Idea Transcript
1206 - Concepts of Physics Friday, November 13th 2009
Notes • Assignment #6 due next Wednesday • Assignment # 7 will be posted later today (or tomorrow) and is due Wednesday, November 25th.
• For both of them you will loose points for unit mistakes!!!
• Also, no pencil for assignments (other than drawings.
Isobaric, Isochoric Process P
Vi
Vf
Pf
Pi V
The plot (left) shows pressure versus volume for an isobaric expansion. Since the pressure is constant, the graph is a horizontal straight line, beginning at the initial volume Vi and ending at the final Vf. In terms of such a plot, the work W = P(Vf - Vi) is the area under the graph, which is the colored rectangle with height P and width Vf - Vi.
An isochoric process occurs at constant volume (see left). A substance is heated in a rigid container, therefore it can’t expand and the volume stays constant. Because the volume is constant, the pressure inside rises and the substance exerts more and more force on the walls. Although enormous forces can be generated in the closed container, no work is done, since the walls don’t move. Since no work is done, the first law of thermodynamics indicates that the heat in an isochoric process serves only to change the internal energy ΔU = Q - W = Q
Example: work and area under P-V graph Determine the work for the process in which the pressure, volume, and temperature of a gas are changed along the straight line from X to Y in the plot below.
Pressure
Y X
The work is given by the area (in color) under the straight line between X and Y. Since the volume increases , work is done by the gas on the surroundings, so the work is positive. The area can be found by counting the squares and multiplying by the area per square. There are 13.5 colored squares in the figure. Each square has an area of (2.0 x 105 Pa)(1.0 x 10-4 m) = 20 J.
Volume
2.0 x 105 Pa
1.0 x 10-4 m3
W = 13.5 square x 20 J/square = 270 J
Adiabatic expansion or compression When a system performs work adiabatically, no heat flows into or out of the system. The picture shows a cylinder filled with a gas surrounded by an insulator - n moles of an ideal gas doing work under adiabatic conditions - expanding quasi statically (very slow) from an initial volume Vi to a final volume Vf. The insulator prevents heat flow, so Q = 0 J. According to the first law of thermodynamics, the change in internal energy is ΔU = Q - W = - W. Since the internal energy of an ideal monatomic gas is U = 3/2nRT, it follows that ΔU = Uf - Ui = 3/2nR(Tf - Ti) Therefore W = 3/2nR(Ti - Tf) When an ideal gas expands it does positive work. Therefore, the term Ti - Tf is also positive, so the final temperature of the gas must be less than the initial temperature. The internal energy of the gas is reduced to provide the necessary energy to do the work, and because the internal energy is proportional to the kelvin temperature, the temperature decreases.
P
The figure shows pressure versus volume (V-P graph) for an adiabatic process. The adiabatic curve (green) intersects the isotherms a the higher initial temperature (Ti = PiVi/(nR)) and the lower final temperature (Tf = PfVf/(nR)). The light blue area marks the work done. The equation that gives the adiabatic (green) curve can be derived using integral calculus. The result is:
PiViγ = PfVfγ
adiabatic expansion or compression
where the exponent γ is the ratio of the specific heat capacities at constant pressure and constant volume, γ = cP/cV. Each point on the adiabatic curve satisfies the relation PV = nRT. Heat is energy that flows from a higher-temperature object to a lower-temperature object because of the temperature difference. It is important to note, that it is not correct to say that a substance contains heat. The substance has internal energy, not heat. The word “heat” is used only when referring to the energy actually in transit from hot to cold.
Table Type of thermal process
Work done
First law of Thermodynamics (ΔU = Q - W)
Isobaric (constant pressure)
W = P(Vf-Vi)
ΔU = Q - P(Vf - Vi)
Isochoric (constant volume)
W=0J
ΔU = Q
Isothermal (constant temperature) W = nRT ln(Vf/Vi) Adiabatic (no heat flow)
W = 3/2nR(Tf - Ti)
0J = Q - nRT ln(Vf/Vi) ΔU = -3/2nR(Tf - Ti)
Specific heat Capacity Greater amounts of heat are needed to raise the temperature of solids or liquids to higher values. More heat is also needed for larger mass of material. The heat supplied or removed in changing the temperature of a substance is referred to as the specific heat capacity of the material c. The heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount ΔT is Q = cmΔT The unit for the specific heat capacity c is J/(kg C).
Example: taking a hot shower Cold water at a temperature of 15 C (c = 4186 J/(kg C) enters a heater, and the resulting hot water has a temperature of 61 C. A person uses 120 kg of hot water in taking a shower. (a) Find the energy meeded to heat the water. (b) Assuming that the utility company charger $0.10 per kilowatt hour for electrical energy, determine the cost of heating the water. The amount Q of heat needed to raise the water temperature can be found form the relation Q = cmΔT, since the specific heat capacity, mass, and temperature change of the water are known. To determine the cost of this energy, we multiply the cost per unit of energy ($0.10 per kilowatt hour) by the amount of energy used, expressed in units kilowatt hours.
