 # HC Verma Solutions Class 12 Chapter 11 Thermal And Chemical Effects Of Current

HC Verma Solutions Class 12 Chapter 11 Thermal and Chemical Effects of Current will help students be more adept and skilled in solving questions that are discussed in the chapter. The solutions have been prepared by our experts and they have provided accurate answers to all the exercises which students can refer and be ready to solve questions at the time of their examinations. This chapter introduces students to a lot of topics related to electric current.

## Main Concepts Related To Thermal And Chemical Effects Of Current

Some of the main topics are;

• Joule’s Law of Heating
• Seebeck Effect
• Peltier Effect
• Thomson Effect
• Electrolysis
• Voltameter and Coulombmeter
• Primary and Secondary Cells

Students will also have to get familiar with mathematical expressions such as I= Q/t where, Q in coulomb and t is in seconds. Complex electric current questions can be solved easily by referring solutions provided here. Class 12 students can further use HC Verma Solutions for various competitive exams like IIT JEE and NEET.

## Class 12 Important Questions In Chapter 11

1. We pass a current through a resistor and as a result its temperature increases. Can you define it as an adiabatic process?

2. Will the electrodes in an electrolytic cell have fixed polarity like a battery? Explain with suitable reasons.

3. When heat is developed in a system it is directly proportional to the current flowing through it. State what kind of heat it is.

(a) Thomson heat (b) Peltier heat (c) Joule heat. (d) It cannot be any of these.

4. What does Faraday constant depend on?

(a) The amount of the electrolyte (b) The current in the electrolyte (c) The amount of charge passed through the electrolyte (d) depends on is a universal constant

5. An iron and copper strip are joined at A. The junction A is maintained at 0°C and the free ends X and Y are maintained at 100°C. Where is the potential difference present?

(a)  between the free ends B and C (b) between the two ends of the copper strip (c) between the two ends of the iron strip (d) between the copper end and the iron end at the junction

## HC Verma Solutions Vol 2 Electric Current Chapter 11

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

Solution:

Current through the wire= i = 2 A

Resistance of the wire = R = 25 Ω

t = 1 min = 60 s

We know Joule’s heating effect, heat developed across the wire,

H = i2 Rt

= 2x2x25x60

= 6000 J

Question 2: A coil of resistance 100 Ω is connected across a battery of emf 6.0V. 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-1, how long will it take to raise the temperature of the coil by 15°C.

Solution:

Resistance of the coil = R = 100 Ω,

Emf of the battery = V = 6 V,

and ∆T = 15°C

we know, Joule’s heating effect heat produced across the coil,

H = V2t/R

Which used to increase the temperature of the coil.

H = c∆T

V2t/R = c∆T

36t/100 = 4×15

t = 166.7 s = 2.8 min

Question 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:

We know, P = V2/R

or R = V2/P

= (25) 2/500

= 125Ω

Now, P = 1000 W

=> R = V2/P = (250) 2/1000 = 62.5Ω

Question 4: A heater coil is to be constructed with a nichrome wire (ρ = 1.0 × 10-8 Ω 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 mm2, 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:

Resistivity = ρ = 1.0 × 10–8 Ω m (given)

We know, P = V2/R

or R = V2/P = (250) 2/500 = 125 Ω

(b) We know resistance R,

R = ρl/A

or l = RA/ρ

= [125×0.5×10-6]/[10-6]

= 62.5 m

(c) Let n be the number of turns in the given coil.

l = 2πr n

where, r = 4.0 mm (Given)

n = l/2πr = 62.5/[2×3.14x4x10-3]

= 2500 (approx)

Question 5: A bulb with rating 250V, 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 mm2. How much power will be consumed by the connecting wires? Resistivity of copper = 1.7 × 10–8 Ω m.

Solution:

We know, R = V2/P = (250)2/100 = 625Ω

Resistance of the copper wire = Rc = ρl/A

= [1.7×10-8x105]/[5×10-6]

= 0.034 Ω

The effective resistance = Reff = R + Rc = 625.034 Ω

The current supplied by the power station by ohm’s law:

i = V/Reff = 220/625.034 A

Find the power supplied to one side of the connecting wire :

P’ = i2 Rc

[Using Joule’s heating effect]

= (220/625.034)2 x 0.034

Total power supplied on both sides:

2P’ = 2 x (220/625.034)2 x 0.034 = 8.4 mW

Question 6: An electric bulb, when connected across a power supply of 220 W, 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:

We know, R = V2/P

= (220)2/60 = 806.67 Ω

The power consumed = P = V’2/R

when the supply drops to V’ = 180 V

P = (180)2/806.67 = 40 W

When supply changes to, V” = 240 V.

