Chapter 16.1 Heating and Chemical Effect of Current PDF

+ Current 101 Heating and Chemical EffectAof – – + 60 – + Joules Heating. E3 When some potential difference V is applied across a resistance R then the work done by the electric field on charge q to flow through the circuit in time t will be W = qV = Vit = i2Rt R i V 2t  Joule. B A R This work appe...

+ Current 101 Heating and Chemical EffectAof – – + 60 – + Joules Heating. E3 When some potential difference V is applied across a resistance R then the work done by the electric field on charge q to flow through the circuit in time t will be W = qV = Vit = i2Rt R i V 2t  Joule. B A R This work appears as thermal energy in the resistor. joules heating. W Vit i 2 Rt V 2t    Cal. This relation is called 4  2 4  2 4  2R J ID Heat produced by the resistance R is H  Some important relations for solving objective questions are as follow : U Condition If R and t are constant D YG H  i 2 and H  V 2 If i and t are constant (series Graph H i (or V) H grouping) HR U If V and t are constant (Parallel grouping) H  R H 1 R ST If V, i and R constant R H Ht t Electric Power. The rate at which electrical energy is dissipated into other forms of energy is called W V2  Vi  i 2 R  electrical power i.e. P  t R 40W 220V (1) Units : It’s S.I. unit is Joule/sec or Watt Bigger S.I. units are KW, MW and HP, remember 1 HP = 746 Watt (2) Rated values : On electrical appliances (Bulbs, Heater ……. etc.) 102 Heating and Chemical Effect of Current Wattage, voltage, ……. etc. are printed called rated values e.g. If suppose we have a bulb of 40 W, 220 V then rated power (PR) = 40 W while rated voltage (VR) = 220 V. It means that on operating the bulb at 220 volt, the power dissipated will be 40 W or in other words 40 J of electrical energy will be converted into heat and light per second. 60 (3) Resistance of electrical appliance : If variation of resistance with temperature is neglected then resistance of any electrical appliance can be calculated by rated power and rated V2 220  220 voltage i.e. by using R  R e.g. Resistance of 100 W, 220 volt bulb is R   484  PR 100 so Pconsumed = V2 Pconsumed (Brightness )   A2  VR PR. If VA <  . PR   VR then Pconsumed V A2  R also ID VR E3 (4) Power consumed (illumination) : An electrical appliance (Bulb, heater, …. etc.) consume rated power (PR) only if applied voltage (VA) is equal to rated voltage (VR) i.e. If VA = we have VR2 R PR so 2 U  110  e.g. If 100 W, 220 V bulb operates on 110 volt supply then Pconsumed     100  25 W  220   D YG Note :  If VA < VR then % drop in output power (PR  Pconsumed )  100 PR  For the series combination of bulbs, current through them will be same so they will consume power in the ratio of resistance i.e., P  R {By P = i2R) while if they are connected in parallel i.e. V is constant so power consumed by them is in the reverse 1 ratio of their resistance i.e. P . R ST U (5) Thickness of filament of bulb : We know that resistance of filament of bulb is given by V2 1 l R  R , also R   , hence we can say that i.e. If rated power of a bulb is A  PR  ( Thickness ) PR R A more, thickness of it’s filament is also more and it’s resistance will be less. V A2 1 P  If applied voltage is constant then P(consumed )  (By ). Hence if different bulbs R R 1 (electrical appliance) operated at same voltage supply then Pconsumed  P R  thickness  R Note :  Different bulbs 25W 220V  Resistance  Thickness of filament  Brightness 100 W 220V 1000 W 220V R25 > R100 > R1000 t1000 > t100 > t40 B1000 > B100 > B25 Heating and Chemical Effect of Current 103 (6) Long distance power transmission : When power is transmitted through a power line of resistance R, power-loss will be i 2 R Now if the power P is transmitted at voltage V Power loss  P  Vi i.e. i  (P / V ) So, P2 R V2 60 Now as for a given power and line, P and R are constant so Power loss  (1 / V 2 ) So if power is transmitted at high voltage, power loss will be small and vice-versa. e.g., E3 power loss at 22 kV is 10–4 times than at 220 V. This is why long distance power transmission is carried out at high voltage. (7) Time taken by heater to boil the water : We know that heat required to raise the temperature  of any substance of mass m and specific heat S is H = m.S. Here heat produced by the heater = Heat required to raise the temp.  of water. for m kg water t t J (m.S. ) p ID p  t = J  m.S.  {J = 4.18 or 4.2 J/cal) 4180 ( or 4200 ) m  p U i.e. D YG Note :  If quantity of water is given n litre then t {S = 1000 cal/kgoC) 4180 (4200 ) n  p Electricity Consumption. (1) The price of electricity consumed is calculated on the basis of electrical energy and not on the basis of electrical power. (2) The unit Joule for energy is very small hence a big practical unit is considered known as kilowatt hour (KWH) or board of trade unit (B.T.U.) or simple unit. U (3) 1 KWH or 1 unit is the quantity of electrical energy which dissipates in one hour in an ST electrical circuit when the electrical power in the circuit is 1 KW thus 1 KW = 1000 W  3600 sec = 3.6  106 J. (4) Important formulae to calculate the no. of consumed units is n  Total watt  Total hours 1000 Concepts  When some potential difference applied across the conductor then collision of free electrons with ions of the lattice result’s in conversion of electrical energy into heat energy  If a heating coil of resistance R, (length l) consumed power P, when voltage V is applied to it then by keeping V is constant if it is cut in n equal parts then resistance of each part will be R n and from Pconsumed  1 R , 104 Heating and Chemical Effect of Current power consumed by each part P'  nP.  Joule’s heating effect of current is common to both ac and dc. Example: 1 60 Example s The approximate value of heat produced in 5 min. by a bulb of 210 watt is (J = 4.2 joule/calorie) [MP PET 2000; MNR 1985] (b) 1,050 (c) 63,000 P  t 210  5  60   15000 Cal 4.2 4.2 (d) 80,000 E3 (a) 15,000 By using H  Example: 2 A heater coil is cut into two parts of equal length and one of them is used in the heater. The ratio of the heat produced by this half coil to that by the original coil is (a) 2 : 1 (c) 1 : 4 (d) 4 : 1 If suppose resistance of the coil is R so resistance of it’s half will be  V 2t 1 H R R H Half R Full R 2   R Half R/2 1 D YG H  H Full  R. Hence by using 2 U Solution : (a) (b) 1 : 2 ID Solution : (a) Note :  In general if coil is divided in n equal parts then heat produced by each part will be n times of the heat produced by coil it self i.e. H = nH Example: 3 If current in an electric bulb changes by 1%, then the power will change by (b) 2% U (a) 1% By using P = i2R  P  i2  Example: 4 Solution : (b) (d) 1 % 2 P i 2  change in power = 2% P i ST Solution : (b) (c) 4% A constant voltage is applied on a uniform wire, then the heat is produced. The heat so produced will be doubled, if (a) The length and the radius of wire are halved are doubled (b) Both length and radius (c) Only the length is doubled doubled (d) Only By using H  l l V 2t and R    A  r2 R heat will be doubled.  H  the radius is V 2 t r 2 r2 H  ; on doubling both r and l l l Heating and Chemical Effect of Current 105 Example: 5 An electric heater of resistance 6 ohm is run for 10 minutes on a 120 volt line. The energy liberated in this period of time is (a) 7.2  10 3 J (c) 43.2  10 4 J (b) 14.4  10 5 J (d) 28.8  10 4 J (120 ) 2  10  60 V 2t  H   14.4  10 5 J 6 R By using H  Example: 6 An electric bulb of 100 W is designed to operate on 220 V. Resistance of the filament is 60 Solution : (b) [EAMCET 1981, 82; MP PMT 1993, 97] (b) 100  (c) 22000  V2 V 2 (220 ) 2 R   484  R P 100 (d) 242  E3 (a) 484  Solution : (a) By using P  Example: 7 An electric bulb is rated 220 V and 100 W. Power consumed by it when operated on 110 volt is (a) 50 W (b) 75 W V   A  VR 2 ID [AFMC 2000; MP PMT 1986, 94; CPMT 1986] (c) 90 W (d) 25 W   110    PR  PConsumed     100  25 W  220   2 By using Pconsumed Example: 8 A 500 watt heating unit is designed to operate from a 115 Volt line. If the line voltage drops to 110 volt, the percentage drop in heat output will be D YG Solution : (c) U Solution : (d) (a) 10.20% (b) 8.1% V By using Pconsumed   A  VR   110    PR  PConsumed     500  456.6 Watt  115   2 So % drop in heat output  (a) Rs. 1.20 So cost  Example: 10 2 PActual  PConsumed (500  456.6)  100   100  8.6 % PActual 500 (b) Rs. 4.00 By using consumed unit (n) or KWH  ST Solution : (d) (d) 7.6% An electric lamp is marked 60 W, 230 V. The cost of 1 kilowatt hour of power is Rs. 1.25. The cost of using this lamp for 8 hours is U Example: 9 (c) 8.6% (c) Rs. 0.25 (d) Rs. 0.60 Total Watt  Total tim e 60  8 12   n 1000 1000 25 12  1.25  0.60 Rs 25 How much energy in Kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days) (a) 1500 (b) 15.000 (c) 15 (d) 150 (50  10 )  (10  30 ) Total Watt  Total tim e  150  n 1000 1000 Solution : (d) By using n  Example: 11 An immersion heater is rated 836 watt. It should heat 1 litre of water from 20o C to 40o C in about (a) 200 sec (b) 100 sec (c) 836 sec (d) 418 sec 106 Heating and Chemical Effect of Current 4180  n   P  t 4180  