Hc Verma II for Class 12 Science Physics Chapter 33 - Thermal And Chemical Effects Of Electric Current

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Page No 217:

Answer:

As a constant potential difference is applied across a bulb, due to Joule's heating effect, the temperature of the bulb increases. As the temperature of the bulb filament increases, its resistance also increases, as resistance R is the function of temperature T. It is given by R = R₀(1+αT). With an increase in the value of resistance, the value of current decreases as $i = \frac{V}{R}$ . Now, the heat generated by the resistance is constantly radiated to the surroundings. Thus, the value of its temperature is maintained and hence its resistance. As a result, current through the bulb filament becomes constant.

Page No 217:

Answer:

For the given time t, let the currents passing through the resistance R₁and R₂ be i₁ and i₂ , respectively.
Applying Kirchoff's Voltage Law to circuit-1, we get:

$ε - i_{1} r - i_{1} R_{1} = 0 \Rightarrow i_{1} = \frac{ε}{r + R_{1}}$
Similarly, the current in the other circuit,
$i_{2} = \frac{ε}{r + R_{2}}$
The thermal energies through the resistances are given by

$i_{1}^{2} R_{1} t = i_{2}^{2} R_{2} t {(\frac{ε}{r + R_{1}})}^{2} R_{1} t = {(\frac{ε}{r + R_{2}})}^{2} R_{2} t \frac{R_{1}}{{(r + R_{1})}^{2}} = \frac{R_{2}}{{(r + R_{2})}^{2}} \frac{(r^{2} + {R_{1}}^{2} + 2 r R_{1})}{R_{1}} = \frac{(r^{2} + {R_{2}}^{2} + 2 r R_{2})}{R_{2}} \frac{r^{2}}{R_{1}} + R_{1} = \frac{r^{2}}{R_{2}} + R_{2} r^{2} (\frac{1}{R_{1}} - \frac{1}{R_{2}}) = R_{2} - R_{1} r^{2} \times \frac{R_{2} - R_{1}}{R_{1} R_{2}} = R_{2} - R_{1} r^{2} = R_{1} R_{2} \Rightarrow r = \sqrt{R_{1} R_{2}}$

Page No 217:

Answer:

No, the rise in the temperature of a resistor on passing current through it is not an adiabatic process. In an adiabatic process, there is no heat exchange between the system and the surroundings. Here, some part of Joule's heat developed inside the resistor increases the temperature of the resistor and the remaining part is dissipated in the surroundings. Thus, the given process cannot be adiabatic.

Page No 217:

Answer:

The battery is doing positive work on a resistor carrying current i. Thus, ∆W is positive. The work done on the resistor is used to increase its thermal energy; thus ∆Q is positive. As the temperature of the resistor rises, ∆U is positive.

Page No 217:

Answer:

The temperature of the hot junction at which the thermo-emf in a thermocouple becomes maximum is called neutral temperature for that thermocouple. For a thermocouple in which the constants a and b have the same sign, the neutral temperature will be less than the temperature of the cold junction of the thermocouple (as $θ_{n} = - \frac{a}{b}$ ).
There will be no neutral or inversion temperature, as the temperature of the hot junction cannot be less than the temperature of the cold junction.

Page No 217:

Answer:

If the inversion temperature and neutral temperature are measured in degree Celsius, then it is correct to say that "inversion temperature is always double the neutral temperature." When temperature is measured in other units, such as Kelvin, then inversion temperature is not the double of neutral temperature.

Page No 217:

Answer:

No, the neutral temperature is not always the arithmetic mean of the inversion temperature and the temperature of the cold junction. That is valid only when the unit of temperature is degree Celsius.

Page No 217:

Answer:

No, the electrodes in an electrolytic cell do not have fixed polarity like that of a battery. If we take an electrolytic cell consisting of the Ag electrodes and the AgNO₃ as electrolyte. When the battery is connected to it, the end to which the positive terminal of the battery is connected is the anode and the end to which the negative terminal is connected is the cathode. ${NO}_{3}^{-}$ ions are deposited at the anode and Ag⁺ ions are deposited at the cathode. When the connection of the electrolytic cell is reversed, the polarities of the electrodes are also reversed.

Page No 217:

Answer:

Yes, the resistance of the electrolyte will decrease with an increase in temperature. This is because when the temperature of an electrolytic solution increases, its viscosity decreases and mobility of the ions in the solution increases.

Page No 217:

Answer:

Plot (a) is the correct option.

