# Thermal and Chemical Effects of Current

## Concept Of Physics

### H C Verma

1   An electric current of 2.0 A passes through a wire of resistance 25 ft. How much heat will be developed in 1 minute ?

##### Solution :

$i = 2A, \qquad r = 25 \Omega,$ $\\$ $t = 1min = 60 sec,$ $\\$ Heat developed $= i^2RT = 2 \times 2 \times 25 \times 60 = 6000J$ $\\$

2   A coil of resistance 100 ft is connected across a battery of emf 6.0 V. Assume that the heat developed in the coil is used to raise its temperature. If the heat capacity of the coil is 4.0 J/K, how long will it take to raise the temperature of the coil by 15°C ?

##### Solution :

$R = 100 \Omega, \qquad E = 6V$ $\\$ Heat Capacity of the coil = $4 J/Kg \qquad \Delta T = 15^{\circ}C$ $\\$ Heat Liberted $\Rightarrow \frac{E^2}{Rt} = 4 J/K \times 15$ $\\$ $\Rightarrow \frac{6 \times 6}{100} \times t = 60 \Rightarrow t = 166.67 sec = 2.8 min$

3   The specification on a heater coil is 250 V, 500 W. Calculate the resistance of the coil. What will be the resistance of a coil of 1000 W to operate at the same voltage ?

##### Solution :

(a) The power consumed by a coil of resistance R when connected across a supply v is $P =\frac{v^2}{R}$ $\\$ $\quad$ The resistance of heater coil is, therefore $R = \frac{v^2}{P} = \frac{(250)^2}{1000} = 62.5 \Omega$ $\\$ (b) If $P = 1000W \quad$ then $\quad R = \frac{v^2}{P}$ = $\frac{(250)^2}{1000}$ = $62.5 \Omega$

4   A heater coil is to be constructed with a nichrome wire $(\rho = 1.0 \times 10^{-6} \Omega-m)$ which can operate at 500 W when connected to a 250 V supply, (a) What would be the resistance of the coil ? (b) If the cross-sectional area of the wire is $0.5 mm ^2$ , what length of the wire will be needed ? (c) If the radius of each turn is $4.0 mm$, how many turns will be there in the coil ?

##### Solution :

$f = 1 \times 10^{-6} \Omega m, \qquad P = 500W, \qquad E =250v$ $\\$ (a) $R = \frac{V^2}{P} = \frac{250 \times 250}{500} = 125 \Omega$ $\\$ (b) $A = 0.5 mm^2 = 0.5 \times 10^{-6} m^2 = 5 \times 10^{-7} m^2$ $\\$ $R = \frac{fI}{A} = I \frac{RA}{f} = \frac{125 \times 5 \times 10^{-7}}{1 \times 10^{-6}} = 625 \times 10^{-1} = 62.5m$ $\\$ (c) $62.5 = 2\pi r \times n, \qquad 62.5 = 3 \times 3.14 \times 4 \times 10^{-3} \times n$ $\\$ $\Rightarrow n = \frac{62.5}{2 \times 3.14 \times 4 \times 10^3} \Rightarrow n = \frac{62.5 \times 10^{-3}}{8 \times 3.14} = 2500 turns$

5   A bulb with rating 250 V, 100 W is connected to a power supply of 220 V situated 10 m away using a copper wire of area of cross-section $5 mm ^2$ . How much power will be consumed by tlje connecting wires ? Resistivity of copper$=1.7 \times 10 ^ {-8} \Omega-m$

##### Solution :

$V = 250V, \qquad P=100W$ $\\$ $R = \frac{v^2}{P} = \frac{(250)^2}{100} = 625 \Omega$ $\\$ Resistance of wire $R =\frac{fI}{A} = 1.7 \times 10^{-8} \times \frac{10}{5 \times 10^{-6}} = 0.034 \Omega$ $\\$ $\therefore$ The effect in resistance $= 625.034 \Omega$ $\\$ $\therefore$ The Current in the conductor $= \frac{V}{R} = \Bigg( \frac{220}{625.034 } \Bigg)A$ $\\$ $\therefore$ The power supplied by one side of connecting wire $= \Bigg( \frac{220}{625.034} \Bigg)^2 \times 0.034$ $\\$ $\therefore$ The total power supplied $= \Bigg( \frac{220}{625.034} \Bigg)^2 \times 0.034 \times 2 = 0.0084 w = 8.4 mw$

6   An electric bulb, when connected across a power supply of 220 V, consumes a power of 60 W. If the supply drops to 180 V, what will be the power consumed ? If the supply is suddenly increased to 240 V, what will be the power consumed ?

