1. ( ) (phase diagram) 1) 10% 300 2) 210 3) (β ) 4) 5) 198.5 6) β 1 2 7) Sn Zn 450 419.6 o C 400 Temperature, T / o C 350 300 250 200 150 Sn 232.0 o C 0.6 14.9 198.5 o C 10 20 30 40 50 60 70 80 90 Concentration of Zinc, c / at.% Zn 1: Sn Zn [1]
I. 1 (Sn) (Zn) 2 2.1 2 T melt 2: ( ) : Gibbs 2.2 3(A) ( )
(A) (B) A B 3: Seebeck (A) (B) ( ) Seebeck Seebeck Seebeck Seebeck 3(B) 2 A B E A ( ) E B ( ) 1K (thermocouple) ( 3(B) A B ) 1 R ( ) CO 1: + R Pt-13%Rh Pt K Ni, Cr Ni T Cu Cu, Ni
2: ( [2] ) mv 0 10 20 30 40 50 60 70 80 90 0 0.000 0.397 0.798 1.203 1.611 2.022 2.436 2.850 3.266 3.681 100 4.095 4.508 4.919 5.327 5.733 6.137 6.539 6.939 7.338 7.737 200 8.137 8.537 8.938 9.341 9.745 10.151 10.560 10.969 11.381 11.793 300 12.207 12.623 13.039 13.454 13.874 14.292 14.712 15.132 15.552 15.974 400 16.395 16.818 17.241 17.664 18.088 18.513 18.938 19.363 19.788 20.214 K ( ) 2 2 0 C 2 1 :? 3 1. 4 ( ) 2. 1
A ~ input 0.000mV AC100V 4: input :? 3. 2 ( ) 5 4. 80 3 2 (Tammann tube) 3 K 2 20 C 2 20 2.436mV(60 C ) 80
5. 4 6. 4 ( ) 7. 8. 4 5(A) 2 0 C 0 C 5(B) 20 C 100 C 1 0mV 3 2 4.095mV (A) V (B) V V o 0 C 1 2 3 V 5: 4
2 0mV 0.798mV 0.798mV 0mV 100 C 4.095 0.798 = 3.297mV 5(A) 5(B) 0 C 4.095 0 0 0.798 = 0.798mV 6: 0 C 2 6 ( ) 400 T = 0.980 T - 0.83 300 200 100 ( 2) 0 0 100 200 300 400 Reference temperature, T/ o C ( Measured temperature, T / o C ) 7: T = 1.020 (T +0.83) 7 ( A)
II. Sn Zn 1 Sn Zn [3] 2 2.1 3 3: No. 1 2 3 4 3 Zn (at.%) 10 15 25 50 Sn (at.%) 90 85 75 50 118.710 65.409 25 g 23.558 g 1.442 g 1.523 g 79 mg 23.55 1.523/1.442 = 24.881 g 25 g : ( ) 8(A) 8(B) 5 5
temperature l (A) (B) time time 8: 2 8(A) 1 l 1 : (T E ) (T L )? Gibbs 2.2 7 8 3 4 T L T E l 232.0 419.6 4: Sn 1 2 3 4 Zn x(at.%zn) 0 10 15 25 50 100 T L ( C) 232.0 419.6 T E ( C) l(min/g)
4 T L 4 T E Duration per weight, l/min g -1 0.10 0.05 0.00 S A B C 0 20 40 60 80 100 Concentration of Zn, c Zn /at.% D Z 9: 0 Zn(Sn) Sn(Zn) S Z 1g l ( 9 ) A B l = 0 S B C D l = 0 Z 3 S Z :?