When current passes through a resistor, the heat produced,
H = I²Rt,
where I = current
R = resistance of the resistor
t = time for which current is flowing
This relation shows that the heat produced for a given time in a resistor varies with the square of current flowing through it. Hence, the plot between H vs I should be a parabola symmetric along the H axis, which is represented by curve a.

Page No 217:

Answer:

Plot (d) is correct.

When current passes through a resistor, the temperature of the resistor increases due to the heat produced in it.
H = i²Rt,
where i = current flowing through the resistor
R = resistance of the resistor
t = time for which the current is flowing

With the increase in the temperature of the resistor, its resistance is also increased. The rate of production of thermal energy in the resistor of the circuit is given by the following relation:
$\frac{d U}{d t} = \frac{d}{d t} (i^{2} R t) = i^{2} R$ ,
where i = current flowing through the resistor
R = resistance of the resistor

Rate of production of thermal energy in the resistor is directly proportional to the resistance. So, due to increase in resistance, $\frac{d U}{d t}$ also increases linearly, which is best represented by plot d.

Page No 218:

Answer:

(b) A is correct but B is wrong.

The value of neutral temperature is constant for a thermocouple. It depends on the nature of materials and is independent of the temperature of the cold junction. Inversion temperature depends on the temperature of the cold junction, as well as the nature of the material.

Page No 218:

Answer:

(c) It cannot be Joule heat.

Joule heat is directly proportional to the square of the current passing through the resistor. Peltier heat is directly proportional to the current passing through the junction.Thomson heat is also directly proportional to the current passing through the section of the wire. Thus, the heat developed can be either Thomson heat or Peltier heat. But it cannot be Joule heat.

Page No 218:

Answer:

(a) due to both A and B

In Seebeck Effect, a temperature difference between two dissimilar electrical conductors produces a potential difference across the junctions of the two different metals. The cause of this potential difference is the diffusion of free electrons from a high electron-density region to a low electron-density region. The free electron-density of the electrons is different in different metals and changes with change in temperature. Hence, both the statements are the causes of Seebeck Effect.

Page No 218:

Answer:

(b) due to A but not due to B

In Peltier Effect, one of the junctions gets heated up and the other cools down when electric current is maintained in a circuit of material consisting of two dissimilar conductors.
This is caused due to the difference in density of free electrons in different metals. When two different metals are joined to form a junction, the electrons tend to diffuse from the side with higher concentration to the side with lower concentration. If current is forced through the junction, positive or negative work is done on the charge carriers, depending on the direction of the current. Accordingly, thermal energy is either produced or absorbed. Thus, Peltier Effect is caused due to A but not due to B.

Page No 218:

Answer:

(c) due to B but not due to A

If a metallic conductor has non-uniform temperature distribution along its length, the density of the free electrons is different for different sections. The electrons diffuse from the sections with higher concentration to those with lower concentration of free electrons. Thus, there is an emf inside the metal that is known as Thomson emf. If a current is forced through the given conductor, positive and negative work is done on the charge carriers, depending on the direction of current. Thus, thermal energy is either produced or absorbed. Thus, the correct cause of the given effect is given by statement B alone.

Page No 218:

Answer:

(c) is a universal constant

Faraday^,s constant is a universal constant. Its value is 9.6845×10⁷ C/kg. It does not depend on the amount of the electrolyte, current in the electrolyte and on the amount of charge passed through the electrolyte.

Page No 218:

Answer:

(a) equal amounts of thermal energy must be produced in the resistors
(d) the temperature may rise equally in the resistors

In a resistor of resistance R, current i is passed for time t then the thermal energy produced in the resistor will be given by
H = i²Rt.
As the resistors are in series, the current through them will be same. Thus, the amount of thermal energy produced in the resistors is same. The rise in the temperature of the resistance will depend on the shape and size of the resistor. Thus, the rise in the temperature of the two resistances may be equal.

Page No 218:

Answer:

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

The copper strip AB and an iron strip AC are joined at A and the junction A is maintained at 0°C and the free ends B and C are maintained at 100°C. In this case, there will be generation of thermo-emf between the points that are at different temperatures. Here, the two ends of the copper, the copper end and the iron end at the junction, the two ends of the iron strip and the free ends B and C are at different temperatures. Hence, there will be potential difference among them.

