6.050J Information and Entropy, Problem Set 10 Solutions

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Department of Mechanical Engineering 6.050J/2.110J Information and Entropy Problem Set 10 Solutions Solution to Problem 1: Entropy Goes Up Solution to Problem 1, part a. The expectation of system energy Es (the expected value of the energy) is calculated by the following formula. Es = � ps,i Es,i (H) i = 0.55 md H − 0.45 md H = 0.1 md H (10–1) Solution to Problem 1, part b. The expectation of the environment energy Ee is found in a similar manner. Ee = � pe,j Ee,j (H) j = 0.25 md H − 0.75 md H = −0.5 md H (10–2) Solution to Problem 1, part c. The environment entropy is calculated via the following formula. Se = kB � � pe,j ln j � = kB 0.75 ln � 1 � pe,j � � �� 1 1 + 0.25 ln 0.75 0.25 = kB (0.75 × 0.2877 + 0.25 × 1.3863) = kB (0.2158 + 0.3466) = 0.5624 kB Solution to Problem 1, part d. The system entropy Ss is calculated in a similar fashion. 1 (10–3) Problem Set 10 Solutions 2 � 1 ps,i i � � � � �� 1 1 = kB 0.45 ln + 0.55 ln 0.45 0.55 Ss = kB � � ps,i ln = kB (0.45 × 0.7985 + 0.55 × 0.5978) = kB (0.3593 + 0.3288) = 0.6881 kB (10–4) Solution to Problem 1, part e. No energy leaves the system and environment combined (by definition) so the expectation of the total energy is just the sum of the expectations of the energy of the system and environment. Et = Es + Ee = 0.1md H − 0.5md H = −0.4md H (10–5) Solution to Problem 1, part f. To find βt we first use Equation 10.20. Note that Ei,j =­2, 0,0, or 2 times md H. 0= � = � (Ei,j − Et ) e−βt Ei,j i,j Ei,j e−βt Ei,j − Et i,j = 2md He � e−βt Ei,j i,j −2md Hβt + 0.4md H � − 2md He2md Hβt e−βt Ei,j i,j = 2md He −2md Hβt − 2md He2md Hβt + 0.4md He2md Hβt + 0.8md H + 0.4md He−2md Hβt = 2.4md He−2md Hβt + 0.8md H − 1.6md He2md Hβt � � = 0.8md H 3e−2md Hβt + 1 − 2e2md Hβt (10–6) We can rearrange this equation into the following form � � 0 = 0.8md H 3e−2md Hβt + 1 − 2e2md Hβt 0 = 3e−2md Hβt + 1 − 2e2md Hβt � �2 0 = 3 + e2md Hβt − 2 e2md Hβt � �2 0 = 2 e2md Hβt − e2md Hβt − 3 (10–7) Letting x = e2md Hβt we have 2x2 − x − 3 = 0 2md Hβt This equation has zeros at x = −1, 1.5. Thus e βt = ln(1.5)/2md H. (10–8) = 1.5, since it must be greater than zero. Therefore Problem Set 10 Solutions 3 Solution to Problem 1, part g. The probabilities are defined as e−βt Ei,j −βt Ei,j i,j e pi,j = � (10–9) Thus e−βt E0,0 = e−(ln(1.5)/2md H)−2md H = 1.5 e −βt E0,1 =1 −βt E1,0 =1 e (10–10) e−βt E1,1 = e−(ln(1.5)/2md H)2md H 1 = 1.5 (10–11) So 1.5 1.5 + 0.66 + 2 1 = 1.5 + 0.66 + 2 1 = 1.5 + 0.66 + 2 0.66 = 1.5 + 0.66 + 2 p0,0 = = 0.36 (10–12) p0,1 = 0.24 (10–13) = 0.24 (10–14) = 0.16 (10–15) p1,0 p1,1 (10–16) Solution to Problem 1, part h. The total entropy is St = kB � � 1 pi � 1 0.36 pi ln i � = kB 0.36 ln � � � + 0.48 ln 1 0.24 � � + 0.16 ln 1 0.15 �� = kB (0.3677 + 0.6851 + 0.2932) = 1.6553kB (10–17) which is higher than the original entropy, 1.2505kB Solution to Problem 1, part i. First let us infer from the four probabilities for the total configuration pt,i,j the probabilities for the two system states ps,i . Problem Set 10 Solutions 4 ps,0 = pt,0,0 + pt,0,1 = 0.36 + 0.24 = 0.60 ps,1 = pt,1,0 + pt,1,1 = 0.24 + 0.16 = 0.40 (10–18) (10–19) Thus the energy is Es = � ps,i Es,i (H) i = 0.40 md H − 0.60 md H = −0.2 md H (10–20) Thus we see that exactly half the energy is in the system. Solution to Problem 1, part j. The system started out with 0.1md H Joules in it, and ended up with −0.2 md H Joules in it. Thus 0.3 md H Joules flowed from the environment to the system. Solution to Problem 2: Energy Conversion (the charge pump) Solution to Problem 2, part a. See Figure 10–3. Solution to Problem 2, part b. See Figure 10–4. Solution to Problem 2, part c. See Figure 10–5. Solution to Problem 2, part d. The formula for the charge on a capacitor is as follows. �0 AV d You start out with a certain amount of charge on the plate: q= q1 = �0 AVlow dmin (10–21) (10–22) Problem Set 10 Solutions 5 Figure 10–3: Charging Cycle and finish with a smaller amount of charge: q2 = �0 AVhigh dmax (10–23) So the charge delivered is: � q0 = q1 − q2 = �0 A Vlow Vhigh − dmax dmin � (10–24) Solution to Problem 2, part e. � Ehigh = Vhigh �0 A Vhigh Vlow − dmax dmin � (10–25) Solution to Problem 2, part f. � Elow = Vlow �0 A Vhigh Vlow − dmax dmin � (10–26) Problem Set 10 Solutions 6 Figure 10–4: Charging Cycle corresponding to user’s guide Solution to Problem 2, part g. Emech = Ehigh − Elow � = �0 A(Vhigh − Vlow ) Vlow Vhigh − dmax dmin � (10–27) Solution to Problem 2, part h. Note that a farad is a coulomb per volt. � q0 = �0 A Vlow Vhigh − dmin dmax � = (8.854 × 10−12 F/m) × (4 × 10−2 m)2 × � 1.5V 9V − −3 0.1 × 10 m 5 × 10−3 m � = 8.854 × 16 × 10−16 (15000 − 1800)FV = 1.869 × 10−10 C (10–28) Problem Set 10 Solutions 7 Figure 10–5: Charging Cycle energy movement The time needed in seconds is numerically equal to the number of charging cycles needed divided by two, which is the desired charge 10−9 coulombs divided by the charge per cycle q0 : 10−9 = 2.675sec 2 × 1.869 × 10−10 It seems that you can comfortably keep up with the power needed by a nanowatt load. (10–29)