CHAPTER – 33
THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT 1.
2.
i = 2 A, r = 25 , t = 1 min = 60 sec 2 Heat developed = i RT = 2 × 2 × 25 × 60 = 6000 J E=6v R = 100 , T = 15°c Heat capacity of the coil = 4 J/k Heat liberate
3.
E2 = 4 J/K × 15 Rt
66 t = 60 t = 166.67 sec = 2.8 min 100
(a) The power consumed by a coil of resistance R when connected across a supply v is P = The resistance of the heater coil is, therefore R = (b) If P = 1000 w
4.
ƒ = 1 × 10
–6
then
m
v2 R
v2 (250)2 = = 125 P 500
v2 (250 )2 = = 62.5 P 1000 P = 500 W E = 250 v R=
2
V 250 250 = = 125 P 500 2 –6 2 –7 2 (b) A = 0.5 mm = 0.5 × 10 m = 5 × 10 m
(a) R =
RA ƒl 125 5 10 7 –1 =l= = = 625 × 10 = 62.5 m A ƒ 1 10 6 –3 62.5 = 3 × 3.14 × 4 × 10 × n (c) 62.5 = 2r × n, R=
62.5
n=
5.
3
n=
2 3.14 4 10 V = 250 V P = 100 w
62.5 10 3 ≈ 2500 turns 8 3.14
v2 (250)2 = = 625 P 100 10 ƒl –8 Resistance of wire R = = 1.7 × 10 × = 0.034 A 5 10 6 The effect in resistance = 625.034 R=
The current in the conductor =
10 cm
V 220 = A R 625.034 2
220 The power supplied by one side of connecting wire = 0.034 625.034 2
220 The total power supplied = 0.034 2 = 0.0084 w = 8.4 mw 625.034 6.
E = 220 v
P = 60 w
2
R=
V 220 220 220 11 = = P 60 3
(a) E = 180 v
P=
V2 180 180 3 = = 40.16 ≈ 40 w R 220 11 33.1
Thermal & Chemical Effects of Electric Current (b) E = 240 v
P=
V2 240 240 3 = = 71.4 ≈ 71 w R 220 11 1% of 220 V = 2.2 v
7.
Output voltage = 220 ± 1%
8.
V2 (220 )2 = = 484 P 100 (a) For minimum power consumed V1 = 220 – 1% = 220 – 2.2 = 217.8 V 217.8 i= 1 = = 0.45 A R 484 Power consumed = i × V1 = 0.45 × 217.8 = 98.01 W (b) for maximum power consumed V2 = 220 + 1% = 220 + 2.2 = 222.2 V 222.2 i= 2 = = 0.459 R 484 Power consumed = i × V2 = 0.459 × 222.2 = 102 W V = 220 v P = 100 w The resistance of bulb R =
R=
9.
V2 220 220 = = 484 P 100
P = 150 w
V=
P = 1000
V = 220 v
Mass of water =
PR =
150 22 22 = 22 150 = 269.4 ≈ 270 v R=
V2 48400 = = 48.4 P 1000
1 1000 = 10 kg 100
Heat required to raise the temp. of given amount of water = mst = 10 × 4200 × 25 = 1050000 Now heat liberated is only 60%. So
V2 T 60% = 1050000 R
(220 )2 60 10500 1 T = 1050000 T = nub = 29.16 min. 48.4 100 6 60 10. Volume of water boiled = 4 × 200 cc = 800 cc T2 = 100°C T2 – T1 = 75°C T1 = 25°C Mass of water boiled = 800 × 1 = 800 gm = 0.8 kg Q(heat req.) = MS = 0.8 × 4200 × 75 = 252000 J. 1000 watt – hour = 1000 × 3600 watt-sec = 1000× 3600 J 252000 No. of units = = 0.07 = 7 paise 1000 3600 (b) Q = mST = 0.8 × 4200 × 95 J
0.8 4200 95 = 0.0886 ≈ 0.09 1000 3600 Money consumed = 0.09 Rs = 9 paise. 11. P = 100 w V = 220 v Case I : Excess power = 100 – 40 = 60 w 60 60 = 36 w Power converted to light = 100 No. of units =
(220)2 = 82.64 w 484 Excess power = 82.64 – 40 = 42.64 w
Case II : Power =
Power converted to light = 42.64
60 = 25.584 w 100 33.2
Thermal & Chemical Effects of Electric Current P = 36 – 25.584 = 10.416