(a) The amount of heat needed to heat the water is: Q = cmΔT = [4186 J/(kg C)](120 kg)(61 C - 15 C) = 2.3 x 107 J (b) The kilowatt hour (kWh) is the unit of energy that utility companies use in your electric bill. To calculate the cost, we need to determine the number of joules in one kilowatt hour. Recall that 1 kilowatt is 1000 watts (1kW = 1000 W), 1 watt is 1 joule per second (1 W = 1 J/s) and 1 hour has 3600 s. Thus 1kWh = (1 kW)(100W/1kW)((1J/s)/1W)(3600s/1h) = 3.60 x 106 J The number for kilowatt hours of energy used to heat the water is (2.3 x 107 J)(1kWh/3.60 x 106 J) = 6.4 kWh Therefore the bill fro the heat is $0.64 (6.4 kWh * $0.10/kWh)
Heat units other than the joule There are a number of heat units in common use. The one you are likely are most familiar with is the calorie and the kilocalorie. One kilocalorie (1 kcal) is defined historically as the amount of heat needed to raise the temperature of 1 kilogram of water by one degree Celsius. With Q = 1.00 kcal, m = 1.00 kg and ΔT = 1.00 C, the equation Q = cmΔT shows that such a definition is equivalent to a specific heat capacity for water of c = 1.00 kcal(kg C). Nutritionists use the word “Calorie” with a capital C, to specify the energy content of foods; this use is unfortunate, since 1 Calorie = 1000 calories = 1 kcal. The British thermal unit (Btu) is another commonly used heat unite and is defined historically as the amount of heat needed to heat 1 pound of water by 1 degree Fahrenheit. Conversion: 1 kcal = 4186 joules
Specific Heat Capacities - part II It is sometimes more convenient to express the amount of material as the number of moles n, rather than the number of kilograms. Therefore, we replace the expression Q = cmΔT with: Q = CnΔT where the capital letter C (instead of the lower case c before) refers to the molar specific heat capacity in units of J/(mol K). In addition, the unit for measuring the temperature change ΔT is K. For gases it is necessary to distinguish between the molar specific heat capacities CP and CV, which apply, respectively, to conditions of constant pressure and constant volume. With the help of the first law of thermodynamics and an ideal gas as an example, it is possible to see why CP and CV differ. To determine the molar specific heat capacities, we must first calculate the heat Q needed to raise the temperature of an ideal gas from Ti to Tf. According to the first law, Q = ΔU + W, we also know that the internal energy of a monatomic ideal gas is U = 3/2nRT. As a result, ΔU = Uf - Ui = 3/2nR(Tf - Ti). When the heating process occurs at constant pressure, the work done is given as W = PΔV. For an ideal gas, PV = nRT, so the work becomes W = nRΔT. On the other hand, when the volume is constant, ΔV = 0 m3, and the work done is zero. The calculation of the heat is summarized below: Q = ΔU + W Q(constant pressure) = 3/2nRΔT + nRΔT = 5/2nRΔT Q(constant volume = 3/2nRΔT + 0 CP = 5/2R and CV = 3/2R --> γ = CP/CV = 5/3 The difference between CP and CV arises because work is done when the gas expands in response to the addition of heat under conditions of constant pressure, whereas no work is done under conditions of constant volume. For a monatomic ideal gas, we get: CP - CV = R
Latent Heat When a substance changes from one phase to another, the amount of heat that must be added or removed depends on the type of material and the nature of the phase change. The heat per kilogram associated with a phase change is referred to as latent heat: HEAT SUPPLIED OR REMOVED IN CHANGING THE PHASE OF A SUBSTANCE The heat Q that must be supplied or removed to change the phase of a mass m of a substance is Q = mL where L is the latent heat of the substance. The SI Unit of latent heat: J/kg The latent heat of fusion Lf refers to the change between solid and liquid phases. The latent heat of vaporization Lv refers to the change between liquid gas phases. The latent heat of sublimation Ls refers to the change between solid and gas phases.