1  (40  20 )  100 sec 836 Solution : (b) By using t  Example: 12 The power of a heater is 500 watt at 800o C. What will be its power at 200o C if   4  10 4 per C (a) 484 W 500 1.08   611 W P2 1.32 A heater of 220 V heats a volume of water in 5 minute time. A heater of 110 V heats the same volume of water in (a) 5 minutes Solution : (d) (c) 10 minutes (d) 20 minutes V 2t. Here volume of water is same. So same heat is required in both cases. R 1 Resistance is also constant so V2t = constant  t  2  V By using H  t 1  V2  t 2  V1 2  5  110  1       t 2  20 min t 2  220  4  2 Water boils in an electric kettle in 15 minutes after switching on. If the length of the heating wire is decreased to 2/3 of its initial value, then the same amount of water will boil with the same supply voltage in [MP PMT 1994] D YG Example: 14 (b) 8 minutes ID Example: 13 P R (1   t 2 ) 500 (1  4  10 4  200 ) V2 1  1  2    P  P2 R1 (1   t1 ) P2 R R (1  4  10 4  800 ) E3  (d) 611 W 60 By using P  i 2 R  (c) 526 W U Solution : (d) (b) 672 W (a) 15 minutes (b) 12 minutes (c) 10 minutes (d) 8 minutes V 2t A l V 2t where R    H . Since volume is constant so H is also constant A l R 2 l1 t2 l2 t2 3   t 2  10 min  so t  l which gives  15 l1 t1 l1 U Solution : (c) By using H  Tricky Example: 1 ST If resistance of the filament increases with temperature, what will be power dissipated in a 220 V – 100 W lamp when connected to 110 V power supply (a) 25 W these Solution : (b) If (b) < 25 W resistance 2 do (c) > 25 W not varies (d) None with of temperature V   110  PConsumed   A . PR     100  25 W. But actually resistance is increasing with  220   VR  temperature so consumed power will be lesser then 25W. 2 Combination of Bulbs (or Electrical Appliances). Heating and Chemical Effect of Current 107 Bulbs (Heater etc.) are in series (1) Total power consumed Ptotal  P1  P2  P3......  Pn 1 1  ...... P1 P2 Pconsumed (Brightness )  V  R  P2 Supply Supply (2) If ‘n’ bulbs are identical, Ptotal  (3) P1 P N 60 Ptotal  (1) Total power consumed P2 P1 (2) If ‘n’ identical bulbs are in parallel. Ptotal  nP 1 Prated i.e. in series combination bulb of lesser wattage will give more bright light and p.d. appeared across it will be more. 1 i.e. in parallel R combination bulb of greater wattage will give more bright light and more current will pass through it. (3) Pconsumed (Brightness )  PR  i  E3 1 Bulbs (Heater etc.) are in parallel ID Some Standard Cases for Series and Parallel Combination. nP hence PP  n2. PS P and then connected in parallel. So PP = n U (1) If n identical bulbs first connected in series so PS  D YG (2) To operate a bulb on voltage which is more then it’s rated voltage, a proper resistance is connected in series with it. e.g. to glow a bulb of 30 W, 6 V with full intensity on 126 volt required series resistance calculated as follows Bulb will glow with it’s full intensity if applied voltage on it is 6 V i.e. 120 V appears across the series resistance R current flows through bulb = current flows through resistanceR i 30 i  5 amp 6 6 V U Hence for resistance V  iR i.e. 120 = 5  R  5  R  R = 24  ST Note :  If you want to learn Short Trick then remember 120 V 126 V  Voperating  VR Series resistance   PR     VR   (3) An electric kettle has two coils when one coil is switched on it takes time t1 to boil water and when the second coil is switched on it takes time t2 to boil the same water. If they are connected in series If they are connected in parallel PP  P1  P2 1 1 1   PS P1 P2  1 1 1   H S / tS H 1 / t1 H 2 / t 2  HP H H  1  2 tp t1 t2 108 Heating and Chemical Effect of Current  H S  H 1  H 2 so t s  t1  t 2  i.e. time taken by combination to boil the same quantity of water t S  t 1  t 2 i.e. time taken by parallel combination to boil H p  H 1  H 2 so 1 1 1   tp t1 t 2 t1t 2 t1  t 2 60 the same quantity of water t p  A B E3 (4) If three identical bulbs are connected in series as shown in figure then on closing the switch S. Bulb C short circuited and hence illumination of bulbs A and B increases C Reason : Voltage on A and B increased. S – ID V + U (5) If three bulbs A, B and C are connected in mixed combination as shown, then illumination of bulb A decreases if either B or C gets fused D YG A Reason : Voltage on A decreases. V B C (6) If two identical bulb A and B are connected in parallel with ammeter A and key K as shown in figure. U It should be remembered that on pressing key reading of ammeter becomes twice. ST A K Reason : Total resistance becomes half. V A B Concepts When a heavy current appliance such us motor, heater or geyser is switched on, it will draw a heavy current from the source so that terminal voltage of source decreases. Hence power consumed by the bulb decreases, so the light of bulb becomes less. r ~ Heater  K Heating and Chemical Effect of Current 109 If the source is ideal i.e. r = 0, there will be no change in the brightness of the bulb. 60  Example s An electric kettle has two heating coils. When one of the coils is connected to ac source the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of E3 Example: 15 water to boil will be Solution : (d) (b) 25 min By using the formula t p  (c) 15 min (d) 8 min ID (a) 4 min [CBSE PMT 2003] t1 t 2 10  40 (as we discussed in theory)  t p   8 min t1  t 2 (10  40 ) U Note :  In this question if coils are connected in series then the time taken by the same quantity of water to boil will be ts = t1 + t2 = 10 + 40 = 50 min If a 30 V, 90 W bulb is to be worked on a 120 V line, a resistance of how many ohms should be connected in series with the bulb D YG Example: 16 (a) 10 ohm Solution : (c)  Voperating  VR By using Series resistance R   PR  R (c) 30 ohm (d) 40 ohm    VR (As we discussed in theory)    (120  30 )  30  30  90 In the circuit shown in figure, the heat produced in 5 ohm resistance is 10 calories per U Example: 17 (b) 20 ohm second. The heat produced in 4 ohm resistance is ST 4 6 (a) 1 cal/sec (b) 2 cal/sec 5 (c) 3 cal/sec (d) 4 cal/sec Solution : (b) Ratio of currents i1 10 2   by using H = i 2 Rt i2 5 1 i2 6 4 Line (2) i  H1  i1  H 2  i2 2  R 10  2  5   1       H 2  2cal / sec R H 1 4   2 2  2 5 i1 Line (1) 110 Heating and Chemical Effect of Current Example: 18 Two heater wires of equal length are first connected in series and then in parallel. The ratio of heat produced in the two cases is [MP PET 2002, 1999; MP PMT 2001, 2000, 1996; AIIMS 2000; MNR 1987; DCE 1997, 94] (a) 2 : 1 60 V 2t 1 H R R Hs Hs R R/2 1   P ;   HP HP 2R Rs 4 E3  If two bulbs of wattage 25 and 100 respectively each rated at 220 volt are connected in series with the supply of 440 volt, then which bulb will fuse (a) 100 watt bulb Solution : (b) (d) 1 : 4 Both the wires are of equal length so they will have same resistance and by using H  Example: 19 (c) 4 : 1 In series V A  (b) 25 watt bulb (c) None of them (d) Both of them 1 i.e. voltage appear on 25W bulb will be more then the voltage appears on PR ID Solution : (d) (b) 1 : 2 100 W bulb. So bulb of 25 W will gets fused. Example: 20 Three equal resistors connected in series across a source of e.m.f. together dissipate 10 D YG U watt. If the same resistors are connected in parallel across the same e.m.f., then the power dissipated will be (a) 10 W Solution : (d) (b) 30 W In series consumed power Ps  [KCET 1999; DCE 1998; CBSE 1998; MP PAT 1996] (c) 10/3 W (d) 90 W P while in parallel consumed power Pp = nP  PP = n2. Ps n  PP  (3)2  10  90W Example: 21 Forty electric bulbs are connected in series across a 220 V supply. After one bulb is fused, U the remaining 39 are connected again in series across the same supply. The illumination will be [Haryana CEE 1996; NCERT 1972] (a) More with 40 bulbs than with 39 ST (c) Equal in both the cases 392 (b) More with 39 bulbs than with 40 (d) In the ratio of 402 : V2. Initially there were 40 bulbs in series so equivalent R resistance was 40 R, finally 39 bulbs are in series so equivalent resistance becomes 39 R. Since resistance decreases so illumination increases with 39 bulbs. Solution : (b) Illumination = PConsumed  Example: 22 Two bulbs of 100 watt and 200 watt, rated at 220 volts are connected in series. On supplying 220 volts, the consumption of power will be (a) 33 watt (b) 66 watt (c) 100 watt (d) 300 watt Heating and Chemical Effect of Current 111 P1 P2 100  200  PConsumed   66 W P1  P2 300 In series PConsumed  Example: 23 Two wires ‘A’ and ‘B’ of the same material have their lengths in the ratio 1 : 2 and radii in the ratio 2 : 1. The two wires are connected in parallel across a battery. The ratio of the heat produced in ‘A’ to the heat produced in ‘B’ for the same time is Resistance R   By using H  l r2 (b) 400 J 2 (c) 25 J (d) 50 J Let resistance of the heating coil be R, when coil cut in two equal parts, resistance of each R R. When these two parts are corrected in parallel, R eq  i.e. resistance 4 2 1 ; Power becomes 4 times i.e. P = 4P = 400 J/sec R D YG becomes, so according to P  Two identical electric lamps marked 500 W, 220 V are connected in series and then joined to a 110 V line. The power consumed by each lamp is (b) 25 W 4 (a) 125 W 4 Solution : (a) r  R RA l 1 1 1  A   B   A      RB 2 2 8 RB l B  rA  A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source. The energy now liberated per second is part will be Example: 25 2  H R V 2t 8  A  B  R HB RA 1 (a) 200 J Solution : (b)  r2 R (d) 8 : 1 ID Example: 24 l (c) 1 : 8 U Solution : (d) (b) 2 : 1 E3 (a) 1 : 2 60 Solution : (b) (c) 225 W 4 (d) 125 W 500 W 220 V Both bulbs are identical so voltage across each bulb will be 55V. 2  125 55 V  55    PR   W   500  4  220   2 55 V 110 V U V Hence power consumed by each bulb is  A  VR 500 W 220 V ST Tricky Example: 2 Electric bulb 50 W – 100 V glowing at full power are to be used in parallel with battery 120 V, 10 . Maximum number of bulbs that can be connected so that they glow in full power is [CPMT 2002] (a) 2 Solution : (c) When (b) 8 bulb glowing at (c) 4 full power, (d) 6 current flows through it i P 50 1 n    amp  i  and voltage across the bulb is 100 V. If suppose n bulbs are n V 100 2 2 connected in parallel with cell as shown in figure then according to the cell equation E  V  i r  120  100  n  10  n i /4. 