The power consumed = P = V’’2/R

P = (180)2/806.67 = 71 W

Question 7: A servo voltage stabilizer 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:

We know, P = V’2/R

The resistance of a bulb that is operated at voltage V and consumes power P

= (220)2/100 = 484 Ω

For minimum power to be consumed, output voltage should be minimum.

The minimum output voltage = V’ = (220 – 2.2)V = 217.8 V

The current through the bulb = i = V’/R = 217.8/484 = 0.45 A

The power consumed by the bulb = P’ = i’ V’ = 0.45 x 217.8 = 98 W

For maximum power to be consumed, output voltage should be maximum.

V” = (220 + 2.2)V = 222.2 V

The current through the bulb = i” = V” R = 222.2 x 484 = 0.459 A

Power consumed by the bulb = P” = i” x V” = 0.459 x 222.2 = 102 W

Question 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:

The resistance of the bulb = R = V2/P

= (220)2/100 = 484 Ω

Power fluctuations, P = 150 W (Given)

Or V2 = RP = 150 x 484

Or V = 269.4 V

Bulb will withstand fluctuations up to 270 V.

Question 9: An immersion heater rated 1000 W, 220 V is used to heat 0.01 m3 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:

Resistance of the immersion heater = R = V2/P

= (220)2/1000 = 48.4 Ω

Heat required to raise the temperature of the given mass of water

Q = msθ

where, Mass of water = m = 1100 × 1000 = 10 Kg

Specific heat of water = s = 4200 Jkg-1 K-1 and

Rise in temperature, θ = 25°C

=> Q = 10 x 4200 x 25

= 1050000 J

Also, we are given that, the heat evolve is only 60%.

Let t be the time taken to raise the temperature of water.

So, v2/R x t x 60% = 1050000 J

=> (220)2/48.4 x t x 60/100 = 1050000 J

or t = 29.17 min

Question 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. 1.00 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:

(a)

We know, heat required for boiling water, Q = ms θ …(1)

Where

m = mass of the water = 800 × 1 = 800 gm = 0.8 Kg

s = specific heat of the water = 4200 Jkg-1 K-1

θ = change in temperature = θ2 − θ1 = 75°C [Given: θ1 = 25°C and θ2 = 100°C]

(1)=> Q = 0.8 x 4200 x 75 = 25200 J

Cost of boiling 4 cups of water = 1/[1000 x 3600] x 252000 = Rs. 0.7

[Because 1000 watt – hour = 1000 × 3600 watt sec.]

(b) using equation (1)

Here

θ = change in temperature = θ2 − θ1 = 95°C [Given: θ1 = 5°C and θ2 = 100°C]

Putting the value in (1), we get

Q = 0.8 x 4200 x 95 = 319200

Cost of boiling 4 cups of water = 1/[1000 x 3600] x 319200 = Rs. 0.09

[Because 1000 watt – hour = 1000 × 3600 watt sec.]

Question 11: The coil of an electric bulb takes 40 watts to start glowing. 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:

Case 1: When supply voltage is 220 V.

Consumed Power = 100 W

Excess power = 100 − 40 = 60 W

Power converted to light = 60% of 60 W = 36 W

case 2: When supply voltage is 200 V.

Power consumed, P = 200/220 x 100 = 82.64 W

Excess power = 82.64 − 40 = 42.64 W

Now,

Power converted to light = 60% of 42.64 W = 25.584 W

% drop in the light intensity = [36 – 25,584]/36 x 100 = 29% (approx)

Question 12: The 2.0 Ω resistor shown in figure is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000 J K–1.

(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 Ω resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes? Solution: From figure, Reff = (6×2)/(6+2) + 1 = (5/2) A

Where Reff is effective resistance of the circuit.

let i be the current passing through the circuit,

i = V/Reff = 6/(5/2( = 12/5 A

From ohm’s law,

i’ x 6 = (i – i’) x 2

Where i’ be the current through 6Ω resistor.