10.416 100 = 28.93 ≈ 29% 36 6 12 5 12 12. Reff = 1 = i= = Amp. 5 / 2 5 8 2 Required % =
6
1 6
12 2 2i 5 24 24 3 8i = i = = Amp 5 58 5 12 3 9 i – i = Amp = 5 5 5 i 6 = (i – i)2 i 6 =
i 2 i-i
9 9 2 15 60 = 5832 5 5 2000 J of heat raises the temp. by 1K 5832 J of heat raises the temp. by 2.916K. (b) When 6 resistor get burnt Reff = 1 + 2 = 3 2
(a) Heat = i RT =
6 = 2 Amp. 3 Heat = 2 × 2 × 2 ×15 × 60 = 7200 J 2000 J raises the temp. by 1K 7200 J raises the temp by 3.6k –6 –6 2 a = – 46 × 10 v/deg, b = –0. 48 × 10 v/deg 13. = 0.001°C 2 –6 –6 2 Emf = aBlAg +(1/2) bBlAg = – 46 × 10 × 0.001 – (1/2) × 0.48 × 10 (0.001) –9 –12 –9 –8 = – 46 × 10 – 0.24 × 10 = – 46.00024 × 10 = – 4.6 × 10 V 2 14. E = aAB + bAB aCuAg = aCuPb – bAgPb = 2.76 – 2.5 = 0.26 v/°C bCuAg = bCuPb – bAgPb = 0.012 – 0.012 vc = 0 –5 E = aAB = (0.26 × 40) V = 1.04 × 10 V 15. = 0°C aCu,Fe = aCu,Pb – aFe,Pb = 2.76 – 16.6 = – 13.8 v/°C 2 BCu,Fe = bCu,Pb – bFe,Pb = 0.012 + 0.030 = 0.042 v/°C i=
a 13.8 = °C = 328.57°C b 0.042 16. (a) 1eq. mass of the substance requires 96500 coulombs Since the element is monoatomic, thus eq. mass = mol. Mass 23 6.023 × 10 atoms require 96500 C Neutral temp. on –
96500 –19 C = 1.6 × 10 C 6.023 10 23 (b) Since the element is diatomic eq.mass = (1/2) mol.mass 23 (1/2) × 6.023 × 10 atoms 2eq. 96500 C 1 atoms require
96500 2 –19 = 3.2 × 10 C 6.023 10 23 17. At Wt. At = 107.9 g/mole I = 0.500 A [As Ag is monoatomic] EAg = 107.9 g 1 atom require =
E 107.9 = = 0.001118 f 96500 M = Zit = 0.001118 × 0.5 × 3600 = 2.01
ZAg =
33.3
Thermal & Chemical Effects of Electric Current 18. t = 3 min = 180 sec w=2g –6 E.C.E = 1.12 × 10 kg/c –3 –6 3 × 10 = 1.12 × 10 × i × 180 i= 19.
3 10 3 1.12 10 6 180
1 10 2 ≈ 15 Amp. 6.72
=
H2 2g 22.4L m = Zit
20. w1 = Zit
E1 w = 1 E2 w2
1L
2 22.4
1 2 2 96500 = = 1732.21 sec ≈ 28.7 min ≈ 29 min. 5T T = 96500 22.4 22.4 5 3 96500 mm 1= 2 1.5 3600 mm = = 26.8 g/mole 2 1.5 3600 3 96500 w 107.9 107.9 3 = 1 w1 = = 12.1 gm 1 26.8 mm 3 2
Thickness = 0.1 mm 21. I = 15 A Surface area = 200 cm , 3 Volume of Ag deposited = 200 × 0.01 = 2 cm for one side For both sides, Mass of Ag = 4 × 10.5 = 42 g
E 107.9 = m = ZIT F 96500 107.9 42 96500 42 = = 2504.17 sec = 41.73 min ≈ 42 min 15 T T = 107.9 15 96500 22. w = Zit 107 .9 2.68 = i 10 60 96500 2.68 965 I= = 3.99 ≈ 4 Amp 107.9 6 2 Heat developed in the 20 resister = (4) × 20 × 10 × 60 = 192000 J = 192 KJ 23. For potential drop, t = 30 min = 180 sec Vi = Vf + iR 12 = 10 + 2i i = 1 Amp ZAg =
107.9 1 30 60 = 2.01 g ≈ 2 g 96500 2 –4 2 24. A= 10 cm × 10 cm –6 t = 10m = 10 × 10 –4 –6 2 –10 –8 3 Volume = A(2t) = 10 × 10 × 2 × 10 × 10 = 2 × 10 × 10 = 2 × 10 m –8 –5 Mass = 2 × 10 × 9000 = 18 × 10 kg –5 –7 W = Z × C 18 × 10 = 3 × 10 × C m = Zit =
q= V=
18 10 5 3 10 7
= 6 × 10
2
W 2 2 = W = Vq = 12 × 6 × 10 = 76 × 10 = 7.6 KJ q
33.4
20
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