III. 1 ( ) 2 3 ( ) 1. 2. 3. ( 80 ) 4. ( ) ( ) 320 320 400 90 320 400 320, 400, 600, 800, 1000, 1500 5. ( ) ( ) Al 2 O 3 ( 0.3µm) 6. 6 6 : (96%) (4%)
7. ( ) 8. (TA) (a) (b) (c) (d) 9. 7 : 3 10 10: 11 2 7 50
11: 10 12 0.95 0.4 population, f 8000 7000 800 600 400 200 0 0.0 0.5 1.0 brightness, B (arbitrary unit) 200 1/2 0.7 11 0.7 1 0.7 1800 12: 10 18 % 8 (lever rule) ( ) 9 8 9 Vegard ( ) Sn Zn Zn Sn 1%
IV. 1 1 2 1 3 1 2 2 (3 ) 4 1 2 3 4 5 3 4 6 1) 2 2) 3) 2 2 4 Q and A
10% 25% [1] H. Okamoto, Phase Diagrams for Binary Alloys, AMS International, Ohio, 2000 [2], (1996 ), [3],, 1991 [4],, 2004 [5],,, 1991
A 1 vs. 5 E = a T +b a b 4 4 0.018 (a 0.0+b) 4.12 (a 100.0+b) 9.34 (a 232.0+b) 17.23 (a 419.6+b) ( ) a b S = { 0.018 (a 0.0+b)} 2 +{4.12 (a 100.0+b)} 2 (1) + {9.34 (a 232.0+b)} 2 +{17.23 (a 419.6+b)} 2 (2) a b S a S a = S b = 0 (3) { = 2 [ 0.018 (a 0.0+b)] 0.0+[4.12 (a 100.0+b)] 100.0 } +[9.34 (a 232.0+b)] 232.0+[17.23 (a 419.6+b)] 419.6 ( ) = 2 9.809 10 3 2.399 10 5 a 7.516 10 2 b = 0 S b { = 2 [ 0.018 (a 0.0+b)]+[4.12 (a 100.0+b)] } +[9.34 (a 232.0+b)]+[17.23 (a 419.6+b)] ( ) = 2 3.067 10 1 7.516 10 2 a 4b = 0 5: ( C) (mv) 0.0-0.018 100.0 4.12 232.0 9.34 419.6 17.23
a b a = 4.099 10 2 b = 3.7 10 2 N x = x 1,x 2,...x N y = y 1,y 2,...y N ( N = 4 x 1 = 0.0 x 2 = 100.0 x 3 = 232.0 x 4 = 419.6 y 1 = 0.018 y 2 = 4.12 y 3 = 9.34 y 4 = 17.23) N S = {y i (ax i +b)} 2 (4) i=1 S a N = 2x i {y i (ax i +b)} i=1 N N N = 2x i y i +2a x 2 i +2b x i = 0 i=1 i=1 i=1 S b N = 2{y i (ax i +b)} i=1 N N N = 2y i +2a x i +2b 1 = 0 i=1 i=1 i=1 a b N N a x 2 i +b x i = i=1 i=1 N N a x i +b 1 = i=1 i=1 N x i y i (5) i=1 N y i (6) i=1 6 : y = ax 2 +bx+c 6: x i x 2 i y i x i y i 1 0.0 0.0-0.018 0.0 2 100.0 1.000 10 4 4.12 4.12 10 2 3 232.0 5.382 10 4 9.34 2.167 10 3 4 419.6 1.761 10 5 17.23 7.230 10 3 751.6 2.399 10 5 30.67 9.809 10 3
B 10 ( ) ( Newton ) q = N(T T R ) (7) q T R N t T q t = mc( T) (8) m, C ( = ) dt dt T t = q mc = N(T T R) mc 1 dt T T R dt = N mc log e (T T R ) 11 h q (8) (9) ( q +h) t = mc T (10) 1 T T R dt dt = N mc 1 T T R h mc ( ) 10 t 1, t 2, t 3... T 1, T 2, T 3... t = (t 1 +t 2)/2 ( ) dt dt = T2 T1 t 2 t 1 11 (K) (9) (11)