=> 6i’ = (12/5) x 2 – 2i’

=> i’ = (3/5) A

and i – i’ = 12/5 = 3/5 = 9/5 A

(a) Heat generated in the 2 Ω resistor

H = (i – i’)2 Rt

= (9/5) 2 x 2 x 15 x 60

= 5832 J

From question, 2000 J of heat raise the temp by 1 K. so, 5832 J of heat raises the temperature of water by (5832/2000) = 2.916 K

(b) When the 6 Ω resistor burn out.

Reff = 1+2 = 3 Ω

And i = (6/3) = 2 A

Where i = current through the circuit

Now, heat generated in 2 Ω resistor = (2)^2 x 2 x 15 x 60 = 7200 J

2000 J of heat raise the temp by 1 K. So, 7200J will raise the temperature by (7200/2000) = 3.6 k

Question 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 bismuch-silver, a = –46 × 10–6 V°C–1 and b = –0.48 × 10–6 V°C–2.

Solution:

emf, E = aθ + 12bθ2

Given:

Difference in temperature of junctions = θ = 0.001°C,

a = − 46 × 10−6 V °C−1

b = − 0.48 × 10−5 V °C−2.

On substituting the given values and solving, we have

E = -4.6 x 10-8 V

Question 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 given below.

 Metal with lead (Pb) a μV/oC b μV/oC Aluminium -0.47 0.003 Bismuth -43.7 -0.47 Copper 2.76 0.012 Gold 2.9 0.0093 Iron 16.6 -0.03 Nickel 19.1 -0.03 Platinum -1.79 -0.035 Silver 2.5 0.012 Steel 10.8 -0.016

Solution:

emf, Ecs = acs θ + 12bcs θ2 …(1)

From table, acs = (2.76 – (-43.7)) μV/oC = 46.46 μV/oC and

bcs = (0.012 – (-0.47)) μV/oC = 0.482 μV/oC

and Difference in temperature, θ = 40°C

Substituting all the values in (1)

(1) => Ecs = 1.04 x 10-5 V

Question 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 (use previous question table).

Solution:

We know, Neutral temperature = θn = -a/b

Using table data,

aCu Fe = aCu Pb – aFe Pb

= 2.76 – 16.6

= -13.84 μV/oC

And

bCu Fe = bCu Pb – bFe Pb

= 0.012 + 0.030

= 0.042 μV/oC2

Now,

Neutral temperature = θn = – aCu Fe/ bCu Fe

= 13.84/0.042

= 330 oC

The inversion temperature is double the neutral temperature, i.e. 659°C.

Question 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) Amount of charge required by 1 equivalent mass of the substance = 96500 C

For monovalent material, equivalent mass = molecular mass

Amount of charge required by 6.023 × 1023 atoms = 96500 C

Amount of charge required by one atom = 96500/[6.023 × 1023] = 1.6 x 10-19 C

(b) For a divalent material:

We know, equivalent mass = (1/2) molecular mass

=> Amount of charge required = 1/2 x 6.023 × 1023 = 96500 C

Amount of charge required by 1 atom = 1.6 x 2 x 10-19 = 3.2 x 10-19 C

Question 17: Find the amount of silver liberated at cathode if 0.500 A of current is passed through AgNO3 electrolyte for 1 hour. Atomic weight of silver is 107.9 g mol–1.

Solution:

Equivalent mass of silver = EAg = 107.9 g

Current passed through AgNO3 electrolyte for 1 hour = 0.005 A

Now, the ECE of silver,

ZAg = EAg/f = 107.9/96500 = 0.001118

From faraday’s law of electrolysis: m = Zit

Where, m = mass of the substance deposited on the electrode

z = electrochemical equivalence

i = current and t = time taken

m = 0.00118 x 0.500 x 3600 = 2.01 g

Question 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 × 10–6 kg C–1.

Solution:

From faraday’s law of electrolysis: m = Zit

Given: Z = 1.12 × 10−6 kg C–1

mass = 3 g = 3 x 10–3 Kg and t = 3 min or 180 s

=> 3 x 10–3 = 1.12 × 10–6 x i x 180

or i = 15 A (approx)

Question 19: Find the time required to liberate 1.0 liter of hydrogen at STP in an electrolytic cell by a current of 5.0 A.

