Massachusetts Insttute of Technology Department of Electrcal Engneerng and Computer Scence Department of Mechancal Engneerng 6.050J/2.0J Informaton and Entropy Sprng 2004 Problem Set 2 Solutons Soluton to Problem : Cheap Heat Soluton to Problem, part a. You must run t n the reverse drecton. Soluton to Problem, part b. The outdoor temperature n Kelvn s 273.5 degrees, and the ndoor temperature s 290.5 degrees Kelvn. Soluton to Problem, part c. To fnd a relatonshp between T,, H c, and H d we take the equaton gven and T ds = p E H E 2 k B T T dt H dh 2 4 but snce ds = 0 the equaton reduces to ntegratng from c to d we have T dt = H dh T T ln T dt = = ln Hd H c Hd H dh H c T = H d H c 2 5 Soluton to Problem, part d. Thus fndng H d we have H d = H c T = 000 273.5 290.5 = 94 A/m 2 6 2 7
Problem Set 2 Solutons 2 Soluton to Problem, part e. The heat extracted from outdoors s Q = S 2 S T 2 8 The work done on the system s the heat pumped to the warm envronment less the heat extracted from the cold envronment The coeffcent of performance then s W = S 2 S T 2 9 Soluton to Problem, part f. η = T T 273.5 = 290.5 273.5 = 6.07 2 0 Agan, to fnd a relatonshp between T,, H a, and H b we take the equaton gven T ds = p E H E 2 k B T T dt H dh 2 but snce ds = 0 the equaton reduces to T dt = H dh ntegratng from a to b we have T2 T T2 ln T dt = = ln T Hb H a Hb H dh H a T = H b H a 2 2 Soluton to Problem, part g. The magnetc feld H a s H a = H b T = 2000 273.5 290.5 = 883 A/m 2 3 2 4
Problem Set 2 Solutons 3 Soluton to Problem, part h. Snce S s constant n ths adabatc leg, dq = 0. To go further you have to calculate the probabltes, snce you need them to fnd the energy E at each of the four corners. You already know the temperature and magnetc feld at each corner, so t s straghtforward to fnd α and then the probabltes usng these equatons from Chapter 2: p = e α e E/k BT α = ln e E/k BT 2 5 2 6 Soluton to Problem, part. For corners a and b: Frst calculate the exponental. p up = e m eff H a/k B T e m eff H a/k B T + e m eff H a/k B T 2 7 e m eff H a/k B T.65 0 24 883 = exp 273.5.38 0 23 2.937 0 2 = exp 3.77 0 2 = exp0.589 =.7894 2 8 Thus....7894 p up,a,b =.7894 +.7894 = 0.762 2 9 p down,a,b = p up = 0.238 2 20 For corners c and d: Frst calculate the exponental... e m eff H a/k B T p up = e m eff H d /k B T e m eff H d /k B T + e m eff H d /k B T 2 2.65 0 24 = exp 94273.5.38 0 23.096 0 2 = exp 3.77 0 2 = exp0.2907 =.337 2 22
Problem Set 2 Solutons 4 Thus... Soluton to Problem, part j..337 p up,c,d =.337 +.337 = 0.643 2 23 p down,c,d = p up = 0.3587 2 24 E a = E p = m eff H a p up,a + m eff H a p down,a = m eff H a p down,a p up,a =.49 0 2 Joules 2 25 E b = E p = m eff H b p up,b + m eff H b p down,b = m eff H b p down,b p up,b =.22 0 2 Joules 2 26 E c = E p = m eff H c p up,c + m eff H c p down,c = m eff H c p down,c p up,c = 3.309 0 2 Joules 2 27 E d = E p = m eff H d p up,d + m eff H d p down,d = m eff H d p down,d p up,d = 3.4 0 2 Joules 2 28 2 29
