33.ELECTRIC CURRENT by Aditya Bhardwaj

CHAPTER – 33

THERMAL AND CHEMICAL EFFECTS OF ELECTRIC CURRENT 1.

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  

E2 = 4 J/K × 15 Rt

66  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

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=

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 = 2r × n, R=

62.5

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

V 220  220 220  11 = =  P 60 3

(a) E = 180 v

V2 180  180  3 = = 40.16 ≈ 40 w R 220  11 33.1

Thermal & Chemical Effects of Electric Current (b) E = 240 v

V2 240  240  3 = = 71.4 ≈ 71 w R 220  11 1% of 220 V = 2.2 v

Output voltage = 220 ± 1%

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 =

V2 220  220 = = 484  P 100

P = 150 w

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 = mst = 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 = mST = 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 % =

1  6 

12  2  2i 5 24 24 3 8i =  i = = Amp 5 58 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. 5T  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

W 2 2 =  W = Vq = 12 × 6 × 10 = 76 × 10 = 7.6 KJ q



33.4