##### Solution :

$E = 220V, \qquad P = 60 W$ $\\$ $R = \frac{V^2}{P}= \frac{220 \times 220}{60} = \frac{220 \times 11}{3} \Omega$ $\\$ (a) $E = 180V, \qquad P = \frac{V^2}{R} = \frac{180 \times 180 \times 3}{220 \times 11} = 40.16 =40w$ $\\$ (b) $E = 240V \qquad P = \frac{V^2}{R} = \frac{240 \times 240 \times 3 }{220 \times 11} = 71.4 = 71W$

7   A servo voltage stabiliser restricts the voltage output to 220 V ± 1%. If an electric bulb rated at 220 V, 100 W is connected to it, what will be the minimum and maximum power consumed by it ?

##### Solution :

Output voltage $= 220 \pm 1\% , \qquad 1\% of 220 V = 2.2V$ $\\$ The resistance of bulb $R = \frac{V^2}{P} = \frac{(220)^2}{100} = 484 \Omega$ $\\$ (a) For minimum power consumed $V_1 = 220 - 1\% = 220 - 2.2 = 217.8$ $\\$ $\therefore i = \frac{V_1}{R} = \frac{217.8}{484} = 0.45A$ $\\$ Power consumed $= i \times V_1 = 0.45 \times 217.8 = 98.01 W$ $\\$ (b) for maximum power consumed $V_2 = 220 + 1\% = 220 + 2.2 = 222.2$ $\\$ $\therefore i = \frac{V_2}{R} = \frac{222.2}{484} = 0.459$ $\\$ Power consumed $= i \times V_2 = 0.45 \times 222.2 = 102 W$ $\\$

8   An electric bulb marked 220 V, 100 W will get fused if it is made to consume 150 W or more. What voltage fluctuation will the bulb withstand ?

##### Solution :

$V = 220V, \qquad P =100W$ $\\$ $R = \frac{V^2}{P} = \frac{220 \times 220 }{100} = 484 \Omega$ $\\$ $P = 150 W \qquad V =\sqrt{PR} = \sqrt{150 \times 22 \times 22} = 22 \sqrt{150} = 269.4 = 270V$

9   An immersion heater rated 1000 W, 220 V is used to heat $0.01 m ^3$of water. Assuming that the power is supplied at 220 V and 60% of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from 15°C to 40°C ?

##### Solution :

$P = 1000, \qquad V =220V \qquad R =\frac{V^2}{P} = \frac{48400}{1000} = 48.4 \Omega$ $\\$ Mass of water $= \frac{1}{100} \times 1000 = 10Kg$ $\\$ Heat required to raise the temp. of given amount of water $= ms \Delta t = 10 \times 4200 \times 25 = 1050000$ $\\$ Now heat liberated is only $60\%$. So $\frac{V^2}{R} \times T \times 60\% = 1050000$ $\\$ $\Rightarrow \frac{(220)^2}{48.4} \times \frac{60}{100} \times T = 1050000 \Rightarrow T = \frac{10500}{6} \times \frac{1}{60} nub = 29.16 min$

10   An electric kettle used to prepare tea, takes 2 minutes to boil 4 cups of water (1 cup contains 200 cc of water) if the room temperature is 25°C. (a) If the cost of power consumption is Rs. 100 per unit (1 unit = 1000 watt-hour), calculate the cost of boiling 4 cups of water, (b) What will be the corresponding cost if the room temperature drops to 5°C ?

