Cover Page: Entropy Summary

Transcription

1 Cover Page: Entropy Summary Heat goes where the ratio of heat to absolute temperature can increase. That ratio (Q/T) is used to define a quantity called entropy. The useful application of this idea shows up as the change in entropy (S) that goes with any heat transfer: S = Q/T Energy flowing as heat from a Hot thing to a Cold thing results in a net increase in entropy: S Tot = S H + S C > 0.0 J/K The Hot thing loses entropy ( S H = -Q H /T H ), but the Cold thing gains more entropy ( S C = +Q C /T C ) than the Hot thing lost, so the net result is an increase in entropy: Heat flowing through engines also must go from Hot (input) to Cold (exhaust). It turns out that the natural entropy requirement limits how much work can be done, because there has to be enough heat in the exhaust (Q C ) to carry the entropy increase. Look at this equation again: Limiting the amount of work output naturally limits the possible efficiency, depending upon heat input temperature (T H ) and waste heat exhaust temperature (T C ). Air conditioners must pass heat energy from low temperature indoors to high temperature outdoors. This is only possible by adding enough mechanical energy (enough work ) to the heat removed from indoors to be sure that the total combined energy going into the hot outdoors carries more entropy out than the entropy taken with the heat from the cold indoors. Of course, the air conditioner input work comes from a motor that consumes electricity. That means that running the air conditioner heats up the environment by an amount that includes Q C from inside the house and W and more Q C of exhausted waste heat from the power plant generating the electricity to run the air conditioner. So you can t win, in the long run, everything heats up more the harder you try to cool it down. Everything: Engines: eff = W / Q H = (Q H Q C ) / Q H where T H > T C AND (Q H /T H ) < (Q C /T C ) (Q H /T H ) LESS than (Q C /T C ) Air Conditioners and Refrigerators: Q H = Q C + W where T H > T C AND (Q H /T H ) > (Q C /T C ) (Q H /T H ) GREATER than (Q C /T C ) 0

2 Here s what Temperature has to do with how well Engines and Air Conditioners do their jobs In the first place, temperature obviously has something to do with heat and very little to do with work. Everyone knows that temperature is used to describe how hot things are. Meanwhile, nobody that I can think of says anything like, I got paid $97 a day to do 78 o F per hour of work at the construction site or Carrying two pianos up three flights of stairs is 148 o F more work than carrying one piano up four flights of stairs. In the second place, although there isn t an obvious direct connection between mechanical work and temperature, there is a connection between heat and work. So, since work is related to heat, and heat is related to temperature, there actually is an indirect relationship between work and temperature. That s the reason for this essay that you re reading. In the third place, air conditioners are just as dependent upon the relationships that link work, heat, and temperature as engines are. This makes it possible to build and operate air conditioners and refrigerators, and can also be used to demonstrate why it can be so expensive to use them. In the First Place, there s Heat Putting a lot of heat energy into something raises its temperature a lot. Putting in a little heat raises the temperature a little, and taking heat out lowers the temperature. So far, so good. Furthermore, some things require more heat than others to gain an equal amount of temperature. That s the concept behind specific heat capacity (the c in Q=mc T). Still good. Now let s consider heat getting from one place to another. Heat travels faster through a big temperature difference than a small one. How fast the heat travels, though, is less important for this lesson than how well the heat travels. How fast heat is conducted through the side of a coffee cup, or into the cooling system of an engine, can be expressed in Joules per second (J/s) according to the conduction formula (Q/t = ka T/L). Like I said, though, that s a separate concept from some kind of description of how well the heat travels. How much heat per second is a good description of how fast heat travels. How much heat per degree turns out to be more like what we need to represent something like how well the heat travels (more like it, but not precisely heat per degree, as explained below). The ratio of heat to temperature (Q/T, not Q/t) is used as a way to keep score in the heat transfer game. This scorekeeping quantity is called entropy. Entropy goes along for the ride with heat every time heat goes from one place to another. The strange truth about entropy that makes it a useful scorekeeper is that whenever heat goes from object A to object B, object B receives more entropy than object A lost. Strange but true, because of the way entropy is defined. Since the temperature of A must be higher than the temperature of B, and temperature is in the denominator of the entropy ratio (Q/T), higher temperature scores lower on the entropy scale, and lower temperature scores higher on the entropy scale. Now here s an important place to be very careful to use Kelvin temperatures. Low temperatures are sometimes labeled as negative, as in -26 o C or 26 o below zero, but that does NOT make the entropy negative. In order to avoid that mistake it is absolutely crucial 1

