 when heat flows out of a system. However, note that in using the conservation of energy equation Heat gain = Heat loss, use ΔT and for a phase transformation use the positive +mL. 3) Example: For water, the melting point is 0^{o}C and L_{f} = 3.33 x 10^{5} j/kg. It means that positive Q = 3.33 x 10^{5} j must be supplied to 1 kg of ice initially at 0^{o}C, to melt it and keep it at temperature 0^{o}C. Or, equivalently, negative Q = 3.33 x 10^{5} j must be taken away from 1 kg of water initially at 0^{o}C, to solidify it and keep it at temperature 0^{o}C. 4) terminology: L_{f} = latent heat of fusion (solid <> liquid) L_{v} = latent heat of vaporization (liquid <> gas)
6. CONSERVATION OF HEAT ENERGY Conservation of heat energy: In an isolated system the net flow of heat between the objects of the system zero. Thus, Heat gain = Heat loss. (In this equation use ΔT)
7. EXAMPLES 1) A 0.05Kg metal at 200 Celsius is dropped into a beaker containing 0.4kg of water at 20 degrees Celsius. If the final equilibrium temperature is 22.4 degrees Celsius, find the specific heat of the metal. Ignore any heat absorbed by the beaker. Use the specific heat of water c = 4186 j/(kg Celsius)
Heat gain = Heat loss => m_{w}c_{w}DT_{w} = m_{metal}c_{metal}DT_{metal}=> => 0.4(4186)(22.4  20) = 0.05c_{metal}(200  22.4) => => 4018.56 = 8.88c_{metal} => c_{metal} = 453j/(kg ^{o}C)
2) What mass of steam initially at 130 ^{o}C is needed to warm 0.2 kg of water in a 0.1 kg glass container from 20 ^{ o}C to 50 ^{o}C? Assume that the final mixture at 50 ^{o}C contains the mass of the steam, which is now of course water. Use c(water) = 4186 j/(kg ^{o}C) c(glass) = 837 j/(kg ^{o}C) c(steam) = 2010 j/(kg ^{o}C) Lv (water) = 2.26x10^{6} j/kg Q=mcDT Q = mL
C_{water} = 4186j/(kg ^{o}C) C_{steam} = 2010j/(kg ^{o}C) C_{glass} = 837j/(kg ^{o}C) L_{vwater} = 2.26x10^{6} j/kg
Heat gain = Heat loss => 0.2(4186)(5020) + 0.1(837)(5020) = = m_{steam}(2010)(130100) + m_{steam}(2.26x10^{6}) + m_{steam}(4186)(10050) => => 27,627 = m_{steam} (60300 + 2.26x10^{6} + 209300) => => m_{steam} = 0.0109 Kg = 1.09x10^{2}Kg
3) An 0.078 kg piece of copper, initially at 290 ^{o}C, is dropped into a 0.245 kg of water contained in a 0.250 kg aluminum calorimeter. The water and calorimeter are initially at 12^{ o}C. What is the final temperature of the system? The following specific heats are given: C_{water} = 4186j/(kg ^{o}C) C_{copper} = 387j/(kg ^{o}C) C_{aluminum} = 900j/(kg ^{o}C)
Heat gain = Heat loss => => 0.245(4186)(T  12) + 0.250(900)(T  12) = 0.078(387)(290  T) => => 1025.57T  12306.84 + 225T  2700 = 8753.94  30.186T => => 1280.756T = 23760.78 => T = 18.6 ^{o}C
4) Heat required to vaporize ice: (See simulation from Program 2 The three phases of water) Find the heat added to the 10^{3}Kg of ice at  30˚C to be raised to 120˚C. The substance is held in a container at constant pressure of one atmosphere. The specific heats of ice is 2090 j/(kg ^{o}C) of water is 4186 j/(kg ^{o}C), of steam is 2010 j/(kg ^{o}C). Also the the latent heat of fusion and of vaporization for water are 3.33x10^{5} j/kg and 2.26x10^{6} j/kg respectively. What is the heat required to be subtracted from the steam at 120˚C so that it becomes ice at 30˚C? The difference in the energies in the various processes is explained by the difference in intermolecular distances (thus forces that water molecules are kept together). Solution: Q= 10^{3} x 2090 x [0(30)] + 10^{3} x 3.33x10^{5} + + 10^{3} x 4186 x (1000) + 10^{3} x 2.26 x 10^{6} + + 10^{3} x 2010x(120100) => => Q = 3.11 x 10^{3} j
Note the process is reversible, so to cool steam of mass 10^{3} Kg from 120˚C to 30˚C we need to remove from it heat of 3.11 x 10^{3} j.
