Rabu, 14 Desember 2011

chapter 20 part II " Thermodinamics "

Chapter 20
Heat and the First Law of Thermodynamics
20.1 Heat and Internal Energy
Internal energy is all the energy of a system that is associated with itsmicroscopiccomponents—atoms and molecules—when viewed from a referenceframe at rest with respect to the center of mass of the system.
Heat is defined as the transfer of energy across the boundary of a systemdue to a temperature difference between the system and its surroundings.
20.2 Specific Heat and Calorimetry

The heat capacity C of a particular sample of a substance is defined as the amountof energy needed to raise the temperature of that sample by 1°C. From this definition,we see that if energy Q produces a changeT in the temperature of a sample, then

20.1 Water, glass, iron. Because water has the highest specificheat (4 186 J/kg .°C), it has the smallest change in temperature.Glass is next (837 J/kg .°C), and iron is last (448 J/kg .°C).
20.2 Iron, glass, water. For a given temperature increase, theenergy transfer by heat is proportional to the specific heat.
Conservation of Energy: Calorimetry
One technique for measuring specific heat involves heating a sample to some knowntemperature Tx, placing it in a vessel containing water of known mass and temperatureTw <Tx, and measuring the temperature of the water after equilibrium has beenreached. This technique is called calorimetry, and devices in which this energy transfer occurs are called calorimeters.
Conservation of energy allows us to write the mathematical representation of this energy statement as
Qcold = - Qhot
The negative sign in the equation is necessary to maintain consistency with our sign convention for heat. Suppose mxis the mass of a sample of some substance whose specific heat we wishto determine. Let us call its specific heat cxand its initial temperature Tx. Likewise, letmw,cw, and Tw represent corresponding values for the water. If Tfis the final equilibriumtemperature after everything is mixed, then from Equation 20.4, we find that theenergy transfer for the water is mwcw(TfT w), which is positive because T f>T w, andthat the energy transfer for the sample of unknown specific heat is mxcx(TfT x),which is negative. Substituting these expressions into Equation 20.5 gives
mwcw(TfT w), = mxcx(TfT x),
Solving for cx gives
cx =   mwcw(TfT w)                                                             mxcx(TfT x)
20.3 Latent Heat
Asubstance 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.
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
The positive sign in Equation 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.

Part A.

Part B.
Part C.

Part D.

Part E.

20.3 The figure below shows a graphical representation ofthe internal energy of the ice in parts A to E as afunction of energy added. Notice that this graphlooksquite different from Figure 20.2—it doesn’thave theflat portions during thephase changes. Regardless ofhow thetemperature is varying inFigure 20.2, theinternalenergy ofthe system simplyincreases linearly with energy input.

20.4 C, A, E. The slope is the ratio of the temperature change to the amount of energy input. Thus, the slopeisproportional to the reciprocal of the specific heat.Water, which has the highest specific heat, has the smallest slope.

Homework
Triangle of the phase differential

20.4 HEAT , AND THE THERMODINAMICS PROCESS
In the macroscopic approach to thermodynamics, we describe the state of a system using such variables as pressure, volume, temperature, and internal energy. As a result, these quantities belong to a category called state variables. For any given configuration of the system, we can identify values of the state variables. It is important to note that a macroscopic state of an isolated system can be specified only if the system is in thermal equilibrium internally. In the case of a gas in a container, internal thermal equilibrium requires that every part of the gas be at the same pressure and temperature. A second category of variables in situations involving energy is transfer variables. These variables are zero unless a process occurs in which energy is transferred across the boundary of the system. Because a transfer of energy across the boundary represents a change in the system, transfer variables are not associated with a given state of the system, but with a change in the state of the system. In the previous sections, we discussed heat as a transfer variable. For a given set of conditions of a system, there is no defined value for the heat. We can only assign a value of the heat if energy crosses the boundary by heat, resulting in a change in the system. State variables are characteristic of a system in thermal equilibrium. Transfer variables are characteristic of a process in which energy is transferred between a system and its environment. In this section, we study another important transfer variable. In this section, we study another important transfer variable for thermodynamic systems—work. Work performed on particles was studied extensively in Chapter 7, and here we investigate the work done on a deformable system—a gas. Consider a gas contained in a cylinder fitted with a movable piston (Fig. 20.3). At equilibrium,