Heat Engines A heat engine is any device that uses heat to perform work. I has three essential features: 1.) Heat is supplied to the engine at a relatively high input temperature from a place called hot reservoir 2.) Part of the input heat is used to perform work by the working substance of the engine, with the material within the engine that actually does the work 3.) The remainder of the input heat is rejected to a place called the cold reservoir, which has a temperature lower than the input temperature The symbol QH is used for the magnitude of the input heat flow (hot side to engine) and QC refers to the magnitude of the rejected heat (engine to cold side) and W stands for the magnitude of the work. Note! These three symbols refer to magnitudes only, without reference to algebraic signs. Therefore, when these symbols appear in an equation, they do not have negative values - ever! The more work a heat engine can produce, that more efficient it is. The efficiency e of a heat engineis defined as the ratio of the work W done by the engine to the input heat QH: e = W/QH An engine must obey the principle of conservation of energy, so QH = W + QC (in absence of other losses. Therefore we can write the efficiency of a heat engine as: e = (QH - QC) / QH = 1 - QC/QH
Example: An automobile engine An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine? Energy conservation indicates that the amount of heat rejected to the cold reservoir is the part of the input heat that is not converted into work. The amount QC rejected is QC = QH - W. We need a value for QH Re-arranging the efficiency definition, we find QH = W/e. Therefore QC = W/e - W = W(1/e -1) = (2510 J)(1/0.220 - 1) = 8900 J
Carnot’s principle and the Carnot Engine The French engineer Sadi Carnot (1796-1832) proposed that a heat engine has maximum efficiency when the processes within the engine are reversible. A reversible process is one in which both the system and its environment can be returned to exactly the states they were in before the process occurred. If you have friction, you loose energy and your process is therefore irreversible. Also spontaneous flow of heat means energy loss and is therefore an irreversible process. Today, the idea that the efficiency of a heat engine is a maximum when the engine operates reversibly is referred to as Carnot’s principle.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND LAW No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures. Furthermore, all reversible engines operating between the same temperatures have the same efficiency. It can be shown that if Carnot’s principle were not valid, it would be possible for heat to flow spontaneously from a cold substance to a hot substance, in violation of the second law of thermodynamics.
An important feature of the Carnot engine is that all input heat QH originates from a hot reservoir at a single temperature TH and all rejected heat QC goes into a cold reservoir at a single temperature TC. Carnot’s principle implies that the efficiency of a reversible engine is independent of the working substance, and therefore can depend only on the temperatures of the hot and cold reservoirs. Since e = 1 - QC/QH This leads to the fact that the ratio of the rejected heat QC and to the input heat QH is: QC/QH = TC/TH where the temperatures TC and TH must be expressed in Kelvins. The efficiency of a Carnot engine eCarnot = 1 - TC/TH This relation give the maximum possible efficiency for a heat engine.
Example: A Tropical Ocean as a Heat Engine Water near the surface of a tropical ocean has a temperature of 298.2 K, whereas water 700 m beneath the surface has a temperature of 280.2 K. It has been proposed that the warm water be used as the hot reservoir and the cool water as the cold reservoir of a heat engine. Find the maximum possible efficiency for such an engine. The maximum possible efficiency is the efficiency that a Carnot engine would have operating between temperatures TH = 298.2 K and TC = 280.2 K. Using TH = 298.2 K and TC = 280.2 K, we find: eCarnot = 1 - TC/TH = 1 - 280.2 K/298.2 K = 0.060 The maximal possible efficiency is only 6.0%. The small efficiency is due to the temperatures between the two reservoirs being so close. But there are limits on how large the efficiency of a heat engine can be. See next example.
Example: natural limits to the Efficiency of a heat engine Consider a hypothetical engine that receives 1000 J of heat as input from a hot reservoir and delivers 1000 J of work, rejecting no heat to a cold reservoir whose temperature is above 0 K. Decide whether this engine violates the first of the second law of thermodynamics, or both.
YOUR TURN ...
Example: natural limits to the Efficiency of a heat engine Consider a hypothetical engine that receives 1000 J of heat as input from a hot reservoir and delivers 1000 J of work, rejecting no heat to a cold reservoir whose temperature is above 0 K. Decide whether this engine violates the first of the second law of thermodynamics, or both. The first law of thermodynamics is an expression of energy conservation. From the point of view of energy conservation, nothing is wrong with an engine that converts 1000 J of heat into 1000 J of work. Energy can be transformed, so the engine does not violate the first law. This engine does, however violate the second law of thermodynamics. Since all of the input heat is converted into work, the efficiency of the engine is 1 (or 100%). But the equation for efficiency (based on the second law) say’s that the maximum possible efficiency is 1 - TC/TH. Since we know TC is above 0 K, it is clear that the ratio TC/TH is greater than zero, which makes the maximum possible efficiency less than 1 (or 100%). Therefore it is the second law of thermodynamics that limits the efficiencies fo heat engines to values less than 100%.