3 to use only absolute temperatures to calculate entropy. So entropy can t really be described as Joules per degree, but instead must be identified as Joules per Kelvin. Here s one more important technicality, besides always using absolute temperatures to calculate entropy. Entropy goes along for the ride whenever and wherever heat goes from one place to another, but entropy is a better indicator of the trip than the destination or the origin. Here s what I mean: We can use the internal energy of an object and the heat added to it to find the new total internal energy, but we don t calculate total entropy in a similar fashion. Heat can be defined technically as the change in internal energy, and there s a similar interpretation for entropy. It s the change in entropy ( S = Q/T) caused by heat transfer that we re interested in, not the total entropy for any object. Here s an example with some numbers: Suppose 1000 J of heat leaves a 400 K box and goes into a 250 K box. The 400 K box loses entropy along with the heat: S H = J / 400 K = -2.5 J/K The 250 K box gains entropy along with the heat: S C = J / 250 K = +4.0 J/K The net result is an increase of entropy: S Tot = +4.0 J/K 2.5 J/K = +1.5 J/K Since heat naturally goes from hot things (high temperature) to cold things (low temperature), the positive total change in entropy ( S Tot > 0.0 J/K) is a natural entropy score for heat transfer. Consider the calculations below for [what does not happen]: Recognize that 1000 J of heat does NOT leave the 250 K box and go into the 400 K box. The 400 K box would gain entropy: S H = J / 400 K = +2.5 J/K of The 250 K box would lose entropy: S C = J / 250 K = -4.0 J/K The net result would be a decrease of entropy: S Tot = -4.0 J/K J/K = -1.5 J/K This illustrates how the entropy score distinguishes between possible and impossible thermodynamic processes. Notice from the first example that it s OK for one part of the system to lose entropy (400 K box lost 2.5 J/K), as long as some other part of the system makes up the difference (250 K box gained 4.0 J/K). Not even 1.0 J of heat leaves the 250 K box and goes into the 400 K box. The 400 K box would gain entropy: S H = +1.0 J / 400 K = J/K The 250 K box would lose entropy: S C = -1.0J / 250 K = J/K The net result would be a decrease of entropy: S Tot = J/K J/K = J/K The overall decrease of entropy (negative TOTAL entropy change) indicates a path of heat transfer that does not work. Everything that IS possible: 2

4 In the Second Place, there s Work Temperature influences heat transfer, but temperature is not a consideration for determining mechanical work. The temperature of mechanical Work is undefined. Heat carries entropy with it. Work carries no entropy. An engine cylinder can absorb heat energy to make a gas expand, pushing a piston to do work. Eventually the piston can t go any further, and the work is finished for the moment, but the remaining heat must be dumped out with the exhaust. Entropy is taken from the heat source with the absorbed heat energy. The work done by the engine delivers energy to the outside world without entropy. That means that the exhaust heat is the only path for the entropy increase required by nature to complete the flow of heat from the high temperature source to the low temperature exhaust. It turns out that the natural entropy requirement limits how much work can be done, because there has to be enough heat in the exhaust (Q C ) to carry the entropy increase. Remember: The work output (mechanical energy) of an engine is the difference between the heat input and the waste heat exhaust (W = Q H Q C ). Since too little exhaust heat can t carry enough entropy to maintain the heat flow, the input and exhaust temperatures (T H & T C ) guarantee a certain amount of waste heat in the exhaust, and therefore set a maximum amount of work output. An upper limit on work output naturally limits the maximum possible efficiency of the engine (eff=w / Q H ), based upon the input temperature (T H ) and the exhaust temperature (T C ). Here s an example with some numbers: Suppose a heat engine draws 1000 J of heat from a 400 K box and dumps exhaust into a 250 K box. If the engine does 200 J of work, then 800 J of heat are wasted in the exhaust. The 400 K box loses entropy along with the heat: S H = J / 400 K = -2.5 J/K The 250 K box gains entropy along with the heat: S C = +800 J / 250 K = +3.2 J/K The net result is an increase of entropy: S Tot = +3.2 J/K 2.5 J/K = +0.7 J/K OK. It might be possible to run this engine at W / Q H = 200 J / 1000 J = 20% efficiency. If you tried to make the engine do 400 J of work, then 600 J of heat would have to be wasted in the exhaust. The 400 K box would have to lose entropy: S H = J / 400 K = -2.5 J/K The 250 K box would have to gain entropy: S C = +600 J / 250 K = +2.4 J/K The net result would be a decrease of entropy: S Tot = +2.4 J/K 2.5 J/K = -0.1 J/K NOT OK. This engine cannot operate at W / Q H = 400 J / 1000 J = 40% efficiency. Engines: eff = W / Q H = (Q H Q C ) / Q H where T H > T C AND (Q H /T H ) < (Q C /T C ) (Q H /T H ) LESS than (Q C /T C ) 3