cupies a volume V and exerts a uniform pressure P on the cylinder’s walls and on the piston. If the piston has a cross-sectional area A, the force exerted by the gas on the piston is F=PA. Now let us assume that we push the piston inward and compress the gasquasi-statically, that is, slowly enough to allow the system to remain essentially in thermal equilibrium at all times. As the piston is pushed downward by an external force F=-F ˆj through a displacement of dr ! dyˆj (Fig. 20.3b), the work done on the gas is, according to our definition of work in Chapter 7,

where we have set the magnitude F of the external force equal to PA because the piston is always in equilibrium between the external force and the force from the gas. For this discussion, we assume the mass of the piston is negligible. Because Ady is the change in volume of the gas dV, we can express the work done on the gas as
If the gas is compressed, dV is negative and the work done on the gas is positive. If the gas expands, dV is positive and the work done on the gas is negative. If the volume remains constant, the work done on the gas is zero. The total work done on the gas as its volume changes from Vi to Vf is given by the integral of Equation 20.7:

The energy transfer Q into or out of a system by heat also depends on the process.
Consider the situations depicted in Figure 20.6. In each case, the gas has the same initial volume, temperature, and pressure, and is assumed to be ideal. In Figure 20.6a, the  gas is thermally insulated from its surroundings except at the bottom of the gas-filled region, where it is in thermal contact with an energy reservoir. An energy reservoir is a source of energy that is considered to be so great that a finite transfer of energy to or from the reservoir does not change its temperature. The piston is held at its initial position by an external agent—a hand, for instance. When the force holding the piston is reduced slightly, the piston rises very slowly to its final position. Because the piston is moving upward, the gas is doing work on the piston. During this expansion to the final volume Vf , just enough energy is transferred by heat from the reservoir to the gas to maintain a constant temperature Ti .
The initial and final states of the ideal gas in Figure 20.6a are identical to the initial and final states in Figure 20.6b, but the paths are different. In the first case, the gas does work on the piston, and energy is transferred slowly to the gas by heat. In the second case, no energy is transferred by heat, and the value of the work done is zero. Therefore, we conclude that energy transfer by heat, like work done, depends on the initial, final, and intermediate states of the system. In other words, because heat and work depend on the path, neither quantity is determined solely by the end points of a thermodynamic process.

20.5 THE FIRST LAW OF THERMODINAMICS
When we introduced the law of conservation of energy in Chapter 7, we stated that the change in the energy of a system is equal to the sum of all transfers of energy across the boundary of the system. The first law of thermodynamics is a special case of the law of conservation of energy that encompasses changes in internal energy and energy transfer by heat and work. It is a law that can be applied to many processes and provides a connection between the microscopic and macroscopic worlds.

where all quantities must have the same units of measure for energy. Equation 20.9 is known as the first law of thermodynamics. One of the important consequences of the first law of thermodynamics is that there exists a quantity known as internal energy whose value is determined by the state of the system. The internal energy is therefore a state variable like pressure, volume, and temperature. When a system undergoes an infinitesimal change in state in which a small amount of energy dQ is transferred by heat and a small amount of work dW is done, the internal energy changes by a small amount dEint. Thus, for infinitesimal processes we can express the first law as

The first law of thermodynamics is an energy conservation equation specifying that the only type of energy that changes in the system is the internal energy E int. Let us investigate some special cases in which this condition exists. First, consider an isolated system—that is, one that does not interact with its surroundings. In this case, no energy transfer by heat takes place and the work done on the system is zero; hence, the internal energy remains constant. That is, because Q = W = 0, it follows that #E int ! 0, and thus Eint, i ! Eint, f . We conclude that the internal energy Eint of an isolated system remains constant. Next, consider the case of a system (one not isolated from its surroundings) that is taken through a cyclic process—that is, a process that starts and ends at the same state. In this case, the change in the internal energy must again be zero, because Eint is a state variable, and therefore the energy Q added to the system must equal the negative of the work W done on the system during the cycle. That is, in a cyclic process,