##### Solution :

Volume of water boiled $= 4 \times 200 cc= 800cc$ $\\$ $T_1 = 25^{\circ}C, \qquad T_2 = 100^{\circ}C \qquad \Rightarrow T_2 -T_1 = 75^{\circ}C$ $\\$ Mass of Water boiled $= 800 \times 1 = 800 gm = 0.8Kg$ $\\$ Q(heat req.) $= MS \Delta \theta = 0.8 \times 4200 \times 75 = 252000J$ $\\$ $1000watt - hour = 1000 \times 3600 watt-sec = 1000 \times 3600 J$ $\\$ No. of units = \frac{252000}{1000 \times 3600} = 0.07 = 7 paise $\\$ (b) $Q = mS \Delta T = 0.8 \times 4200 \times 95 J$ $\\$ No. of units $= \frac{0.8 \times 4200 \times 95 }{1000 \times 3600} = 0.0886 = 0.09$ $\\$ MOney consumed $= 0.09 Rs = 9$ paise

11   The coil of an electric bulb takes 40 watts to start giowing. If more than 40 W is supplied, 60% of the extra power is converted into light and the remaining into heat. The bulb consumes 100 W at 220 V. Find the percentage drop in the light intensity at a point if the supply voltage changes from 220 V to 200 V.

##### Solution :

$P = 100 W \qquad V =220V$ $\\$ Case I : Excess power $= 100 -40 =60w$ $\\$ Power converted to light $= \frac{60 \times 60}{100} = 36 w$ $\\$ Case II : Power $= \frac{(220)^2}{484} = 82.64 w$ $\\$ Excess power $= 82.64 - 40 42.64 w$ $\\$ Power converted to light $= 42.64 \times \frac{60}{100} = 25.584 w$ $\\$ $\Delta P = 36 - 25. 584 = 10.416$ $\\$ Required $\% = \frac{10.416}{36} \times 100 = 28.93 = 29 \%$

12   The $2.0 \Omega$ resistor shown in figure (33-E1) is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000 J/K. (a) If the circuit is active for 15 minutes, what would be the rise in the temperature of the water ? (b) Suppose the 6'0 ft resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes ?

##### Solution :

$R_{eff} = \frac{12 }{8 } + 1 = \frac{5}{2} \qquad i = \frac{6}{(5/2)} = \frac{12}{5} Amp$ $\\$ $i' 6= (i-i')2 \Rightarrow i'6 = \frac{12}{50} \times 2-2i$ $\\$ $8i' = \frac{24}{5} \Rightarrow i'= \frac{24}{5 \times 8} = \frac{3}{5} Amp$ $\\$ $i-i' = \frac{12}{5} - \frac{3}{5} = \frac{9}{5} Amp$ $\\$ (a) Heat $= i^2 RT = \frac{9}{5} \times \frac{9}{5} \times 2 \times 15 \times 60 = 5832$ $\\$ 2000 J of heat raises the temp. by 1K $\\$ 5832 J of heat raises the temp. by 2.916K $\\$ (b) When $6 \Omega$ resistor get burnt $R_{eff} = 1 + 2 = 3\Omega$ $\\$ $i = \frac{6}{3 } = 2 Amp$ $\\$ Heat $= 2 \times 2 \times 2 \times 15 \times 60 = 7200 J$ $\\$ 2000 J raises the temp. by 1K $\\$ 7200 J raises the temp. by 3.6K

13   The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0°C and 0.001°C. Find the thermo-emf (Seebeck emf) developed. For bismuth- silver, $a = - 4.6 \times 10^{-6}V/deg$ and $b = -0.48 \times 10^{-6} V/deg^ 2$.

##### Solution :

$\theta = 0.001 ^{\circ}C, \qquad a =-46 \times 10^{-6}v/deg, \qquad b =-0.48 \times 10^{-6}v/deg,$ $\\$ $Emf = a_{BIAg} \theta + (1/2) b_{BIAg} \theta^2 = -4.6 \times 10^ -6 \times 0.001 - (1/2) \times 0.48 \times 10^{-6} (0.001)^2$ $\\$ $= -46 \times 10^{-9} - 0.24 \times 10^{-12} = -46.00024 \times 10^{-9} = -4.6 \times 10^{-8}V$

14   Find the thermo-emf developed in a copper-silver thermocouple when the junctions are kept at 0°C and 40°C. Use the data in table (33.1).

