Massachusetts Institute of TechnologyDepartment of Electrical Engineering and Computer ScienceDepartment of Mechanical Engineering6.050J/2.110J Information and Entropy Spring 2003Issued: May 5, 2003 Problem Set 11 Due: May 9, 2003Problem 1: Cheap HeatThe heat engine described in the Chapter 13 of the notes catches your fancy because it seems to be versatile.You don’t need a heat engine but you do want to heat your house more efficiently. You figure that you canmake the device work as a heat pump, transferring energy from the lower temperature outdoors to yourhigher temperature living room. The cycle of the heat engine is pictured here (the cycle of your heat pumpmay be different but has the same rectangular shape).Figure 11–1: Heat Engine CycleYou decide to simplify the design by using only one dipole, so there are only two states, called up anddown, with energiesState Energyup −mdHdown mdHTable 11–1: System Parameterswhere md= 1.165 × 10−29Joule-meters/ampere.In a heat pump the idea is to take heat from the low-temperature (T1) environment (in this case theoutdoors) into the working material (in this case the magnetic dipole) in order to heat the material insidethe high-temperature environment. You decide to run the system using a rectangular path similar to theone pictured above.1Problem Set 11 2a. To use the device as a heat pump, you can either run it as shown above or in the reverse direction.Which should you choose? Hint: a reversible heat engine, when run backwards, can serve as eithera refrigerator or a heat pump.Your living room is at room temperature (17 degrees Celsius). You want your device to work when thetemperature outdoors is freezing (0 degrees Celsius).b. Express the indoor and outdoor temperature in Kelvin.Continuing your design, you want to calculate how much energy can be delivered to the living room. Youfigure that an inexpensive permanent magnet could be moved toward and away from the dipole, and its fieldat the dipole could be made as large as 3000 amperes/meter. You decide to run the cycle as follows:Point a b c dH (A/m) ? 3000 1000 ?Table 11–2: More ParametersNow you want to calculate the work and heat into the system during each of the four legs of the cycle.On each leg, the change in energy is related to the work and heat as dE = dw + dq. According to Chapter13, the work and heat are related to changes dH and dS as dw = (E/H) dH and dq = TdS. The way thecycle is defined it is relatively easy to find dq since on each leg either S or T is constant.c. In order to analyze this cycle you must first determine the magnetic field Haand Hdat pointsa and d. Focus first on the leg c to d, where dS = 0 and dT = T2− T1. This equation fromChapter 13 is helpful:T dS = Xipi(Ei(H) − E)2!1kBT1TdT −1HdH(11–1)Use this equation to determine a relationship among T1, T2, Hc, and Hd.d. From this relationship find Hd.The “coefficient of performance” of a heat pump is defined as the ratio of heat released to the hotenvironment to the work done on the system, in our case by the magnetic field. The larger this coefficient,the less power is needed for the same amount of heating. Surprisingly, the coefficient of performance canbe found without actually solving for the amount of energy converted. T he co effic ient is therefore universal,not depending on the details of the refrigerator but only on the two temperatures T1and T2.e. Find the coefficient of performance of this heat pump.f. Now focus on the leg b to a. Determine a relationship among T1, T2, Ha, and Hb.g. Calculate the magnetic field Ha.h. Next you want to know the work and heat into the system during the four legs of the cycle. Firstconsider the top leg, d to c. What is the heat into the system dq on that leg?To go further you have to calculate the probabilities, since you need them to find the energy E at each ofthe four corners. You already know the temperature and magnetic field at each corner, so it is straightforwardto find α and then the probabilities using these equations from Chapter 13:pi= e−αe−Ei/kBT(11–2)α = ln Xie−Ei/kBT!(11–3)Problem Set 11 3Because the magnetic energy is so small compared with thermal energy kBT , the probabilities are allvery close to 0.5. You may find it necessary to retain a lot of significant figures, or else use a suitableapproximation.i. Find the two probabilities pupand pdownat each of the four corners of the cycle. Note thatfor adiabatic legs, the probabilities do not change so for example they are the same at pointsa and b. Hint: you can check your results with these approximate answers – at point a, pup=0.5 + 4.3664 × 10−6and at point c, pdown= 0.5 − 1.455 × 10−6.j. Find the energy E =PiEipifor each of the four corners.k. Find the difference between the two entropies, S2− S1. Since S1and S2are very nearly equal,you will have to keep many significant figures before taking the difference between them.l. Find the heat dq for each of the four legs (you have already found one of these). Hint: the heatis zero for two of the legs.m. Find the work dw for each of the four legs by calculating the change in energy and subtractingthe heat.n. Find the net work into the system in one cycle, starting at point a and ending up at the samepoint. Hint: it is pos itive, meaning that energy actually is put in as work.o. Calculate the ratio of heat out at the higher temperature T2to work in during one cycle. Comparethis to the coefficient of performance calculated earlier.p. How many cycles would this machine require to heat one gram of air one degree Celsius? (Assumethat the specific heat of air is approximately 0.715 Joules per degree Celsius per gram.)q. You want to speed up the machine by using more dipoles. You have heard that Avogadro’snumber 6.02 × 1023is the numb er of molecules in a reasonable size chunk of material, so youdecide to use that many dipoles. Ass uming you can get it all working, how many cycles wouldyour new machine need to heat the same mass the s ame amount?Problem 2: Information is CoolAn exciting application of thermodynamics is in the study of biological systems. The energy cycles thatwe have seen in class are the starting point for modelling processes that occur throughout nature, whetherit be the dynamics of a muscle cell or an animal’s thermal regulation facilities. For this problem we willexamine forms of energy loss and storage in the human body. Please note that the unit Calorie, used toindicate the energy