On a PV diagram, a cyclic process appears as a closed curve. (The processes described in
Figure 20.5 are represented by open curves because the initial and final states differ.) It can be shown that in a cyclic process, the net work done on the system per cycle equals the area enclosed by the path representing the process on a PV diagram.

20.6 SOME APLLICATIONS OF THE FIRST LAW OF THERMODINAMICS
The first law of thermodynamics that we discussed in the preceding section relates the changes in internal energy of a system to transfers of energy by work or heat. In this section, we consider applications of the first law to processes through which a gas is taken. As a model, we consider the sample of gas contained in the piston–cylinder apparatus in Figure 20.7. This figure shows work being done on the gas and energy transferring in by heat, so the internal energy of the gas is rising. In the following discussion of various processes, refer back to this figure and mentally alter the directions of the transfer of energy so as to reflect what is happening in the process.
Before we apply the first law of thermodynamics to specific systems, it is useful to
first define some idealized thermodynamic processes. An adiabatic process is one during which no energy enters or leaves the system by heat—that is, Q = 0. An adiabatic process can be achieved either by thermally insulating the walls of the system, such as the cylinder in Figure 20.7, or by performing the process rapidly, so that there is negligible time for energy to transfer by heat. Applying the first law of thermodynamics to an adiabatic process, we see that :
From this result, we see that if a gas is compressed adiabatically such that W is positive, then Eint is positive and the temperature of the gas increases. Conversely, the temperature of a gas decreases when the gas expands adiabatically.
The process described in Figure 20.6b, called an adiabatic free expansion, is unique. The process is adiabatic because it takes place in an insulated container. Because the gas expands into a vacuum, it does not apply a force on a piston as was depicted in Figure 20.6a, so no work is done on or by the gas. Thus, in this adiabatic process, both Q ! 0 and W ! 0. As a result, #Eint ! 0 for this process, as we can see from the first law. That is, the initial and final internal energies of a gas are equal in an adiabatic free expansion. As we shall see in the next chapter, the internal energy of an ideal gas depends only on its temperature. Thus, we expect no change in temperature during an adiabatic free expansion. This prediction is in accord with the results of experiments performed at low pressures. (Experiments performed at high pressures for real gases show a slight change in temperature after the expansion. This change is due to intermolecular interactions, which represent a deviation from the model of an ideal gas.)
A process that occurs at constant pressure is called an isobaric process. In Figure 20.7, an isobaric process could be established by allowing the piston to move freely so that it is always in equilibrium between the net force from the gas pushing upward and the weight of the piston plus the force due to atmospheric pressure pushing downward. In Figure 20.5, the first process in part (a) and the second process in part (b) are isobaric. In such a process, the values of the heat and the work are both usually nonzero. The work done on the gas in an isobaric process is simply :
where P is the constant pressure.
A process that takes place at constant volume is called an isovolumetric process. In Figure 20.7, clamping the piston at a fixed position would ensure an isovolumetric process. In Figure 20.5, the second process in part (a) and the first process in part (b) are isovolumetric. In such a process, the value of the work done is zero because the volume does not change. Hence, from the first law we see that in an isovolumetric process, because W = 0,
This expression specifies that if energy is added by heat to a system kept at constant
volume, then all of the transferred energy remains in the system as an increase in its internal energy. For example, when a can of spray paint is thrown into a fire, energy enters the system (the gas in the can) by heat through the metal walls of the can. Consequently, the temperature, and thus the pressure, in the can increases until the can possibly explodes.
A process that occurs at constant temperature is called an isothermal process. In Figure 20.7, this process can be established by immersing the cylinder in Figure 20.7 in an ice-water bath or by putting the cylinder in contact with some other constant-temperature reservoir. A plot of P versus V at constant temperature for an ideal gas yields a hyperbolic curve called an isotherm. The internal energy of an ideal gas is a function of temperature only. Hence, in an isothermal process involving an ideal gas, E int = 0. For an isothermal process, then, we conclude from the first law that the energy transfer Q must be equal to the negative of the work done on the gas—that is, Q =-W. Any energy that enters the system by heat is transferred out of the system by work; as a result, no change in the internal energy of the system occurs in an isothermal process.
Isothermal Expansion of an Ideal Gas
Suppose that an ideal gas is allowed to expand quasi-statically at constant temperature. This process is described by the PV diagram shown in Figure 20.8. The curve is a hyperbola (see Appendix B, Eq. B.23), and the ideal gas law with T constant indicates that the equation of this curve is PV = constant. Let us calculate the work done on the gas in the expansion from state i to state f. The work done on the gas is given by Equation 20.8. Because the gas is ideal and the process is quasi-static, we can use the expression PV = nRT for each point on the path. Therefore, we have