Problem Set 2 Solutons 5 Soluton to Problem, part k. therefore S = k B Soluton to Problem, part l. p ln p = k B p up,a,b ln p up,a,b + p down,a,b ln p down,a,b = 7.573 0 24 2 30 S 2 = k B p ln p = k B p up,c,d ln p up,c,d + p down,c,d ln p down,c,d = 9.00 0 24 2 3 S 2 S =.432 0 24 Joules/Kelvn 2 32 dq ba = T ds = 0 Joules 2 33 dq ad = T ds = S S 2 = 4.42 0 22 Joules 2 34 dq dc = T ds = 0 Joules 2 35 dq cb = T ds = T S 2 S = 3.898 0 22 Joules 2 36 2 37
Problem Set 2 Solutons 6 Soluton to Problem, part m. dw ba = de ba dq ba = E b E a 0 = 7.2 0 23 Joules dw ad = de ad dq ad = E a E d dq ad = 4.234 0 22 Joules dw dc = de dc dq dc = E d E c 0 =.95 0 23 Joules dw cb = de cb dq cb = E c E b dq cb = 5.003 0 22 Joules 2 38 Soluton to Problem, part n. The work s the sum of the prevous. 2.44 0 23 Joules 2 39 Soluton to Problem, part o. 6 2 40 Ths s close to the coeffcent. Soluton to Problem 0, part. The number of Joules requred to heat one gram of ar one degree s Soluton to Problem, part p. 0.75.277 0 3 = 5.59 030 cycles 2 4.82 0 2 cycles 2 42 Soluton to Problem 2: Informaton s Cool Soluton to Problem 2, part a. 75 Calores/hour 4.868 0 3 Joules/Calore 3600 sec/hour = 87.225 Joules/sec 2 43 People don t lght up lke lghtbulbs because the energy they expend s dstrbuted about the whole body, not concentrated on a mcroscopc flament.
Problem Set 2 Solutons 7 Soluton to Problem 2, part b. A person needs to consume 75 Calores tmes 24 hours, or Soluton to Problem 2, part c. 75 Calores/hour 24 hours/day = 800 Calores/day 2 44 Mark burns an extra 825 Calores per week than a person who does no exercse. Paul consumes an extra 260 Calores every tme he jobs, and thus burns only 45 Calores more than a person who does no exercse. Paul thus gans, n a year: John gans: Thus Paul gans: And John gans: Soluton to Problem 2, part d. 780 Calores/week 52 weeks/year = 40560 Calores/year 2 45 825 Calores/week 52 weeks/year = 42900 Calores/year 2 46 40560Calores 4.868 0 3 Joules/Calore 33. 0 6 Joules/kg fat 42900Calores 4.868 0 3 Joules/Calore 33. 0 6 Joules/kg fat The amount of heat the room s losng, n Watts, s: = 5.3 kg fat 2 47 = 5.42 kg fat 2 48 5000 0 3 Joules/hour 3600 sec/hour = 388 Watts 2 49 If the temperature s to reman constant, the students and professor must produce the same amount of energy 200 + 20A + 80S = 388 Watts 2 50 where A = Awake and S = Sleepng. If the lecture has 25 students, then the sum of A and S equals 25 and so 200 + 20A + 8025 A = 388 2200 + 40A = 388 A = 20.3 students 2 5 We conclude that the temperature wll always ncrease, snce even wth all the students asleep the amount of heat produced s greater than the heat dsspaton. whch means that 20 students must be awake, 3 students asleep, and one student drftng n and out of conscousness, hs head bobbng forward, wakng hmself up every so often, for an average of 66% of the tme awake, 33% asleep.
Problem Set 2 Solutons 8 Soluton to Problem 2, part e. The number of Calores consumed n rasng 335 ml of water to body temperature 37 degrees Celsus s 0.355 Lter Calores/Lter/degree C 37degrees C = 3 Calores 2 52 Only 7% of the Calores are consumed rasng the rootbeer to body temperature. So Paul s argument s not correct.