Solution:

We know, Mass of 1 liter hydrogen = m = 2/22.4 g

Current = i = 5A (Given)

From faraday’s law of electrolysis: m = Zit

2/22.4 = (5t)/96500

=> t = 29 min (Approx)

Question 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 2A is maintained for 1.50 hours. It is found that 1.00g 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 mol–1.

Solution:

We know, equivalent mass = (1/3) x Atomic weight

E.C.E of the salt = Z = [Equivalent mass]/96500

= [Atomic weight]/[3×96500]

Given: mass of salt deposited = m = 1 g; current = i = 2A and t (time) = 1.5 h = 5400 sec

(a) We know, m = Zit [ faraday’s law of electrolysis]

1 x 10-3 = [Atomic weight]/[3×96500] x 2 x 5400

or Atomic Weight = 26.8 x 10-3 kg/mole

(b) Using relation between equivalent mass and mass deposited on plates:

E1/E2 = m1/m2

=> 26.8/[3×107.9] = 1/m2

=> m2 = 12.1 g

Question 21: A brass plate having surface area 200 cm2 on one side is electroplated with 0.10 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–1.

Solution:

Surface area of the plate = 200 cm2

Current = i = 15 A

Thickness of silver deposited = 0.1 mm = 0.01 cm

Volume of Ag deposited on one side = 200 × 0.01 cm3 = 2 cm3

Specific gravity of silver = 10.5 and

Atomic weight = 107.9 g mol–1

Now,

Mass of silver deposited = m = volume of Ag deposited on both sides x specific gravity x 100

= 4 x 10-3 x 10.5 x 1000

= 42 kg

Therefore, m = 42 kg

Again, using formula, m = Zit

=> 42 = ZAg x 15 x t

=> t = 2504.17 s = 42 min

Question 22: Figure 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 20Ω resistor during this period. Atomic weight of silver is 107.9 g mol–1. Solution:

m = 2.68 g and t = 10 minutes = 600 s

Using the formula, m = Zit

2.68 x 10-3 = [107×10-3]/96500 x i x 600

=> i = 3.99 or 4 A

Heat developed in the 20 Ω resistor from Joule’s heating effect, H = i2 Rt

=> H = (4)2 x 20 x 600 = 192000 J = 192 kJ

Question 23: The potential difference across the terminals of a battery of emf 12 V and internal resistance 2 Ω drops to 10 V when it is connected to a silver voltmeter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9 g mol-1.

Solution:

Emf of battery, E = 12 V

Voltmeter = V = 10 V and battery’s internal resistance = r = 2 Ω

By Kirchhoff’s Law in the circuit, E = V + ir

=> i = (E-V)/r = (12-10)/2 = 1 A

Again, we know, m = Zit

=> m = (107.9)/96500 x 1 x 0.5 x 3600

= 2.01 g

Question 24: A plate of area 10 cm2 is to be electroplated with copper (density 9000 kg m-3) to a thickness of 10 micrometers 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 × 10-7 kg C-1 and specific heat capacity of water = 4200 J kg-1 K-1.

Solution:

ECE of copper = 3 × 10-7 kg C-1 and specific heat capacity of water = 4200 J kg-1 K-1.

Surface area of the plate = A = 10 cm2 = 10 × 10−4 m2

Thickness of copper deposited = t = 10 μm = 10−5 m

Density of copper = 9000 kg/m3

Volume of copper deposited = V = A x (2t)

=> V = 10 x 10-4 x 2 x 10 x 10-6 = 2 x 10-8 m3

Again, mass of copper deposited = m = Volume x Density

= 2 x 10-8 x 9000

=> m = 18 x 10-5 Kg

Using formula, m = ZQ

Where, m = mass of the substance

Q = charge

Z = Electrochemical equivalent

Using values, we have

18 x 10-5 = 3 x 10-7 x Q

=> Q = 6 x 102 C

Now,

Energy spent by the cell = Work done by the cell

=> W = VQ = 12 x 6 x 102 = 72 x 102 = 7.2 kg

=> W = 7.2 kg

Let ∆θ = rise in temperature of water. When this energy used to heat 100 g of water

=> 7.2 x 103 = 100 x 10-3 x 4200 x ∆θ

=> ∆θ = 17 K