2 n i/ n i i/ n 1 2 n 120 V, 10  60 112 Heating and Chemical Effect of Current E3 Thermo Electric Effect of Current ID If two wires of different metals are joined at their ends so as to form two junctions, then the resulting arrangement is called a “Thermo couple”. Seebeck Effect. (1) Definition : When the two junctions of a thermo couple are maintained at different U temperatures, then a current starts flowing through the loop known as thermo electric current. The potential difference between the junctions is called thermo electric emf which is of the order of a few micro-volts per degree temperature difference (V/oC). D YG i Fe Hot Cu Cold G Cu ST U (2) Origin of thermo emf : The density of free electrons in a metal is generally different from the density of free electrons in another metal. When a metal is brought into intimate contact (say by soldering) with other metal, the electrons tend to diffuse from one metal to another, so as to equalise the electron densities. As an illustration, when copper is brought into intimate contact with iron, the electrons diffuse from iron to copper. But this diffusion cannot go on continuously because due to diffusion, the potential of copper decreases and the potential of iron increases. In other words, iron becomes positive with respect to copper. This is what stops further diffusion. In the case of thermocouple whose junctions are at the same temperature, the emf’s at the junctions will be equal in magnitude but opposite in direction. So, the net emf for the whole of thermocouple will be zero. Let us now consider the case when the temperature of one junction of the thermocouple is raised. Raising the temperature of one junction will affect the electron density in the two metals differently. Moreover, the transfer of electrons at the junction will be easier than the transfer of electrons at the cold junction. Due to both these reasons, the emf’s at the two junctions will be different. This produces a net emf in the thermocouple. This emf is known as Seebeck emf. Heating and Chemical Effect of Current 113 (3) Seebeck series : The magnitude and direction of thermo emf in a thermocouple depends not only on the temperature difference between the hot and cold junctions but also on the nature of metals constituting the thermo couple. Seebeck arranged different metals in the decreasing order of their electron density. Some of the metals forming the series are as below. 60 Sb, Fe, Ag, Au, Sn, Pb, Cu, Pt, Ni, Bi E3 (i) About magnitude thermo emf : Thermo electric emf is directly proportional to the distance between the two metals in series. Farther the metals in the series forming the thermo couple greater is the thermo emf. Thus maximum thermo emf is obtained for Sb-Bi thermo couple. (ii) Direction of thermo electric current : If a metal occurring earlier in the series is ID termed as A and the metal occurring later in the series is termed as B, then the rule for the direction of conventional current in thermocouple made of elements A and B is ABC. That is, at the cold junction current will flow from A to B. e.g. in Fe-Cu thermocouple, at the cold junction current flows from A to B that is from Fe to Cu. At the hot junction, the current flows from Cu to Fe. This may be remembered easily by the hot coffee. U (4) Law of thermoelectricity (i) Law of successive temperature : If initially temperature limits of the cold and the hot D YG junction are t1 and t2, say the thermo emf is E tt12. When the temperature limits are t2 and t3, then say the thermo emf is E tt23 then E tt12  E tt23  E tt13 where E tt13 is the thermo emf when the temperature limits are E tt13 (ii) Law of intermediate metals : Let A, B and C be the three metals of Seebeck series, where B lies between A and C. According to this law, E AB  E BC  E CA U When tin is used as a soldering metal in Fe-Cu thermocouple then at the junction, two different thermo couples are being formed. One is between iron and tin and the other is between tin and copper, as shown in figure (i) ST Now iron is thermoelectrically more positive as compared to tin and tin is more positive with respect to copper (the element which occurs earlier in the seebeck series gets positively charged on losing the electrons at the junction), so as clear from the figure below, the thermo emf’s of both the thermocouples shown in the figure (ii) are additive Cu Sn Cu E Sn Sn E Fe Fe (i) (ii ) 114 Heating and Chemical Effect of Current  If soldering metal in a thermocouple is an intermediate metal in the series then thermo emf will not be affected. 60 It is also clear from the above discussions that if the soldering metal does not lie between two metals (in Seebeck series) of thermocouple then the resultant emf will be subtractive. (5) Effect of temperature on thermo emf : In a thermocouple as the temperature of the E3 hot junction increases keeping the cold junction at constant temperature (say 0 oC). The thermo emf increases till it becomes maximum at a certain temperature. (i) Thermo electric emf is given by the equation E   t  1  t2 2 ID where  and  are thermo electric constant having units are volt/oC and volt/oC2 respectively (t = temperature of hot junction). U (ii) The temperature of hot junction at which thermo emf becomes maximum is called neutral temperature (tn). Neutral temperature is constant for a thermo couple (e.g. for Cu-Fe, tn = 270oC) (iii) Neutral temperature is independent of the temperature of cold junction. D YG (iv) If temperature of hot junction increases beyond neutral temperature, thermo emf start decreasing and at a particular temperature it becomes zero, on heating slightly further, the direction of emf is reversed. This temperature of hot junction is called temperature of inversion (ti). (v) Graphical representation of thermo emf U t t (a) t n  i c 2 Thermo emf (b) Graph is parabolic (c) For E to be maximum (at t = tn) ST dE   0 i.e.  +  tn = 0  t    dt O tn ti Temperature of hot junction (6) Thermo electric power : The rate of change of thermo emf with the change in the temperature of the hot junction is called thermoelectric power. It is also given by the slope of parabolic curve representing the variation of thermo emf with temperature of the hot junction, as discussed in previous section. It is observed from the above graph that as temperature of hot junction increases from that of the cold junction to the neutral junction, though the thermo emf is increasing but the slope of the graph, that is the rate of change of thermo emf with temperature of hot junction is Heating and Chemical Effect of Current 115 decreasing. Note that, at the neutral temperature, the thermo emf is maximum but the slope i.e. the thermoelectric power is zero. 60  dE  The thermo electric power   is also called Seebeck coefficient. Differentiating both  dt  dE d 1 sides of the equation of thermo emf with respect to t, we have P   ( t   t 2 )  dt dt 2 P    t The equation of the thermo electric power is of the type y  mx  c, so the graph of thermo E3 electric power is as shown below. P  ID Slope  t U Peltier Effect. D YG (1) If a current is passed through a junction of two different metals, the heat is either evolved or absorbed at the junction. This effect is known as Peltier effect. It is the reverse of Seebeck effect. Before going into the detailed explanation, we will first revise an important concept about absorption and evolution of energy when electric charge is made to pass through two points having some potential difference. When a positive charge flows from high potential to low potential, it releases energy and when positive charge flows from low potential to high potential it absorbs energy. (2) Explanation of Peltier effect : In the light of above statement it can be seen that if U current is made to flow in Fe-Cu thermocouple by connecting it to a battery then the junction at which current goes from Fe to Cu becomes hot because here positive charge is flowing from high ST potential to low potential, so energy