Massachusetts Institute of TechnologyDepartment of Electrical Engineering and Computer ScienceDepartment of Mechanical Engineering6.050J/2.110J Information and Entropy Spring 2004Issued: May 3, 2004 Problem Set 12 Due: May 7, 2004Problem 1: Cheap HeatThe heat engine described in the Chapter 12 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 12–1: Heat Engine CycleYou have devised a way to make a large number of dipoles behave in unison, like a single dipole with amuch large magnetic moment. Looking in your base ment, you find you have 105dipoles at your disposal,each with a dipole mome nt of a single electron, 1.165 × 1029Joule-meters/Ampere. Thus there are twostates, with energiesState Energyup −meffHdown meffHTable 12–1: System Parameterswhere meff= 1.165 × 10−24Joule-meters/ampere is the effective dipole moment of this array.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 12 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 temperatures 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 2000 amperes/meter. You decide to run the cycle as follows:Point a b c dH (A/m) ? 2000 1000 ?Table 12–2: 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 Chapter12, 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 12 is helpful:T dS = Xipi(Ei(H) − E)2!1kBT1TdT −1HdH(12–1)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. T he 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 12:pi= e−αe−Ei/kBT(12–2)α = ln Xie−Ei/kBT!(12–3)Problem Set 12 3i. Find the two probabilities pupand pdownat each of the four corners of the cycle. Note that foradiabatic legs, the probabilities do not change so for example they are the same at points a andb.j. Find the energy E =PiEipifor each of the four corners.k. Find the difference between the two entropies, S2− S1.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.)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 content of foods you consume, is actually a kilocalorie which is equal to 4.1868 × 103Joules. The approximate basal metabolic rate of an average 70 kg adult male is 75 Calories/hour (that is,the body dissipates 75 Calories per hour to maintain its normal body temperature while performing normalbodily tasks). What a person doesn’t take in through diet will ultimately be burned from the body’s storeof fat, and when a person takes in more Calories than is needed the excess energy is normally stored as fat.One kilogram of fat corresponds to 33.1 × 106Joules of metabolized