5 In the Third Place, Work can drive heat backwards but ONLY IF Temperature differences naturally drive heat from hot to cold, along with an increase in entropy. Engines use this natural entropy pressure to keep heat flowing from input to exhaust while the engine redirects a limited amount of the energy to do some mechanical work. Refrigerators and air conditioners rely upon the fact that the natural tendency to increase entropy can be more powerful than the natural tendency to go from hot to cold. Consider what air conditioning does: Heat is driven from the low temperature indoors to the high temperature outdoors. This, of course, is contrary to what the heat would do spontaneously, if we left it alone. All by itself, the heat would naturally flow from the high temperature outdoors into the cool interior of the house, warming the house and increasing the entropy. But we don t just leave the heat alone to follow its natural tendencies. It turns out that it is possible to make the heat go outdoors into the higher temperature, as long as we make sure that it takes plenty of entropy along with it. It s not easy, but it is possible. Think of it in terms of a heat engine running backwards. Instead of taking in Q H to put out W and exhaust Q C, now we ll be using W as INPUT to COLLECT Q C and exhaust the combined total of W and Q C as Q H. The 20% efficient engine described in the Second Place example does not run backwards to feed 800 J of heat from 250 K up to 400 K, but what about the failed 40% efficiency engine? Here s the example with some numbers: If we reverse everything in the engine concept, then we can collect 600 J at 250 K ( removing 2.4 J/K of entropy) and use 400 J of work to move the whole 1000 J outdoors, taking 2.5 J/K of entropy with it, as follows: The 250 K box ( indoors ) would lose entropy along with the heat: S C = -600 J / 250 K = -2.4 J/K The 400 K box ( outdoors ) would gain entropy along with the heat: S H = J / 400 K = +2.5 J/K The net result would be an increase of entropy: S Tot = +2.5 J/K 2.4 J/K = +0.1 J/K At least, theoretically it could happen. In practical terms, it takes more work energy input than that. It also requires a gas that can absorb heat at low temperature, condense into a liquid under the pressure provided by the work energy input, and then evaporate at a high temperature to exhaust the unwanted heat outdoors. In more realistic terms, a real air conditioner really can move heat from a 295 K house (72 o F) out into a 305 K environment (90 o F) if you have enough work energy input to spend to be sure that the entropy going into the hot outdoors is greater than the entropy removed with the heat from the cold indoors. Air Conditioners and Refrigerators: Q H = Q C + W where T H > T C AND (Q H /T H ) > (Q C /T C ) (Q H /T H ) GREATER than (Q C /T C ) 4

6 Homework Problems: 1. Consider an engine that takes heat from an 855 K input and exhausts waste heat to the environment at 425 K. How much work output would this engine deliver if it took 6275 J from the 855 K input, and generated an overall total entropy increase of 1.75 J/K? 2. Consider an engine that takes heat from a 705 K input and exhausts waste heat at 315 K. What would be the efficiency of this engine if it dumped J of waste heat as exhaust, and generated an overall total entropy increase of 12.3 J/K? 3. Maximum efficiency for a heat engine matches minimum entropy increase. Using zero as a theoretical minimum entropy increase, calculate a theoretical maximum efficiency for an engine that took 6275 J of input from an 855 K source and exhausted its waste heat at 425 K. 4. An air conditioner maintains an indoor temperature of 295 K (72 o F) against an outdoor temperature of 315 K (108 o F). If it generates an overall total entropy increase of 235 J/K and removes 7.5 x 10 5 J of heat from the house, how much energy must be provided to operate the air conditioner? 5. Consider an air conditioner that generates 3.04 J/K per second of overall entropy increase while operating on a day when it is 308 K (90 o F) outside. a) If it maintains an indoor temperature of 299 K (78 o F) while using 1.05 x 10 3 W of power to operate the air conditioner, expelling 4.80 x 10 3 J per second to the outdoors, how much heat is being removed from the house each second? b) How much input power would be required to operate the air conditioner if it were required to remove the same amount of heat from the indoors as in part a), but now to maintain an indoor temperature of 295 K (72 o F)? 5