is released. Remember that, in iron-copper thermocouple, the polarity of the contact potential at each junction is such iron is at higher potential. Similarly the junction where current flows from Cu to Fe becomes colder because at this junction current is flowing from negative to positive potential, so energy is absorbed. Thus it is observed that on application of potential difference in a thermocouple temperature difference is automatically created. The amount of heat absorbed at cold junction is equal to the heat released at hot junction. – + Cu – + Fe 116 Heating and Chemical Effect of Current (3) Peltier co-efficient () : Heat absorbed or liberated at the junction is directly proportional to the charge passing through the junction i.e. H  Q  H = Q ; where  is called Peltier co-efficient. It’s unit is J/C or volt. (i) If Q = 1 then H   i.e. Peltier co-efficient of a junction is defined as heat absorbed or 60 liberated at the junction when a unit quantity of electric charge flows across the junction (H is also known as Peltier emf). (ii) Relation between  and absolute temperature : Suppose the temperature of the cold E3 junction is T and that of the hot junction is T + dT and let dE be the thermo emf produced, then dE dE it is found that   T  T  S ; where T is in Kelvin and  P  Seebeck coefficient S dT dT ID (iii) -depends on : (a) Temperature of junction (b) Difference in electron density of the two metal used in thermocouple. (iv) Comparison between Joule and Peltier effect Joules effect Peltier’s Effect (a) In peltier’s effect energy is released at one junction and absorbed at the other junction. U (a) In joule’s effect energy is only released. (b) Heat produced depends upon i1,  junction at which the heat is released or absorbed changes when the direction of current changes. (c) It is identically produced by ac or dc (c) In Peltier’s effect if ac is passed, at the U D YG (b) Heat produced depends upon i2, so, heat is always released, whether i is positive or negative. ST (d) Joules effect is irreversible. (e) In Joule’s effect heat is released throughout the length of wire. same junction heat is released when current flows in one direction and absorbed when the direction of current reverses. The net amount of heat released or absorbed at a junction is therefore zero. Thus, Peltier’s effect cannot be observed with ac. (d) Peltier effect is reversible, complimentary is Seebeck effect. its (e) In this effect heat is released or absorbed only at the junctions. Thomson’s Effect. (1) Definition : In Thomson’s effect we deal with only metallic rod and not with thermocouple as in Peltiers effect and Seebeck’s effect. (That’s why sometimes it is known as homogeneous thermo electric effect. When a current flows thorough an unequally heated metal, there is an absorption or evolution of heat in the body of the metal. This is Thomson’s effect. (2) Types of Thomson’s effect Heating and Chemical Effect of Current 117 (i) Positive Thomson’s effect In positive Thomson’s effect it is found that hot end is at high potential and cold end is at low potential. e.g. Cu, Sn, Ag, Sb (ii) 60 Element’s occurring before lead in Seebeck series are called thermoelectrically negative but this does not mean that their Thompson effect is negative. Negative Thomson’s effect E3 In the elements which show negative Thomson’s effect, it is found that the hot end is at low potential and the cold end is at higher potential e.g. Fe, Co, Bi (3) Thomson’s co-efficient : In Thomson’s effect it is found that heat released or absorbed is proportional to Q i.e. H  Q  H  σQ Δθ where  = Thomson’s coefficient. It’s unit is Joule/coulomboC or volt/oC and  = temperature difference. ID (i) If Q = 1 and  = 1 then σ  H so the amount of heat energy absorbed or evolved per second between two points of a conductor having a unit temperature difference, when a unit current is passed is known as Thomson’s co-efficient for the material of a conductor. constant  dS dt dE dT so d2E also dT 2 dS d 2 E  dS  hence   T     T   ; where  = Thermo electric 2 dT dT  dT  D YG Seebeck co-efficient S  U (ii) It can be proved that Thomson co-efficient of the material of conductor   T Application of Thermo Electric Effect. (1) To measure temperature : A thermocouple is used to measure very high (2000 oC) as well as very low (– 200oC) temperature in industries and laboratories. The thermocouple used to measure very high temperature is called pyrometer. U (2) To detect heat radiation : A thermopile is a sensitive instrument used for detection of heat radiation and measurement of their intensity. It is based upon Seebeck effect. Sb Heat radiations ST A thermopile consists of a number of thermocouples of Sb-Bi, all connected in series. T1 G Bi T2 This instrument is so sensitive that it can detect heat radiations from a match stick lighted at a distance of 50 metres from the thermopile. 118 Heating and Chemical Effect of Current (3) Thermoelectric refrigerator : The working of thermo-electric refrigerator is based on Peltier effect. According to Peltier effect, if current is passed through a thermocouple, heat is absorbed at one junction and is evolved at the other junction of the thermocouple. If on the whole, the heat is absorbed, then the thermocouple acts as thermoelectric refrigerator. It’s 60 efficiency is small in comparison to conventional refrigerator. (4) Thermoelectric generator : Thermocouple can be used to generate electric power using Seebeck effect in remote areas. It can be achieved by heating one junction in a flame of kerosene oil lamp and keeping the other junction at room or atmospheric temperature. The Concepts E3 thermo emf so developed is used to operate radio receivers or even radio transmitters. The emf developed in a thermo couple is rather small i.e. of the order of a few V/oC.  A current is passed in a thermocouple formed with dissimilar metals whose one junction is heated and other is cooled. If 1 and 2 are the Peltier co-efficient of cold and hot junction respectively then the net emf across the junction is proportional to (2 – 1) ID  The smallest temperature difference that can be measured with a combination of a thermocouple of thermo e.m.f. 30 V per degree and a galvanometer of 50 ohm resistance capable of measuring a minimum current of 3 × 10–7 ampere is D YG Example: 26 U Example s (a) 0.5 degree (b) 1.0 degree (c) 1.5 degree (d) 2.0 degree Solution : (a) By using E = a  i R = a  3  10 Example: 27 The expression for thermo e.m.f. in a thermocouple is given by the relation E  40   20 , where  is the temperature difference of two junctions. For this, the neutral temperature will be [AMU (Engg.) 2000] –7 U (a) 100o C     = 0.5 degree 2 (b) 200o C (c) 300o C Comparing the given equation of thermo e.m.f. with E =  t +  1. By using t n    t n  400 o C.  10 (d) 400o C 1  t 2 we get   40 and 2 ST Solution : (d)  50 = 30  10 –6   Example: 28 One junction of a certain thermoelectric couple is at a fixed temperature Tr and the other junction is at temperature T. The thermo electromotive force for this is expressed by 1   E  K (T  Tr ) T0  1 (T  Tr ) at temperature T = T0 , the thermo electric power will be 2 2   (a) 1 2 KT0 (b) KT0 Solution : (a) As we know thermo electric power S  putting T  1 1 T0 we get S  KT0. 2 2 (c) 1 2 KT02 (d) 1 2 K(T0  Tr ) 2 dE. Hence by differentiating the given equation and dT Heating and Chemical Effect of Current 119 Example: 29 The cold junction of a thermocouple is maintained at 10 o C. No thermo e.m.f. is developed when the hot junction is maintained at 530o C. The neutral temperature is (a) 260o C Solution : (b) (b) 270o C (c) 265o C Given t c  10 o C and t i  530 o C hence by using t n  (d) 520o C ti  tc  t n  270 o C 2 60 Example: 30 The thermo emf develops in a Cu-Fe thermocouple is 8.6 V/oC. It temperature of cold junction is 0oC and temperature of hot junction is 40oC then the emf obtained shall be Solution : (a) (b) 3.44 V (c) 3.44 V By using thermo emf e = a where a  8.6 V o So e = 8.6  10–6  40 = 344 V = 0.344 mV. C (d) 3.44 mV and  = temperature difference = 40oC E3 (a) 0.344 mV (a) 65 V (b) 60 V ID Example: 31 A thermo couple develops 200 V between 0oC and 100oC. If it develops 64 V and 76 V respectively between (0oC – 32oC) and (32oC – 70oC) then what will be the thermo emf it develops between 70oC and 100oC (c) 55 V (d) 50 V A thermo couple is formed by two metals X and Y metal X comes earlier to Y in Seebeck series. If temperature of hot junction increases beyond the temperature of inversion. Then direction of current in thermocouple will so D YG Example: 32 U 100 100 70 100 Solution : (b) By using e 100  200  64  76  e 70  e 70  60 V  e 032  e 32  e 70 0 (a) X to Y through cold junction (b) X to Y through hot junction (c) Y to X through cold junction (d) Both (b) and (c) In the normal condition current flows from X to Y through cold. While after increasing the temperature of hot junction beyond temperature of inversion. The current is reversed i.e. X to Y through hot junction or Y to X through cold junction. Example: 33 Peltier co-efficient of a