##### Solution :

$E = a_{AB}\theta + b_{AB} \theta^2 \qquad a_{CuAg} = a_{CuPb} - b_{AgPb} = 2.76 -2.5 = 0.26 \mu v/^{\circ}C$ $\\$ $b_{CuAg} = b_{CuPb} - b_{AgPb} = 0. 012 = 0.012 \mu vc = 0$ $\\$ $E = a_{AB} \theta =(0.26 \times 40 )\mu V = 1.04 \times 10^{-5}V$

15   Find the neutral temperature and inversion temperature of copper-iron thermocouple if the reference junction is kept at 0°C. Use the data in table (33.1)

##### Solution :

$\theta = 0^{\circ} C$ $\\$ $a_{Cu,Fe} = a_{Cu,Pb} - a_{Fe,Pb} = 2.76 - 16.6 = - 13.8 \mu v/^{\circ}C$ $\\$ $B_{Cu,Fe} = b_{Cu,Pb} - b_{Fe,Pb} = 0.012 + 0.030 = 0.042 \mu v/^{\circ}C^2$ $\\$ Neutral Tmp. on $-\frac{a}{b} = \frac{13.8}{0.042}^{\circ}C = 328.57 ^{\circ}C$

16   Find the charge required to flow through an electrolyte to liberate one atom of (a) a monovalent material and (b) a divalent material.

##### Solution :

(a) 1eq. mass of the substance requires 96500 coulombs $\\$ Since the element is monoatomic, thus eq. mass = mol. Mass $\\$ $6.023 \times 10^{23}$ atoms reuire $96500 C$ $\\$ 1 atoms require $\frac{96500 }{6.023 \times 10^{23}} C = 1.6 \times 10^{-19}C$ $\\$ (b) Since the element is diatomic eq.mass = (1/2) mol.mass $\\$ $\therefore (1/2) \times 6.023 \times 10^{23} \quad atoms \quad 2eq \quad 96500C$ $\\$ $\Rightarrow 1 \quad atom \require \quad = \frac{96500 \times 2}{6.023 \times 10^{23}} = 3.2 \times 10^{-19}C$

17   Find the amount of silver liberated at cathode if 0.500 A of current is passed through $AgNO _a$ electrolyte for 1 hour. Atomic weight of silver is 107.9 g/mole.

##### Solution :

At Wt. At = 107.9 g/mole $\\$ $I = 0.500 A$ $\\$ $E_{Ag} = 107.9 A \qquad$ [ As Ag is monoatomic] $\\$ $Z_{Ag} = \frac{E}{f} = \frac{107.9 }{96500} = 0.001118$ $\\$ $M = Zit = 0.001118 \times 0.5 \times 3600 = 2.01$ $\\$

18   An electroplating unit plates 3.0 g of silver on a brass plate in 3.0 minutes. Find the current used by the unit. The electrochemical equivalent of silver is $1.12 \times 10 ^ {-6} kg/C$.

##### Solution :

$t = 3 min = 180sec, \qquad w = 2g$ $\\$ E.C.E. $= 1.12 \times 10^{-6} Kg/c$ $\\$ $\Rightarrow 3 \times 10^{-3} = 1.12 \times 10^{-6} \times i \times 180$ $\\$ $\Rightarrow i = \frac{3 \times 10^{-3}}{1.12 \times 10^{-6} \times 180} = \frac{1}{6.72} \times 10^2 = 15 Amp$

19   Find the time required to liberate 1 litre of hydrogen at STP in an electrolytic cell by a current of 5.0 A.

##### Solution :

$\frac{H_2}{22.4L} \rightarrow 2g, \qquad 1L \rightarrow \frac{2}{22.4}$ $\\$ $m = Zit , \qquad \frac{2}{22.4} = \frac{1}{96500} \times 5 \times T \quad \Rightarrow T= \frac{2}{22.4} \times \frac{96500}{5} = 1732.21 sec = 28.7min =29 min$ $\\$

20   Two voltameters, one having a solution of silver salt and the other of a trivalent-metal salt, are connected in series and a current of 2 A is maintained for 1'50 hours. It is found that 1.00 g of the trivalent-metal is deposited. (a) What is the atomic weight of the trivalent metal ? (b) How much silver is deposited during this period ? Atomic weight of silver is 107.9 g/mole.