Page No 218:

Answer:

(a) there will be no neutral temperature
(b) there will be no inversion temperature

The temperature of the hot junction at which the thermo-emf in the thermocouple becomes maximum is called neutral temperature for that thermocouple. The signs of the constants a and b are same. Therefore from the relation, $θ_{n} = - \frac{a}{b},$ the neutral temperature will be less than the temperature of the cold junction of thermocouple.
Hence, there will be no neutral or inversion temperature, as the temperature of the hot junction cannot be less than the temperature of the cold junction.

Page No 218:

Answer:

(c) The rate of liberation of material will remain the same.

In an electrolytic cell, both the electrodes are made of the same material. Thus, on reversing the terminals of the battery, the direction of the flow of charges will be reversed, but the rate of the electrolysis will remain the same.

Page No 218:

Answer:

(a) the nature of the material

The electrochemical equivalent of a substance is the ratio of the relative atomic mass of the substance to its valency. Thus, it is only dependent on the nature of the material.

Page No 218:

Answer:

Given:
Current through the wire, i = 2 A
Resistance of the wire, R = 25 Ω
Time taken, t = 1 min = 60 s
Heat developed across the wire,
H = i²Rt
= 2 × 2 × 25 × 60
= 100 × 60 J = 6000 J

Page No 218:

Answer:

Given:
Resistance of the coil, R = 100 Ω,
Emf of the battery, V = 6 V,
Change in temperature, ∆T = 15°C
Heat produced across the coil,
$H = \frac{V^{2}}{R} t$
This heat produced is used to increase the temperature of the coil.
$\Rightarrow H = c ∆ T \Rightarrow \frac{V^{2}}{R} t = c ∆ T \Rightarrow \frac{36}{100} t = 4 \times 15 \Rightarrow t = \frac{6000}{36} = 166.7 s = 2.8 \min$

Page No 218:

Answer:

Let R be the resistance of the coil.
The power P consumed by a coil of resistance R when connected across a supply V is given by
$P = \frac{V^{2}}{R} \Rightarrow R = \frac{V^{2}}{P} \Rightarrow R = \frac{{(250)}^{2}}{500} = 125 Ω$

Now, P = 1000 W
$\Rightarrow R = \frac{V^{2}}{P} = \frac{{(250)}^{2}}{1000} = 62.5 Ω$

Page No 219:

Answer:

(a) Let R be the resistance of the coil.
The power P consumed by a coil of resistance R when connected across a supply V is given by
$P = \frac{V^{2}}{R} \Rightarrow R = \frac{V^{2}}{P} \Rightarrow R = \frac{{(250)}^{2}}{500} = 125 Ω$

(b) We know:
$R = ρ \frac{l}{A} \Rightarrow l = \frac{R A}{ρ} \Rightarrow l = \frac{125 \times 0.5 \times 10^{- 6}}{10^{- 6}} = 62.5 m$

(c) Let n be the number of turns in the coil. Then,
$l = 2 π r n \Rightarrow n = \frac{l}{2 π r} \Rightarrow n = \frac{62.5}{2 \times 3.14 \times 4 \times 10^{- 3}} \approx 2500$

Page No 219:

Answer:

Let R be the resistance of the bulb. If P is the power consumed by the bulb when operated at voltage V, then
$R = \frac{V^{2}}{P} = \frac{{(250)}^{2}}{100} = 625 Ω$
Resistance of the copper wire,
$R_{c} = ρ \frac{l}{A} = \frac{1.7 \times 10^{- 8} \times 10}{5 \times 10^{- 6}} = 0.034 Ω$
The effective resistance,
$R_{e f f} = R + R_{c} = 625.034 Ω$
The current supplied by the power station,
$i = \frac{V}{R_{e f f}} = \{\frac{220}{625.034}\} A$
The power supplied to one side of the connecting wire,
$P' = i^{2} R_{c} = {(\frac{220}{625.034})}^{2} \times 0.034$
The total power supplied on both sides,
$2 P' = {(\frac{220}{625.034})}^{2} \times 0.034 \times 2 = 0.0084 W = 8.4 mW$

Page No 219:

Answer:

The resistance of a bulb that consumes power P and is operated at voltage V is given by
$R = \frac{V^{2}}{P} = \frac{220 \times 220}{60} = 806.67 Ω$

(a) Now the supply drops to V' = 180 V.
So, the power consumed,
$P' = \frac{V'^{2}}{R} = \frac{{(180)}^{2}}{806.67} = 40 W$

(b) Now the supply increases to V" = 240 V. Therefore,
$P " = \frac{{V^{''}}^{2}}{R} = 71 W$

Page No 219:

Answer:

Output voltage, V = 220 V ± 1% = 220 V ± 2.2 V
The resistance of a bulb that is operated at voltage V and consumes power P is given by
$R = \frac{V^{2}}{P} = \frac{(220)^{2}}{100} \Rightarrow R = \frac{48400}{100} = 484 Ω$

(a) 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,
$i' = \frac{V'}{R} = \frac{217.8}{484} = 0.45 A$
Power consumed by the bulb, P' = i' × V'
                                                    = 0.45 × 217.8 = 98.0 W

(b) For maximum power to be consumed, output voltage should be maximum. The maximum output voltage,
V" = (220 + 2.2) V
     = 222.2 V
The current through the bulb,
$i " = \frac{V "}{R} = \frac{222.2}{484} = 0.459 A$
Power consumed by the bulb,
P" = i" × V"
    = 0.459 × 222.2 = 102 W

Page No 219:

Answer:

Given that the operating voltage is V and power consumed is P.
Therefore, the resistance of the bulb,
$R = \frac{V^{2}}{P} = \frac{(220 \times 220)}{100} = 484 Ω$
The power fluctuation, p = 150 W. So, the voltage fluctuation that the bulb can withstand,
$v = \sqrt{p R} = \sqrt{150 \times 484} = 269.4 V = 270 V$
The bulb will withstand up to 270 V.

Page No 219:

Answer:

Given the operating voltage V and power consumed P, the resistance of the immersion heater,
$R = \frac{V^{2}}{P} = \frac{{(220)}^{2}}{1000} = 48.4 Ω$
Mass of water, m = $\frac{1}{100}$ × 1000 = 10 Kg
Specific heat of water, s = 4200 Jkg^{$-$ 1} K^{$- 1$}
Rise in temperature, θ = 25°C
Heat required to raise the temperature of the given mass of water,
Q = msθ = 10 × 4200 × 25 = 1050000 J
Let t be the time taken to increase the temperature of water. The heat liberated is only 60%. So,
$(\frac{V^{2}}{R})$ × t × 60% = 1050000 J
$\Rightarrow \frac{(220)^{2}}{48.4} \times t \times \frac{60}{100} = 1050000$
⇒ t = 29.17 minutes

Page No 219:

Answer:

Time taken to boil 4 cups of water, t = 2 minutes
Volume of water boiled = 4 × 200 cc = 800 cc
Initial temperature, θ₁ = 25°C
Final temperature, θ₂ = 100°C
Change in temperature, θ = θ₂ − θ₁ = 75°C
Mass of water to be boiled, m = 800 × 1 = 800 gm = 0.8 Kg
Heat required for boiling water,
Q = msθ = 0.8 × 4200 × 75 = 252000 J

We know:
1000 watt - hour = 1000 × 3600 watt sec.
∴ Cost of boiling 4 cups of water $= \frac{1}{1000 \times 3600} \times 252000$
= Rs. 0.7

(b) Initial temperature, θ₁ = 5°C
Final temperature, θ₂ = 100°C
Change in temperature, θ = θ₂ − θ₁ = 95°C
Q = msθ = 0.8 × 4200 × 95 = 319200
∴ Cost of boiling 4 cups of water $= \frac{1}{1000 \times 3600} \times 319200$
= Rs 0.09

Page No 219:

Answer:

Case-I : When the supply voltage is 220 V.
Power consumed by the bulb = 100 W
Excess power = 100 − 40 = 60 W
Power converted to light = 60% of 60 W = 36 W

Case-II : When the supply voltage is 200 V.
Power consumed = $\frac{200}{220} \times 100$ = 82.64 W
Excess power = 82.64 − 40 = 42.64 W
Power converted to light = 60% of 42.64 W = 25.584 W

Percentage drop in light intensity,
$p = \frac{36 - 25.584}{36} \times 100 \Rightarrow p = 28.93 \approx 29 %$

Page No 219:

Answer:

The effective resistance of the circuit,
$R_{e f f} = (\frac{6 \times 2}{6 + 2}) + 1 = \frac{5}{2} A$
Current i through the circuit,
$i = \frac{V}{R_{e f f}} = \frac{6}{5 / 2} = \frac{12}{5} A$

Let i' be the current through the 6 Ω resistor. Then,
i' × 6 = (i − i') × 2
$\Rightarrow 6 i' = (\frac{12}{5}) \times 2 - 2 i' \Rightarrow 8 i' = \frac{24}{5} \Rightarrow i' = \frac{24}{(5 \times 8)} = \frac{3}{5} A \Rightarrow i - i' = \frac{12}{5} - \frac{3}{5} = \frac{9}{5} A$