thermo couple is 2 nano volts. How much heat is developed at a junction if 2.5 amp current flows for 2 minute U Solution : (d) (a) 6 ergs (b) 6  10–7 ergs (c) 16 ergs (d) 6  10–3 erg ST Solution : (a) H   i t  (2  10 9 )  2.5  (2  60)  6  10 7 J  6 erg Example: 34 A thermo couple develops 40 V/kelvin. If hot and cold junctions be at 40oC and 20oC respectively then the emf develops by a thermopile using such 150 thermo couples in series shall be (a) 150 mV Solution : (d) (b) 80 mV (c) 144 mV (d) 120 mV The temperature difference is 20oC = 20 K. So that thermo emf developed E = a  40 V K  20 K  800 V. Hence total emf = 150  800 = 12  104 V = 120 mV 120 Heating and Chemical Effect of Current Chemical Effect of Current 60 Current can produce or speed up chemical change, this ability of current is called chemical effect (shown by dc not by ac). E3 When current is passed through an electrolyte, it dissociates into positive and negative ions. This is called chemical effect of current. Important Terms Related to Chemical Effect. These liquids which do not allow current to pass through them are called insulators (e.g. vegetable oils, distilled water etc.) while the liquids which allows the current to pass through them but do not dissociates into ions are called good conductors (e.g. Hg etc.) U Note :  ID (1) Electrolytes : The liquids which allows the current to pass through them and also dissociates into ions on passing current through them are called electrolytes e.g. solutions of salts, acids and bases in water, etc.  Solutions of cane sugar, glycerin, alcohol etc. are examples of non-electrolytes. D YG (2) Electrolysis : The process of decomposition of electrolyte solution into ions on passing the current through it is called electrolysis. Electric current Electroly te Positive ions (called cations) Decompose d into Negative ions (called Anions) Note :  Practical applications of electrolysis are Electrotyping, extraction of metals from the U ores, Purification of metals, Manufacture of chemicals, Production of O2 and H2, Medical applications and electroplating. ST  Electroplating : It is a process of depositing a thin layer of one metal over another metal by the method of electrolysis. The articles of cheap metals are coated with precious metals like silver and gold to make their look more attractive. The article to be electroplated is made the cathode and the metal to be deposited in made the anode. A soluble salt of the precious metal is taken as the electrolyte. (If gold is to be coated then auric chloride is used as electrolyte). (3) Electrodes : Two metal plates which are partially dipped in the electrolyte for passing the current through the electrolyte. Anode : Connected to positive terminal of battery Cathode : Connected to negative terminal of battery (4) Voltameter : The vessel in which the electrolysis is carried out is called a voltameter. It contains two electrodes and electrolyte. It is also known as electrolytic cell. Anode + Cathod – e 60 Heating and Chemical Effect of Current 121 E3 (5) Equivalent weight : The ratio of the atomic weight of an element to its valency is defined as it’s equivalent weight. (6) Types of voltameter : Voltameter is divided mainly in following types Ag voltameter Water voltameter In copper voltameter, electrolyte is solution of copper e.g. CuSO4, CuCl2, Cu(NO3)2 etc. Cathode may be of any material, but anode In silver voltameter electrolyte is a solution of silver, e.g. AgNO3. Cathode may be of any metal but anode must be of silver. The dissociation In water voltameter the electrolyte used is acidic water, because it is much more conducting than that of pure water. So acid CH2SO4 must be of copper. reaction is as follows increases the concentration of free ions in the solution. The electrodes are made of platinum, because it does not dissolve into electrolyte and does not react with the CuSO4 Cu++ + SO4– – Cu++ moves towards cathode and takes 2 electron to become neutral cathode and deposited on Cu++ + 2e Cu U SO4– – moves towards anode and looses 2 electrons their. ST Copper is deposited on the cathode and an equivalent amount of copper is lost by the anode, but the concentration Rh of copper sulphate solution remains the same. In this process, A two C electrons per Cu reaction are active and valence plat Cu two. solution es is also of copper atom Cu lost U AgNO3 Ag+ + NO3– The silver dissolves from the anode gets deposited on the cathode. During this process, the concentration of the electrolyte remains unchanged. In this process one electron per reaction is active and valence of Ag atom is also one. D YG CuSO4 in water dissociates as follows ID Cu-voltameter Cu deposited  – + Anode Rh Ag Cathod e AgNO3 solution products of electrolysis. When current flows through the electrolyte, hydrogen gas is collected in the tube placed over the cathode (– ve electrode) and oxygen is collected in the tube placed over the anode (+ve electrode). Hydrogen and oxygen are O2 H2 liberated in the proportional in which they are found in water i.e. the volume ratio of hydrogen and oxygen is 2 : 1. + – – A + Rh 60 122 Heating and Chemical Effect of Current Faraday’s Law of Electrolysis. (1) First law : It states that the mass of substance deposited at the cathode during electrolysis is directly proportional to the quantity of electricity (total charge) passed through E3 the electrolyte. Let m be the mass of the substance liberated, when a charge q is passed through the electrolyte. Then, according to the Faraday’s first law of electrolysis m  q or m  zq , where the ID constant of proportionality z is called electrochemical equivalent (E.C.E.) of the substance. If a constant current i is passed through the electrolyte for time t, then the total charge passing through the electrolyte is given by q  i t U Therefore we have m  zit. If q = 1 coulomb, then we have m = z  1 or z = m D YG Hence, the electrochemical equivalent of substance may be defined as the mass of its substance deposited at the cathode, when one coulomb of charge passes through the electrolyte. S.I. unit of electrochemical equivalent of a substance is kilogram coulomb–1 (kg-C–1). (2) Second law : If same quantity of electricity is passed through different electrolytes, masses of the substance deposited at the respective cathodes are directly proportional to their chemical equivalents. Let m be the mass of the ions of a substance liberated, whose chemical equivalent is E. U Then, according to Faraday’s of electrolysis, m  E or m = constant  E or Note  Chemical equivalent E also known as equivalent weight in gm i.e. Atomic mass ( A) Valance (V ) ST E : m  constant E (3) Relation between chemical equivalent and electrochemical equivalent : Suppose that on passing same amount of electricity q through two different electrolytes, masses of the two substances liberated are m1 and m2. If E1 and E2 are their chemical equivalents, then from m E Faraday’s second law, we have 1  1 m2 E2 Further, if z1 and z2 are the respective electrochemical equivalents of the two substances, m1 z then from Faraday’s first law, we have m 1  z 1 q and m 2  z 2 q   1 m2 z2 Heating and Chemical Effect of Current 123 So from above equation z1 E  1 z2 E2   z 2  z1  zE E2 E1 (4) Faraday constant : As we discussed above E  z  E  Fz As z  m E and z  Q F E A  F VF 60 ‘F’ is proportionality constant called Faraday’s constant.  z E m  hence if Q = 1 Faraday then E  m i.e. If F Q (from I law) so electricity supplied to a voltameter is 1 Faraday then amount of substance liberated or electricity is required. Note :   1 Faraday = 96500 C  Also F = Ne given mass  valency atomic mass ID Remember Number of gm equivalent  E3 deposited is (in gm) equal to the chemical equivalent. e.g. to deposit 16 gm O2; 2 Faraday {where N = Avogrado number) U Electro Chemical Cell. D YG It is an arrangement in which the chemical energy is converted into electrical energy due to chemical action taking place in it. The total amount of energy that can be provided by this cell is limited and depends upon the amount of reactants. Electro chemical cells are of two types. (1) Primary cell : Is that cell in which electrical energy is produced due to chemical energy. In the primary cell, chemical reaction is irreversible. This cell can not be recharged but the chemicals have to be replaced after a long use examples of primary cells; Voltaic cell, Daniel cell, Leclanche cell and Dry cell etc. U (i) Voltaic cell + Zn – ST Cu Electrolyt e Dil. H2SO4 Cu Polarisation Local action Zn Positive electrode – Cu rod Negative electrode – Zn rod Electrolyte Emf – dil. H2SO4 – 1.08 V Positive electrode – Cu rod Negative electrode – Zn rod Main chemical reactions H2SO4 2H+ + SO4– – Zn Zn++ + 2e– (ii) Daniel cell C – Cu Crystal s Zn Rod A+ Cu Pot CuSO4 Solution (Depolariser) Porous pot Dil. H2SO4 (electrolyte ) Electrolyte Emf – – dil. H2SO4 1.1 V 124 Heating and Chemical Effect of Current Main chemical reactions Zn + H2SO4 ZnSO4 + 2H+ + 2e– 2H+ + CuSO4 H2SO4 + Cu++ Lechlanche cell C – A Glass pot + Graphite rod Porous pot MnO2 + charcoal dust (depolariser) NH4Cl