##### Solution :

$w_1 = Zit, \qquad \Rightarrow 1 = \frac{mm}{3 \times 96500} \times 2 \times 1.5 \times 3600 \Rightarrow mm = \frac{ 3 \times 96500}{2 \times 1.5 \times 3600} = 26.8 g/mole$ $\\$ $\frac{E_1}{E_2} = \frac{w_1}{w_2} \Rightarrow \frac{107.9}{\Bigg( \frac{mm}{3} \Bigg)} =\frac{w_1}{1} \Rightarrow w_1 = \frac{107.9 \times 3}{26.8} = 12.1gm$

21   A plate of area $10 cm^2$ is to be electroplated with copper $(density 9000 kg/m ^3 )$ to a thicknes of 10 micrometres on both sides, using a cell of 12 V. Calculate the energy spent by the cell in the process of deposition. If this energy is used to heat 100 g of water, calculate the rise in the temperature of the water. ECE of copper $- 3 \times 10^{-7} kg/C$ and specific heat capacity of water $- 4200 J/kg-K.$

##### Solution :

$A = 10cm^2 \times 10^{-4}cm^2$ $\\$ $t = 10m = 10 \times 10^{-6 }$ $\\$ Volume $= A(2t) = 10 \times 10^-4 \times 2 \times 10 \times 10^{-6} = 2 \times 10^2 \times 10^{-10} = 2 \times 10^{-8} m^3$ $\\$ Mass $=2 \times 10^{-8} \times 9000 = 10 \times 10^{-5} Kg$ $\\$ $W = Z \times C \Rightarrow 18 \times 10^{-5} = 3 \times 10^{-7} \times C$ $\\$ $\Rightarrow q = \frac{17 \times 10^{-5}}{3 \times 10^{-7}} = 6 \times 10^2$ $\\$ $V = \frac{W}{q} \Rightarrow W = Vq = 12 \times 6 \times 10^{2} = 76 \times 10^2 = 7.6 KJ$

22   The potential difference across the terminals of a battery of emf 12 V and internal resistance 2$\Omega$ drops to 10 V when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9 g/mole

##### Solution :

For potential drop, t = 30 min = 180 sec $\\$ $V_1 = V_f +iR \Rightarrow 12 = 10 + 2i \Rightarrow i = 1Amp$ $\\$ $m = Zit = \frac{107.9}{96500} \times 1 \times 30 \times 60 = 2.01g = 2g$

23   Figure (33-E2) shows an electrolyte of AgCl through which a current is passed. It is observed that 2.68 g of silver is deposited in 10 minutes on the cathode. Find the heat developed in the 20 $\Omega$ resistor during this period. Atomic weight of silver is 107.9 g/mole.

##### Solution :

$w =Zit$ $\\$ $2.68 = \frac{107.9}{96500} \times i \times 10 \times 60$ $\\$ $\Rightarrow I = \frac{2.68 \times 965}{107.9 \times 6 } = 3.99 = 4Amp$ $\\$ Heat developed in the $20 \Omega$ resister $= (4)^2 \times 20 \times 10 \times 60 = 192000J = 192 KJ$

24   A brass plate having surface area 200 $cm ^2$on one side is electroplated with 0 1 0 mm thick silver layers on both sides using a 15 A current. Find the time taken to do the job. The specific gravity of silver is 10.5 and its atomic weight is 107.9 g/mol.

##### Solution :

$I = 15A, \qquad Surface \quad area = 200 cm^2, \qquad Thickness =0.1 mm$ $\\$ Volume of Ag deposited $= 200 \times 0.01 = 2cm^3$ for one side. $\\$ For both sides, Mass of Ag $= 4 \times 10.5 = 42g$ $\\$ $Z_{Ag} = \frac{E}{F} = \frac{107.9}{96500} \qquad m =ZIT$ $\\$ $\Rightarrow 42 = \frac{107.9}{96500} \times 15 \times T \quad \Rightarrow T = \frac{42 \times 96500}{107.9 \times 15} = 2504.17 sec = 411.73 min = 42 min$