(a) Heat generated in the 2 Ω resistor,
H = (i - i')²Rt
$\Rightarrow H = (\frac{9}{5}) \times (\frac{9}{5}) \times 2 \times 15 \times 60 = 5832 J$
The heat capacity of the calorimeter together with water is 2000 J K−1. Thus, 2000 J of heat raise the temp by 1 K.
∴ 5832 J of heat raises the temperature by $\frac{5832}{2000}$ = 2.916 K
(b) When the 6 Ω resistor gets burnt, the effective resistance of the circuit,
R_eff = 1 + 2 = 3 Ω
Current through the circuit,
i = $\frac{6}{3}$ = 2 A
Heat generated in the 2 Ω resistor = (2)²× 2 × 15 × 60 = 7200 J
2000 J raise the temperature by 1 K.
∴ 7200 J raise the temperature by $\frac{7200}{2000}$ = 3.6 K .

Page No 219:

Answer:

Given:
Difference in temperature, θ = 0.001°C,
a = − 46 × 10⁻⁶ V °C⁻¹
b = − 0.48 × 10⁻⁵ V °C⁻².
Emf, E = aθ + $\frac{1}{2}$ bθ²
⇒ E = (− 46 × 10⁻⁶) × (0.001) $- \frac{1}{2}$ × (0.48 × 10⁻⁶ )× (0.001)²
       = − 46 × 10⁻⁹ − 0.24 × 10⁻¹²
       = − 46.0024 × 10⁻⁹
       = − 4.6 × 10⁻⁸ V

Page No 219:

Answer:

Difference in temperature, θ = 40°C
Emf, E_cs = a_csθ + $\frac{1}{2}$ b_csθ²    ...(1)
a_cs = [2.76 − (−43.7) μV
      = 46.46 μV/°C
b_cs = [0.012 − (−0.47) μV/°C
      = 0.482 μV/°C²
Putting this value in eq. (1), we get:
E_cs = 46.46 × 10⁻⁶ × 40 + $\frac{1}{2}$ × 0.482 × 10⁻⁶ × (40)²
      = 1.04 × 10⁻⁵ V

Page No 219:

Answer:

Neutral temperature,
$θ_{n} = - \frac{a}{b}$
$a_{C u F e} = a_{C u P b} - a_{F e P b} = 2.76 - 16.6 = 13.84 μV ° C^{- 1}$
$b_{C u F e} = b_{C u P b} - b_{F e P b} = 0.012 + 0.030 = 0.042 μV ° C^{- 2}$

Thus, the neutral temperature,
$θ_{n} = \frac{- a_{C u F e}}{b_{C u F e}} = \frac{13.84}{0.042} = 329.52 ° C = 330 ° C$
The inversion temperature is double the neutral temperature, i.e. 659 $°$ C.

Page No 219:

Answer:

(a) Amount of charge required by 1 equivalent mass of the substance = 96500 C
For a monovalent material,
equivalent mass = molecular mass
⇒ Amount of charge required by 6.023 × 10²³ atoms = 96500 C
∴ Amount of charge required by 1 atom = $\frac{96500}{6.023 \times 10^{23}} = 1.6 \times 10^{- 19} C$

(b) For a divalent material,
equivalent mass = $\frac{1}{2}$ molecular mass
⇒ Amount of charge required by $\frac{1}{2}$ × 6.023 × 10²³ = 96500 C
∴ Amount of charge required by 1 atom = 1.6 × 2 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ C

Page No 219:

Answer:

Equivalent mass of silver, E_Ag = 107.9 g (∵ Ag is monoatomic)
The ECE of silver,
$Z_{A g} = \frac{E_{A g}}{f} = \frac{107.9}{96500} = 0.001118$
Using the formula, m = Zit, we get:
m = 0.00118 × 0.500 × 3600
= 2.01 g
So, 2.01 g of silver is liberated.