Solution (Electrolyte) – Carbon rod Negative electrode – Zn rod 60 Zn rod Positive electrode Electrolyte solution Emf – – E3 (iii) NH4Cl 1.45 V Main chemical reactions Zn + 2NH4Cl 2NH3 + ZnCl2 + 2H+ + 2e– (iv) Dry cell ID 2H+ + 2MnO2 Mn2O3 + H2O + 2 unit of positive charge Positive electrode – Graphite rod Zn pot U C +A D YG Porous pot Negative electrode – Carbon rod with brass cap – Zn vessel Electrolyte – Paste of NH4Cl and saw dust MnO2 + charcoal dust (depolariser) NH4Cl Solution (Electrolyte) Emf – Main chemical reactions 1.5 V – Similar to Leclanche cell ST U (2) Secondary cell : A secondary cell is that cell in which the electrical energy is first stored up as a chemical energy and when the current is taken from the cell, the chemical energy is reconverted into electrical energy. In the secondary cell chemical reaction are reversible. The secondary cells are also called storage cell or accumulator. The commonly used secondary cells are In charged Lead accumulator + – + – Ni(OH) 2 Glass vessel PbO2 Pb dil. H2SO4 Positive electrode Alkali accumulator Perforated lead plates coated with PbO2 Fe(OH)2 Perforate d steel grid KOH 20% + Li(OH), 1% Perforated steel plate coated with Ni(OH)4 Heating and Chemical Effect of Current 125 Negative electrode Perforated lead plates coated with pure lead Perforated steel plate coated with Fe During charging Chemical reaction Chemical reaction At cathode : PbSO4 + 2H H2SO4 + 2e Pb + – Fe(OH)2 + 2OH+ – 2e– Ni(OH)4 At anode : PbSO4 + SO4– 2H2SO4 – At cathode : + 2H2O – 2e– PbO2 + Fe(OH)2 + 2K+ + 2e– Fe + 2KOH Emf of cell : When cell is fully charge then E = 1.36 volt E3 Specific gravity of H2SO4 increases and when specific gravity becomes 1.25 the cell is fully charged. At anode : 60 + Emf of cell : When cell is full charged them E = 2.2 volt Chemical reaction Chemical reaction At cathode : Pb + SO4– – – 2e– PbSO4 At anode : Fe + 2OH– – 2e– Fe(OH)2 At anode : Specific gravity of H2SO4 decreases and when specific gravity falls below 1.18 the cell requires recharging. Ni(OH)4 + 2K+ + 2e– Ni(OH)2 + 2KOH U PbO4 + 2H+ – 2e– + H2SO4 PbSO2 + 2H2O D YG Emf of cell : When emf of cell falls below 1.9 volt the cell requires recharging. Efficiency At cathode : ID During discharging 80% Emf of cell : When emf of cell falls below 1.1 V it requires charging. 60% ST U (3) Defects In a primary cell : In voltaic cell there are two main defects arises. Local action : It arises due to the presence of impurities of iron, carbon etc. on the surface of commercial Zn rod used as an electrode. The particles of these impurities and Zn in contact with sulphuric acid form minute voltaic cell in which small local electric currents are set up resulting in the wastage of Zn even when the cell is not sending the external current. Removal : By amalgamating Zn rod with mercury (i.e. the surface of Zn is coated with Hg). Polarisation : It arises when the positive H2 ions which are formed by the action of Zn on sulphuric acid, travel towards the Cu rod and after transferring, the positive charge converted into H2 gas atoms and get deposited in the form of neutral layer of a gas on the surface of Cu rod. This weakens the action of cell in two ways. Removal : Either by brushing the anode the remove the layer or by using a depolariser (i.e. some oxidising agent MnO2, CuSO4 etc which may oxidise H2 into water). Note :  The end point voltage of dry cell is 0.8 V. Concepts  Electrolysis takes place for dc and low frequency ac, as at high frequency, due to inertia (i.e. mass) ions cannot follow the frequency of ac.   Electrolytes are less conducting then the metallic conductors because ions are heavier than electrons. If  is the density of the material deposited and A is the area of deposition then the thickness (d) of the layer of 126 Heating and Chemical Effect of Current the material deposited in electroplating process is d  m A  Zi t A ; where m = deposited mass, Z = electro chemical equivalent, i = electric current. 60 Example s In an electroplating experiment, m gm of silver is deposited when 4 ampere of current flows for 2 minute. The amount (in gm) of silver deposited by 6 ampere of current for 40 second will be [MNR 1991; MP PET 2002] (a) 4 m (b) m/2 (c) m/4 (d) 2m m it m 4  2  60 Solution : (b) By using m  zit  1  1 1    m2  m / 2 m 2 i2 t2 m2 6  40 E3 Example: 35 When a copper voltameter is connected with a battery of emf 12 volts, 2 gms of copper is deposited in 30 minutes. If the same voltameter is connected across a 6 volt battery, then the mass of copper deposited in 45 minutes would be (a) 1 gm (b) 1.5 gm (c) 2 gm (d) 2.5 gm m1 V1 t1 zVt 2 12  30     m 2  1.5 gm By using m  zi t   m2 V2 t 2 m2 6  45 R U Example: 39 D YG U ID Example: 36 A current of 16 ampere flows through molten NaCl for 10 minute. The amount of metallic sodium that appears at the negative electrode would be (a) 0.23 gm (b) 1.15 gm (c) 2.3 gm (d) 11.5 gm 23 A Solution : (c) By using m = zit  it  m   16  10  60  2.3 gm 1  96500 VF Example: 37 For depositing of 1 gm of Cu in copper voltameter on passing 2 amperes of current, the time required will be (For copper Z = 0.00033 gm/C) (a) Approx. 20 minutes(b) Approx. 25 minutes (c) Approx. 30 minutes (d)Approx. 35 minutes Solution : (b) By using m = zit  1 = 0.00033  2  t  t = 1515.15 sec  25 min. Example: 38 Two electrolytic cells containing CuSO4 and AgNO3 respectively are connected in series and a current is passed through them until 1 mg of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately (Atomic weights of copper and Silver are respectively 63.57 and 107.88) (a) 1.7 mg (b) 3.4 mg (c) 5.1 mg (d) 6.8 mg m E 1 63.57 / 2 31.7    m 2  3.4 mg Solution : (b) By using 1  1  m2 E2 m 2 107.88 / 1 107.88 ST Solution : (b) Example: 40 Silver and copper voltameter are connected in parallel with a battery of e.m.f. 12 V. In 30 minutes, 1 gm of silver and 1.8 gm of copper are liberated. The energy supplied by the battery is (ZCu = 6.6 × 10–4 gm/C and ZAg = 11.2 × 10–4 gm/C) (a) 24.13 J Solution : (a) (b) 2.413 J (c) 0.2413 J By using m  z i t, for Ag voltameter 1 = 11.2  10 –4 For Cu voltameter 1.8 = 6.6  10 –4 Agamp. volta  i1  30  60  i1 = 0.5 i 1  i2  30  60  i2 = 1.5 amp Main current i = i1 + i2 = 1.5 + 0.5 = 2A. So energy supplied = Vi = 12  2 = 24 J Example: 41 (d) 2413 J i i2 Cu volta 12 V Amount of electricity required to pass through the H2O voltameter so as to liberate 11.2 litre of hydrogen will be Heating and Chemical Effect of Current 127 (a) 1 Faraday 60 ID Total surface area of a cathode is 0.05 m2 and 1 A current passes through it for 1 hour. Thickness of nickel deposited on the cathode is (Given that density of nickel = 9 gm/cc and it’s ECE = 3.04  10–4 gm/C) (a) 2.4 m (b) 2.4 cm (c) 2.4 m (d) None of these Mass deposited = density  volume of the metal m=Ax …… (i) Hence from Faraday first law m = Zit ……(ii) So from equation (i) and (ii) Zit =   Ax  Example: 44 Resistance of a voltameter is 2 , it is connected in series to a battery of 10 V through a resistance of 3 . In a certain time mass deposited on cathode is 1 gm. Now the voltameter and the 3 resistance are connected in parallel with the battery. Increase in the deposited mass on cathode in the same time will be (a) 0 (b) 1.5 gm (c) 2.5 gm (d) 2 gm Remember mass of the metal deposited on cathode depends on the current through the voltameter and not on the current supplied by the battery. Hence by using m = Zit, we can say m Parallel iParallel 5   m Parallel   1  2.5 gm. m Series iSeries 2 U Solution : (b) Zit 3.04  10 4  10 3  1  36 ......   2.4  10 6 m  2.4 m A 9000  0.05 D YG x U Solution : (c) (d) 3 Faraday We know that 1 gm of hydrogen is equal to 1 gm equivalent wt. of hydrogen. It means that 11.2 litre of hydrogen at NTP represents 1 gm equivalent of hydrogen, so for liberation it requires 1 Faraday electricity. Amount of electricity required to liberate 16 gm of oxygen is 1 (a) 1 Faraday (b) 2 Faraday (c) Faraday (d) 3 Faraday 2 Given mass 16 Number of gm equivalent  = 2. Hence 2 Faraday electricity is  gm equivalent weight 16 / 2 needed. Example: 43 (c) 2 Faraday E3 Solution : (b) 1 Faraday 2  11. 2   11. 2  Mass of hydrogen in 11.2 litres of hydrogen    M     2  1 gm  22. 4   22.4  Solution : (a) Example: 42 (b) Hence increase in mass = 2,5 – 1 = 1.5 gm ST 3 i1 2 2 Volta i2 Volta i1  10  2A 5 i2  3 10V 10  5A 2 10V Tricky Example: 3 In a copper voltameter, the mass deposited in 30 s is m gram. If the current time graph is as shown in the figure, the e.c.e. of copper, in gm/coulomb, will be (a) m m (b) 2 (c) 0.6 m i (mA) 100 mA 10 20 30 t (sec) 128 Heating and Chemical Effect of Current (d) 0.1 m 1 (10  30 )  100  10  3  2 Coulomb 2 m m From Faraday's first law m = zit  z .  it 2 ST U D YG U ID E3 60 Solution : (b) Area of the given curve on x-axis  i t  ST U D YG U ID E3 60 Electrostatics 103 + Heating and Chemical Effect of Current 127 A – – + – + Assignmen 60 tJoule's Heating A 220 V, 1000 W bulb is connected across a 110 V mains supply. The power consumed will be (a) 1000 W (b) 870  (b) 40 W 2 D YG (b) 4 : 3 (b) 60 W (b) 2 : 3 (b) 5 W U ST i 2 l r 2   P0   V  (d) P   0  P0  V  (c) 2 : 1 (d) 5 : 2 (c) 100 W (d) Equal in all bulbs [BHU 1999; KCET (Engg./Med.) 2001] (c) 3 : 4 (d) 4 : 3 (c) 20 W (d) None of these  l  (b) i 2  2   r  2 (c) i 2 l  / r (d) i l  / r If 2.2 KW power is transmitted through a 10 ohm line at 22000 V, the power lost in the form of heat will be [MP PET/P (a) 0.1 W 11.  