Temperature, T / o C 300 250 200 150 cooling curve log derivative 0.25 0.20 0.15 0.10 0.05 0.00 Logarithmic derivative (min -1 ) 100 0 10 20 30 40 50 60 70 Time, t/min 13: ( ) ( ) ( ) -0.05 13 22min ( 22 44min) 0.014min 1 ( ) (9) 45min 52min (7) (T F (m F C F )
C [4] I 30cm 1cm 1cm 10 1mm (JIS ) 21cm ( 14 ) 1mm 20.98cm 1mm 20.99cm 20.98cm 20.9 4 8 1 20.98cm 20.97cm 20.99cm 20.9 3 20.98±0.02cm 4 20.9786cm ( ) 1mm 21.00cm 21cm 10cm 20cm 22cm ( ) 5800K 5.8 10 3 K 10 29.61cm 29.61 20.98 = 621.2178cm 2 2178? 29.61cm, 20.98cm 29.60 29.62cm, 20.97 20.99cm S true 29.60 20.97 < S true < 29.62 20.99 620.712 < S true < 621.7238 621.2178 621 2 ±5 3 text 20 21 14:
3 621.2cm 2 ( 20.98cm 20.9786cm 621.2178cm 2 ) 0.31cm? m n (m < n) m / / 18 1.59mm ( 1 ) 0.31cm=3.1mm 3.1+1.59 = 4.69mm 1.59 5 3.1 1 4.69 6 6378km 3 1 3.1mm (=0.0000031km) 6378+0.0000031 = 6378.0000031km 6378km 1 6378km 3 (1.521 10 8 km) (1.471 10 8 km) 1.521 10 8 1.471 10 8 = 0.050 10 8 km 3 1? a, b δa, δb S δs S +δs S 1+ δs S = (a+δa)(b+δb) ab = ab ab + a δb + δa b ab ab = 1+ δa a + δb b (2 ) δs S = δa a + δb b ( ) 5% 2% 5%+2% = 7%
D A 15 A 5 12 Electromotive force, E / V 4.2 4.0 3.8 3.6 (a) water (heating) 3.4 0 20 40 60 80 100 Time, t / s Electromotive force, E / V 11.0 10.5 10.0 9.5 9.0 (b) Tin (cooling) 0 200 400 600 Time, t / s Electromotive force, E / V 19.0 18.0 17.0 (c) Zinc (cooling) 16.0 0 200 400 600 Time, t / s 15: (a) (b) (c) ( ) 15 15(b) 9.40mV 9.27mV E Sn = 9.34±0.07mV 7 7 4? 7: ( C) (mv) (mv) 0-0.018 0.02 100 4.12 0.01 232 9.34 0.07 419 17.23 0.02 12
Gauss [5] ( ) µ ( ) σ x x+δx P(x)dx = 1 ( ) exp (x µ)2 2πσ 2σ 2 ( ) E = at +b ( ) E = 232.0a+b 9.34mV ( ) 1 [9.34 (232.0a +b)]2 exp 2π 0.7 2 0.7 2 7 7 ( 1 P = exp (E ice at ice b) 2 ) 2πσice 2σ 2 ice ( 1 exp (E water at water b) 2 ) 2πσwater 2σ 2 water ( 1 exp (E ) tin at tin b) 2 2πσtin 2σ 2 tin ( 1 exp (E ) zinc at zinc b) 2 2πσzinc 2σzinc 2 a b P = { ( 1 exp (E i (at i +b)) 2 )} i 2πσi 2σi 2 (12) { ( 1 = } exp ) (E i (at i +b)) 2 2πσi 2σ 2 (13) i i i S(a, b; T i, E i, σ i ) = i [E i (at i +b)] 2 σ 2 i A a b n E i T i σ 2 i i=1 n E i σ 2 i=1 i = a = a n T 2 i σ 2 i i=1 n T i σ 2 i=1 i n T i +b σ 2 i=1 i n 1 +b σ 2 i=1 i 7 a = 0.0411 b = 0.00492 ( 16)
20 Electromotive force, E / mv 15 10 5 0 0 100 200 300 400 500 Temperature, T / o C 16: 4 ( σ i ) A