Page No 219:

Answer:

Given:
Mass of silver deposited, m = 3 g
Time taken, t = 3 min. = 180 s
E.C.E. of silver, Z = 1.12 × 10⁻⁶kg C⁻¹

Using the formula, m = Zit, we get:
$3 \times 10^{- 3} = 1.12 \times 10^{- 6} \times i \times 180$
$\Rightarrow i = \frac{3 \times 10^{- 3}}{1.12 \times 10^{- 6} \times 180} \Rightarrow i = \frac{1}{6.72} \times 10^{2} \Rightarrow i = 14.89 \approx 15 A$

Page No 219:

Answer:

Let the required time be t.
Mass of 1 litre hydrogen,
$m = \frac{2}{22.4} g$
Using the formula, m = Zit, we get:
$\frac{2}{22.4} = \frac{1 \times 5 \times t}{96500} \Rightarrow t = \frac{2 \times 96500}{22.4 \times 5} \Rightarrow t = 1723.21 s = 28.7 minutes \approx 29 minutes$

Page No 219:

Answer:

Given:
Mass of salt deposited, m = 1 g
Current, i = 2 A
Time, t = 1.5 hours = 5400 s

For the trivalent metal salt:
Equivalent mass = $\frac{1}{3}$ Atomic weight
The E.C.E of the salt,
$Z = \frac{Equivalent mass}{96500} = \frac{Atomic weight}{3 \times 96500}$

(a) Using the formula, m = Zit, we get:
$1 \times 10^{- 3} = \frac{Atomic weight}{3 \times 96500} \times 2 \times 5400 \Rightarrow Atomic weight = \frac{3 \times 96500 \times 10^{- 3}}{2 \times 5400} = 26.8 \times 10^{- 3} kg / mole \Rightarrow Atomic weight = 26.8 g / mole$

(b) Using the relation between equivalent mass and mass deposited on plates, we get:
$\frac{E_{1}}{E_{2}} = \frac{m_{1}}{m_{2}} \Rightarrow \frac{26.8}{3 \times 107.9} = \frac{1}{m_{2}} \Rightarrow m_{2} = 12.1 g$

Page No 219:

Answer:

Given:
Current, i = 15 A
Surface area of the plate = 200 cm²,
Thickness of silver deposited= 0.1 mm = 0.01 cm
Volume of Ag deposited on one side = 200 × 0.01 cm³ = 2 cm³
∴ Volume of Ag deposited on both side = 4 cm³
Mass of silver deposited,
m = Volume × Specific gravity × 1000 = 4 × 10^-3 × 10.5 ×1000 = 42 kg
Using the formula, m = Zit, we get:
42 = Z_Ag × 15 × t
$\Rightarrow t = \frac{42 \times 96500}{107.9 \times 15} s \Rightarrow t = 2504.17 s = 42 minutes$

Page No 219:

Answer:

Given:
Mass of silver deposited, m = 2.68 g
Time, t = 10 minutes = 600 s
Using the formula, m = Zit, we get:
$2.68 \times 10^{- 3} = \frac{107.9 \times 10^{- 3}}{96500} \times i \times 600 \Rightarrow i = \frac{2.68 \times 96500}{107.9 \times 600} \Rightarrow i = 3.99 = 4 A$

Heat developed in the 20 Ω resistor,
$H = i^{2} R t \Rightarrow H = {(4)}^{2} \times 20 \times 600 \Rightarrow H = 192000 J = 192 kJ$

Page No 219:

Answer:

Let i be the current through the circuit.
Emf of battery, E = 12 V
Voltage drop across the voltameter, V = 10 V
Internal resistance of the battery, r = 2 Ω
Applying Kirchoff's Law in the circuit, we get:
$E = V + i r \Rightarrow i = \frac{E - V}{r} = \frac{12 - 10}{2} = 1 A$
Using the formula m = Zit, we get:
$m = \frac{107.9}{96500} \times 1 \times 0.5 \times 3600 = 2.01 g$

Page No 219:

Answer:

Surface area of the plate, A = 10 cm² = 10 × 10⁻⁴ m²
Thickness of copper deposited, t = 10 μm = 10⁻⁵m
Density of copper = 9000 kg/m³
Volume of copper deposited, V = A(2t)
V = 10 × 10⁻⁴ × 2 × 10 × 10⁻⁶
   = 2 × 10² × 10⁻¹⁰
   = 2 × 10⁻⁸ m³
Mass of copper deposited, m = Volume × Density = 2 × 10⁻⁸ × 9000
⇒ m = 18 × 10⁻⁵ kg

Using the formula, m = ZQ, we get:
18 × 10⁻⁵ = 3 × 10⁻⁷ × Q
⇒ Q = 6 × 10² C
Energy spent by the cell = Work done by the cell
⇒W = VQ
       = 12 × 6 × 10²
       = 72 × 10² = 7.2 kJ
Let ∆θ be the rise in temperature of water. When this energy is used to heat 100 g of water, we have:
7.2 × 10³ = 100 × 10⁻³ × 4200 × ∆θ
⇒ ∆θ = 17 K

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