V (c) P    V0 A current i passes through a wire of length l, radius of cross-section r and density . The rate of heat generation is [AMU (Med.) 1999] (a) 10. 2   P0   Two wires A and B of same material and mass have their lengths in the ratio 1 : 2. On connecting them to the same source, the rate of heat dissipation in B is found to be 5W. The rate of heat dissipation in A is (a) 10 W 9.  V (b) P    V0 If two electric bulbs have 40W and 60W rating at 220 V, then the ratio of their resistances will be (a) 3 : 2 8. (d) 10 W Three bulbs of 40W, 60W and 100W are arranged in series with 220V. Which bulb has minimum resistance[AFMC 2001 (a) 40 W 7. (c) 20W Two wires have resistance of 2 and 4 connected to same voltage, ratio of heat dissipated at resistance is [UPSEAT 20 (a) 1 : 2 6. (d) 780  An electric bulb is designed to draw power P0 at voltage V0. If the voltage is V it draws a power P. Then V  (a) P   0  P0  V  5. (c) 807  An electric bulb marked 40W and 200V, is used in a circuit of supply voltage 100V. Now its power is (a) 100 W 4. (d) 250 W An electric bulb is rated 60W, 220V. The resistance of its filament is (a) 708  3. (c) 500 W ID 2. (b) 750 W U 1. E3 Basic Level (b) 1 W (c) 10 W (d) 100 W The rated powers of two instruments working at 220 V is 200 W and 100W respectively. If their resistances are R1 and R2 respectively, then [NCERT 1980; CPMT 1991, 97] (a) R1 = 4R2 12. (b) R1 = 2R2 (c) R2 = 2R1 (d) R2 = 4R1 The heating element of an electric toaster has a resistance of 22  and is connected to an ordinary house lighting circuit of 110 V. If J = 4.2 J/cal, the heat generated in 1 minute is (a) 26.19 cal (b) 130.95 cal (c) 7857 cal (d) 2310 cal 128 Heating and Chemical Effect of Current Electric power is transmitted over long distances through conducting wires at high voltage because (a) High voltage travels faster (c) Power loss is less high voltage The ratio of diameters of two similar copper wires of equal length is 1 : 2. A constant current is passed through them when connected in series. The ratio of heat produced in the two will be (a) 1 : 2 15. (b) 1 : 3 (c) 4 : 1 (d) 1 : 5 Two identical batteries, each of e.m.f. 2 V and internal resistance 1.0 ohm are available to produce heat in an external resistance R = 0.5 ohm by passing a current through it. The maximum Joulean power that can be developed across R using these batteries is (a) 1.28 W 16. 60 14. (b) Power loss is large (d) Generator produces electrical energy at a very (b) 2.0 W (c) 8 W 9 (d) 3.2 W E3 13. A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. If both length and radius of the wire are halved then the heat developed in the same duration will become (b) Twice (c) One-fourth (d) Same ID (a) Half Advance Level Time taken by a 836 W heater to heat one litre of water from 10 o C to 40 o C is (a) 150 sec (a) Rs. 10 19. (b) 100 sec (c) 50 sec (d) 200 sec An electric lamp is marked 60 W, 230 V. The cost of a KW × hour of power is Rs. 1.25. The cost of using this lamp 8 hrs a day for 30 days is D YG 18. [AIEEE 2004] U 17. (b) Rs. 16 (c) Rs. 18 (d) Rs. 24 A steel wire has a resistance twice that of an aluminium wire. Both of them are connected with a constant voltage supply. More heat will be dissipated in (a) Steel wire when both are connected in series (b) Steel wire when both are connected in parallel (c) Aluminium wire when both are connected in series (d) Aluminium wire when both are connected in parallel Which of the following plots may represent the thermal energy produced in a resistor in a given time as a function of the electric current U 20. U d ST (a) a (b) b c b a (c) c i (d) d 21. An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from room temperature 20o C? The temperature of boiling water is 100o C (a) 6.4 minutes 22. (b) 6.3 minutes (c) 12.6 minutes (d) 12.8 minutes The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated 220 V and 100 W is connected (220  0.8) V source, then the actual power would be (a) 100  0.8 W (b) 100  (0.8)2 W (c) Between 100  0.8 W and 100 W (d) Between 100  (0.8)2 W and 100  0.8 W Heating and Chemical Effect of Current 129 According to Joule’s law, if the potential difference across a conductor having a material of specific resistance remains constant, then the heat produced in the conductor is directly proportional to (a)  (c) 2 : 1 (b) 250 s (d) 1 : 2 (c) 245 s (d) 247 s A house is fitted with 10 lamps of 60W each, 10 fans consuming 0.5 A each and an electric kettle of resistance 110 . If the energy is supplied at 220 V and costs 50 paise per KWh. The electric bill for 10 days, if all appliances are used for 6 hours daily will be approx. Rs. (b) 64 (c) 68 (d) 70 A constant current i is passed through a resistor. Taking the temperature coefficient of resistance into account, indicate which of the plots shown in figure best represents the rate of production of thermal energy in the resistor U dU dt (b) b (c) c (d) d D YG (a) a 28. 1  A heating coil of 2000 W is immersed in an electric kettle. The time taken in raising the temperature of 1 litre of water from 4oC to 100o C will be – (Only 80% part of the thermal energy produced is used in raising the temperature of water (a) 60 27. (d)  60 (b) 2 : 1 (a) 252 s 26. 1 A 100 W bulb and a 25 W bulb are designed for the same voltage. They have filaments of the same length and material. The ratio of the diameter of the 100 W bulb to that of the 25 W bulb is (a) 4 : 1 25. (c) E3 24. (b) 2 ID 23. d c b a t A dwelling house is installed with 15 lamps, each of resistance 103  and 4 ceiling fans each driven by 1/8th horse-power motor. If the lamps and fans are run on an average for 6 hours daily, then the number of B.O.T. units consumed by lamps in a month of 31 days will be (Given supply voltage – 220 V) (a) 135 (c) 165 (d) 180 A person decides to use his bath-tub water to generate electric power to run a 40 W bulb. The bath-tub is located at a height of 10 m from the ground and it holds 20 litres of water. He installs a water driven wheel generator on the ground. The rate at which he should drain the water from the bath tub to light the bulb and U 29. (b) 150 ST the time he keeps the bulb on will be respectively – (The efficiency of the generator is 90%) (g = 9.8 m/s2) (a) 0.345 kg/ s, 441 s 30. (b) 40 kg/ s, 100 s (c) 0.454 kg/ s, 441 s (d) None of these A current enters at a point in a solid metallic sphere and leaves from exactly opposite point. Heat produced in it will be (a) Uniform throughout (b) Maximum at the point of entrance and exist (c) Maximum in the perpendicular diameter plane (d) Minimum at the point of entry and exist Grouping of Electrical Appliances Basic Level 130 Heating and Chemical Effect of Current 31. When three identical bulbs of 60 W, 200 V rating are connected in series at a 200 V supply, the power drawn by them will be [CBSE PMT 2004; MP PET 2003] (b) 20 W (b) R / 2 (c) R (d) 2R E3 Two 220 V, 100 W bulbs are connected first in series and then in parallel. Each time the combination is connected to 220 V ac supply line. The power drawn by the combination in each case respectively will be (a) 100 W, 50 W 34. (d) 180 W In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be (a) R / 4 33. (c) 60 W 60 (a) 10 W 32. (b) 200 W, 150 W (c) 50 W, 200 W (d) 50 W, 100 W A wire when connected to 220V mains supply has power dissipation P1. Now the wire is cut into two equal (a) 1 (b) 4 (b) p/n D YG (a) P1  P2 watt (c) P1 P2 watt P1  P2 (d) P1  P2 watt P1 P2 (c) 7 min 30 sec (d) 2 min 30 sec (b) 5 5 , 2 2 (c) 2 5 , 5 2 (d) 2 2 , 5 5 ST Two resistances are connected in series across a battery and consume a power P. If these are connected in parallel the power consumed will be (b) 4P (c) 2P (d) P/4 A uniform wire connected across a supply produces heat H per second. If the wire is cut into n equal parts and all the parts are connected in parallel across the same supply, the heat produced per second will be (a) 41. P1 P2 W (b) 5 min U 5 2 , 2 5 (a) P 40. (b) [MP PMT 2001; MP PET 2002] Two bulbs of 500 W and 200 W are manufactured to operate on 220 V line. The ratio of heat produced in 500 W and 200 W, in two cases, when firstly they are joined in parallel and secondly in series will be (a) 39. (d) np An electric kettle has two heating coils. When one coil is used, water in the kettle boils in 5 minutes, while when second coil is used, same water boils in 10 minutes. If the two coils, connected in parallel are used simultaneously, the same water will boil in time [MP PET 2001] (a) 3 min 20 sec 38. (c) p Two electric bulbs rated P1 watt V volts and P2 watt V volts are connected in parallel and V volts are applied to it. The total power will be 37. (d) 3 n identical bulbs, each designed to draw a power p from a certain voltage supply, are joined in series across that supply. The total power which they will draw is (a) p/n2 36. (c) 2 U 35. ID pieces which are connected in parallel to the same supply. Power dissipation in this case is P2. Then P2 : P1 is H n2 (b) n 2 H (c) nH (d) H n In a house having 220 V line, following appliances are operating (i) 60 W bulb (ii) 1000 W heater and (iii) a 40 W radio set. The current passing through fuse for this line will be (a) 3 A 11 (b) 2 A 11 60 W 1000W 40W Fuse 220V Heating and Chemical Effect of Current 131 (c) 5 A (d) 6 A 42. 