Example: a Heat Pump An ideal Carnot heat pump is used to heat a house to a temperature of TH = 294 K. How much work must be done by the pump to deliver QH = 3350 J of heat into the house when the outdoor temperature TC is (a) 273 K and (b) 251 K? The conservation of energy (QH = W + QC) applies to the heat pump. Thus, the work can be determined from W = QH - QC, provided we can obtain a value for QC, the heat taken by the pump from the outside. To determine QC, we use the fact that the pump is a Carnot heat pump an operates reversible. Therefore, the relation QC/QH = TC/TH applies. Solving for QC, we obtain QC = TC/TH * QH and therefore: W = -TC/TH * QH + QH = QH (1 - TC/TH) (a) At an indoor temperature TH = 294 K and an outdoor temperature of TC = 273 K, the work needed is: W = QH (1 - TC/TH) = (3350 J)(1 - 273K/294K) = 240 J (b) similar (you can do it at home) we obtain W = 479 J Note! The coefficient of performance of a heat pump is the ratio of the heat QH delivered into the house to the W required to deliver it: coefficient of performance: QH/W
Entropy A Carnot engine has the maximum possible efficiency, because the processes occurring within are reversible. Irreversible processes, such as friction, cause real engines to operate at less than maximum efficiency, for they reduce our ability to use heat to perform work. Lets make an extreme example, imagine that a hot object is placed in thermal contact with a cold object, so heat flows spontaneously, and hence irreversibly, from hot to cold. Eventually both objects will reach the same temperature, and TC = TH. A Carnot engine using these two objects as heat reservoirs is unable to do work, because the efficiency of the engine is zero [eCarnot = 1 - TC/TH) = 0]. In general irreversible processes cause us to lose some, but not necessarily all of the ability to perform work. This partial loss can be expressed in terms of a concept called entropy. We start from the relation QC/QH = TC/TH that applied to a Carnot engine. This equation can be rearranged as QC/TC = QH/TH, which focuses attention on the heat Q divided by the Kelvin temperature T. The quantity Q/T is called the change in the entropy ΔS: ΔS = (Q/T)R where the subscript R refers to reversible, T must be expressed in K. Entropy has the SI unit J/K. Entropy, like the internal energy, is a function of the state or condition to the system. If we look at a Carnot engine, we can define ΔSH = -QH/TH as the change in entropy in the hot reservoir (the minus sign is needed to indicate a decrease in entropy) and ΔSC = QC/TC as the amount of change (increase) of entropy in the cold reservoir. Total change is then: ΔSC + ΔSH = QC/TC - QH/TH = 0 This can be generalized to the fact that reversible processed do not alter the total entropy of the universe, which is one of the formulation we had for the second law of thermodynamics on Wednesday.
Example: The Entropy of the Universe Increases We have two reservoirs (hot and cold) that are connected via a copper rod. 1200 J are spontaneously flowing through the copper rod from the hot reservoir at 650 K to the cold reservoir at 350 K. Determine the amount by which this irreversible process changes the entropy of the universe, assuming that no other changes occur The hot-to-cold heat flow is irreversible, so the relation ΔS = (Q/T) is applied to a hypothetical process whereby the 1200 J of heat is taken reversibly form the hot reservoir and added reversibly to the cold reservoir. The total entropy change of the universe is the algebraic sum of the entropy changes for each reservoir: ΔS(universe) = -(1200J)/(650 K) + (1200 J)(350 K) = 1.6 J/K The irreversible process causes the entropy of the universe to increase by 1.6 J/K. THE SECOND LAW OF THERMODYNAMICS The total entropy of the universe does not change when a reversible process occurs (ΔSuniverse = 0 J/K) and does not increase when an irreversible process occurs (ΔSuniverse > 0 J/K). When an irreversible process occurs and the entropy of the universe increases, the energy available for doing work decreases.
Example: Order to Disorder Find the change in entropy that results when a 2.3 kg block of ice melts slowly (reversibly) at 273 K. Since the phase change occurs reversible at a constant temperature, the change in entropy can be found by using ΔS = (Q/T)R where Q is the heat absorbed by the melting ice. This heat can be determined by using the relation Q = mLf, where m is the mass and Lf = 3.35 x 105 J/kg is the latent heat of fusion of water. We find that the change in entropy is ΔS = Q/T = mLf/T = {(2.3 kg)(3.35 x 105 J/kg)} / {273 K} = 2.8 x 103 J/K
The Third Law of Thermodynamics To the zeroth, first, and second laws of thermodynamics we add the third (and last) law. The third law of thermodynamics indicates that it is impossible to reach a temperature of absolute zero. THE THIRD LAW OF THERMODYNAMICS It is not possible to lower the temperature of any system to absolute zero ( T = 0K) in a finite number of steps. This law, like the second law can be expressed in a number of ways. This law is needed to explain a number of experimental observations, that cannot be explained by the other laws of thermodynamics. A detailed discussion is beyond the scope of this course, we just state it here for completeness.