Numerically, this work W equals the negative of the shaded area under the PV curveshown in Figure 20.8. Because the gas expands, Vf ( Vi and the value for the work done on the gas is negative, as we expect. If the gas is compressed, then Vf ' Vi and the work done on the gas is positive.

20.7 Energy Transfer Mechanisms
In Chapter 7, we introduced a global approach to energy analysis of physical processes through Equation 7.17, #Esystem ! T, where T represents energy transfer. Earlier in this chapter, we discussed two of the terms on the right-hand side of this equation, work and heat. In this section, we explore more details about heat as a means of energy transfer and consider two other energy transfer methods that are often related to temperature changes—convection (a form of matter transfer) and electromagnetic radiation.
Thermal conduction
The process of energy transfer by heat can also be called conduction or thermal conduction. In this process, the transfer can be represented on an atomic scale as an exchange of kinetic energy between microscopic particles—molecules, atoms, and free electrons—in which less-energetic particles gain energy in collisions with more energetic particles. For example, if you hold one end of a long metal bar and insert the other end into a flame, you will find that the temperature of the metal in your hand soon increases. The energy reaches your hand by means of conduction. We can understand the process of conduction by examining what is happening to the microscopic particles in the metal. Initially, before the rod is inserted into the flame, the microscopic particles are vibrating about their equilibrium positions. As the flame heats the rod, the particles near the flame begin to vibrate with greater and greater amplitudes. These particles, in turn, collide with their neighbors and transfer some of their energy in the collisions. Slowly, the amplitudes of vibration of metal atoms and electrons farther and farther from the flame increase until, eventually, those in the metal near your hand are affected. This increased vibration is detected by an increase in the temperature of the metal and of your potentially burned hand.
Note that has units of watts when Q is in joules and t is in seconds. This is not surprising because P is power—the rate of energy transfer by heat. For a slab of infinitesimal thickness dx and temperature difference dT, we can write the law of thermal conduction where the proportionality constant k is the thermal conductivity of the material and |dT/dx | is the temperature gradient (the rate at which temperature varies with position).

Home Solution
In engineering practice, the term L/k for a particular substance is referred to as the R value of the material. Thus

At one time or another, you probably have warmed your hands by holding them over an open flame. In this situation, the air directly above the flame is heated and expands. As a result, the density of this air decreases and the air rises. This hot air warms your hands as it flows by. Energy transferred by the movement of a warm substance is said to have been transferred by convection.
1. ISOBAR                                         P = K
the first law of thermodinamics :     ∆U = Q-W
W = P (∆V)
∆U = Q-P (∆V)

2. ISOKHORIK                              V = K
∆U = Q-W
W = P (∆V)
W = P ( 0 )
∆U = Q

∆U = Q-W
∆U = O-W
∆U = -W

4. ISOTERMAL                            T = K
∆U = 3/2 nR∆T = O
∆U = Q-W = O
W = Q