60 Advance Level A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source, as shown in the figure. Now W1, W2 and W3 are the output powers of the bulbs B1, B2 and B3, respectively. Then B2 B1 (a) W1 >W2 = W3 E3 B3 (b) W1 >W2 > W3 (c) W1 E2 (d) Data is not sufficient to predict it In a given thermocouple the temperature of the cold junction is 20 o C while the neutral temperature is 270 o C. (c) 500 o C (b) 520 o C (a) 540 o C 58. ID What will be the temperature of inversion (d) 420 o C Consider the following two statements A and B, and identify the correct choice of given answers. U [EAMCET (Med.) 2000] A. Thermo e.m.f. is minimum at neutral temperature of a thermocouple D YG B. When two junctions made of two different metallic wires are maintained at different temperatures, an electric current is generated in the circuit. (a) A is false and B is true (b) A is true and B is false 59. Two different metals are joined end to end. One end is kept at constant temperature and the other end is heated to a very high temperature. The graph depicting the thermo e.m.f. is E E (a) (b) t (a) 1360 C (d) t t [EAMCET (Med.) 1999] o (b) 650 C o (c) 340 C (d) 170oC The temperature of the cold junction of thermocouple is 0o C and the temperature of hot junction is T oC. The emf is E = 16 T – 0.04 T2 V. The temperature of inversion is (a) 200o C 62. (c) t U o 61. E E For a thermocouple if the cold junction is maintained at 0oC the inversion temperature is 680oC. Its Neutral temperature is ST 60. (c) Both A and B are false (d) Both A and B are true (b) 400o C (c) 100o C In a Cu-Fe thermo couple the battery current is driven from Cu-Fe through J2. Then (a) J2 should heat’s up (b) J2 should cool down (c) J1 should cool down (d) 300o C 134 Heating and Chemical Effect of Current (d) Both J1 and J2 either heat’s up or cool down depending upon the direction of current Advance Level (a) 20o C (d) 8o C Thomson coefficient of a conductor is 10 V / K. The two ends of it are kept at 50o C and 60o C respectively. Amount of heat absorbed by the conductor when a charge of 10 C flows through it is (a) 1000 J 65. (c) 12o C E3 64. (b) 16o C 60 The thermo e.m.f. of a thermocouple is 25 V / oC at room temperature. A galvanometer of 40 ohm resistance capable of detecting current as low as 10–5 A is connected with the thermocouple. The smallest temperature difference that can be detected by this system is (b) 100 J (c) 100 mJ (d) 1 mJ A thermo couple of resistance 2.6  is in series with a meter of resistance 7.4 . It can produce 10 V/oC difference between junctions. The meter reads 10 mV when a junction is at 0oC and the other junction is in molten metal. Temperature of molten metal is (a) 1350o C ID 63. (b) 1500o C (c) 1000o C (d) 1850o C Chemical Effect of Current U Basic Level The electrochemical equivalent of a metal is 3.3  10 7 kg / Coulomb. The mass of the metal liberated at the D YG 66. cathode when a 3 A current is passed for 2 seconds will be (a) 6.6  10 7 kg 67. U (b) 0.180 g [AIEEE 2003] (c) 0.141 g (d) 0.126 g ST (b) 6 min 42 sec (c) 4 min 40 sec (d) 5 min 50 sec On passing 96500 coulomb of charge through a solution CuSO 4 the amount of copper liberated is (a) 64 gm 70. (d) 1.1  10 7 kg If 96500 coulombs of electricity liberates one gram equivalent of any substance, the time taken for a current of 0.15 amperes to deposit 20mg of copper from a solution of copper sulphate is (Chemical equivalent of copper = 32) [Kerala (Engg.) 2002] (a) 5 min 20 sec 69. (c) 19.8  10 7 kg The negative Zn pole of a Daniell cell sending a constant current through the circuit, decreases in mass by 0.13 g in 30 minutes. If the electrochemical equivalent of Zn and Cu are 32.5 and 31.5 respectively, the increase in the mass of the positive Cu pole in this time is (a) 0.242 g 68. (b) 9.9  10 7 kg (b) 32 gm (c) 32 kg (d) 64 kg The electrochemical equivalent of a material in an electrolyte depends on (a) The nature of the material electrolyte (c) The amount of charge passed through electrolyte (b) The current through the (d) The amount of material present in electrolyte Heating and Chemical Effect of Current 135 Two electrolytic cells containing CuSO4 and AgNO3 respectively are connected in series and a current is passed through them until 1 mg of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately [Atomic weights of copper and silver are respectively 63.57 and 107.88] (a) 1.7 mg (c) 0.378 g (d) 2.27 g A steady current of 5 amps is maintained for 45 minutes. During this time it deposits 4.572 gms of zinc at the cathode of a voltameter. E.C.E. of zinc is (b) 3.387  10–4 C/gm (c) 3.384  10–3 gm/C (d) 3.394  10–3 C/gm 965 C charge deposits 1.08 gm of silver when passed through silver nitrate solution. What is the equivalent weight of silver (b) 10.8 (c) 1.08 (d) None of these D YG If in a voltaic cell 5 gm of zinc is consumed, then we get how many ampere hours. (Given that e.c.e. of Zn is 3.387 × 10–7 kg/C) (a) 2.05 77. E3 (b) 0.227 g (a) 108 76. (d) 10.8 g The electro-chemical equivalent of magnesium is 0.126 mg/C. A current of 5 A is passed in a suitable solution for 1 hour. The mass of magnesium deposited will be (a) 3.387  10–4 gm/C 75. (c) 5.4 g 60 (b) 0.09 g (a) 0.0378 g 74. (d) 6.8 mg If nearly 10+5 coulomb liberate 1 gm equivalent of aluminium, then the amount of aluminium (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 ampere will be (a) 0.6 g 73. (c) 5.1 mg ID 72. (b) 3.4 mg U 71. (b) 8.2 (c) 4.1 (d) 5 × 3.387 × 10–7 During the electrolysis, it is the (a) Electronic conduction every where (b) Ionic conduction every where (c) Ionic conduction inside and electronic conduction outside the voltmeter (d) Electronic conduction inside and ionic conduction outside the voltameter During electrolysis of acidulated water, volumes of H2 and O2 are in the ratio of U 78. (b) 1 : 2 ST (a) 1 : 1 79. (d) 8 : 1 Advance Level A silver voltameter of resistance 2 ohm and a 3 ohm resistor are connected in series across a cell. If a resistance of 2 ohm is connected in parallel with the voltameter, then the rate of deposition of silver (a) Decreases by 25% 80. (c) 2 : 1 (b) Increases by 25% (c) Increases by 37.5% (d) Decreases by 37.5% If 100 KWh of energy is consumed at 33 V in a copper voltameter, the mass of copper liberated is (Given e.c.e. of copper = 3.3× 10–7 kg/.C) (a) 1.65 kg (b) 1.8 kg (c) 3.3 kg (d) 3.6 kg 136 Heating and Chemical Effect of Current A current of 1.5 A flows through a copper voltameter. The thickness of copper deposited on the electrode surface of area 50 cm2 in 20 minutes will be (Density of copper = 9000 kg/m3 and e.c.e. of copper = 0.00033 g/C) (a) 2.6 × 10–5 m (c) – 0.03 A (d) – 0.01 A A silver and a copper voltmeters are connected across a 6 V battery of negligible resistance. In half an hour, 1 gm of copper and 2 gm of silver are deposited. The rate at which energy is supplied by the battery will approximately be (Given E.C.E. of copper = 3.294 × 10–4 g/C and E.C.E. of silver = 1.118 × 10–3 g/C) (b) 32 W (c) 96 W (d) 16 W Area of a electrode is 32 cm2. It is to be coated with Cu. Density of Cu is 9000 kg/m2, thickness of Cu deposited on each side of the rectangular cathode is 0.01 mm. Energy spent by a battery of emf 10 V is (ECE of Cu is 3.2  10–4 gm/C) (a) 18 J (b) 1800 J (c) 18 kJ (d) 180 kJ A charged capacitor of 5  10–2 F capacity is discharged through a resistor R of 20  and a Cu voltmeter of internal resistance 30  connected in series. If 4.62  10–6 kg Cu is deposited, the heat generated in the resistor R will be (E.C.E. of Cu = 3.3  10–7 kg/C) (a) 200 J (b) 784 J (c) 830 J ST U (d) 2000 J D YG U 85. 60 (b) + 0.02 A (a) 64 W 84. (d) 1.3 × 10–4 m An ammeter, suspected to give inaccurate reading, is connected in series with a silver voltameter. The ammeter indicates 0.54 A. A steady current passed for one hour deposits 2.0124 gm of silver. If the e.c.e. of silver is 1.118 × 10–3 gmC–1, then the error in ammeter reading is (a) + 0.04 A 83. (c) 1.3 × 10–5 m E3 82. (b) 2.6 × 10–4 m ID 81. 20  i 30  Charged capacitor 136 Electrostatics A + – – + – + 60 Answer Sheet Assignment (Basic & Advance Level) 2 3 4 5 6 7 8 9 10 11 12 13 14 d c d b c c a a a a c c c c 21 22 23 24 25 26 27 28 29 30 31 b c d b a b d a c b b 41 42 43 44 45 46 47 48 49 50 51 c d b b b c d a b d c 61 62 63 64 65 66 67 68 69 a b b d a c d b b 81 82 83 84 85 c a d c b 16 17 18 19 20 b a a c a, d a 33 34 35 36 37 38 39 40 a c b b a a a b b 52 53 54 55 56 57 58 59 60 d c b c c b a d c ID 32 71 72 73 74 75 76 77 78 79 80 a b c d a a c c c d d U 70 D YG U ST 15 E3 1