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#### Page No 217:

#### Question 1:

If a constant potential difference is applied across a bulb, the current slightly decreases as time passes and then becomes constant. Explain.

#### 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* = *R*_{0}(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:

#### Question 2:

Two unequal resistances, *R*_{1} and *R*_{2}_{,} are connected across two identical batteries of emf *ε* and internal resistance *r* (figure). Can the thermal energies developed in *R*_{1} and *R*_{2} be equal in a given time? If yes, what will be the condition?

Figure

#### Answer:

For the given time *t*, let the currents passing through the resistance *R*_{1}_{ }and *R*_{2} be *i*_{1} and *i*_{2} , respectively.

Applying Kirchoff's Voltage Law to circuit-1, we get:

$\epsilon -{i}_{1}r-{i}_{1}{R}_{1}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=\frac{\epsilon}{r+{R}_{1}}$

Similarly, the current in the other circuit,

${i}_{2}=\frac{\epsilon}{r+{R}_{2}}$

The thermal energies through the resistances are given by

${i}_{1}^{2}{R}_{1}t={i}_{2}^{2}{R}_{2}t\phantom{\rule{0ex}{0ex}}{\left(\frac{\epsilon}{r+{R}_{1}}\right)}^{2}{R}_{1}t={\left(\frac{\epsilon}{r+{R}_{2}}\right)}^{2}{R}_{2}t\phantom{\rule{0ex}{0ex}}\frac{{R}_{1}}{{\left(r+{R}_{1}\right)}^{2}}=\frac{{R}_{2}}{{\left(r+{R}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{\left({r}^{2}+{{R}_{1}}^{2}+2r{R}_{1}\right)}{{R}_{1}}=\frac{\left({r}^{2}+{{R}_{2}}^{2}+2r{R}_{2}\right)}{{R}_{2}}\phantom{\rule{0ex}{0ex}}\frac{{r}^{2}}{{R}_{1}}+{R}_{1}=\frac{{r}^{2}}{{R}_{2}}+{R}_{2}\phantom{\rule{0ex}{0ex}}{r}^{2}\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)={R}_{2}-{R}_{1}\phantom{\rule{0ex}{0ex}}{r}^{2}\times \frac{{R}_{2}-{R}_{1}}{{R}_{1}{R}_{2}}={R}_{2}-{R}_{1}\phantom{\rule{0ex}{0ex}}{r}^{2}={R}_{1}{R}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{{R}_{1}{R}_{2}}$

#### Page No 217:

#### Question 3:

When a current passes through a resistor, its temperature increases. Is it an adiabatic process?

#### 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:

#### Question 4:

Apply the first law of thermodynamics to a resistor carrying a current *i*. Identify which of the quantities ∆*Q*, ∆*U* and ∆*W* are zero, positive and negative.

#### 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:

#### Question 5:

Do all thermocouples have a neutral temperature?

#### 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 ${\theta}_{n}=-\frac{a}{b}$).

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:

#### Question 6:

Is inversion temperature always double the neutral temperature? Does the unit of temperature have an effect in deciding this question?

#### 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:

#### Question 7:

Is neutral temperature always the arithmetic mean of the inversion temperature and the temperature of the cold junction? Does the unit of temperature have an effect in deciding this question?

#### 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:

#### Question 8:

Do the electrodes in an electrolytic cell have fixed polarity like a battery?

#### 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 AgNO_{3} 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. ${\mathrm{NO}}_{3}^{-}$ 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:

#### Question 9:

As temperature increases, the viscosity of liquids decreases considerably. Will this decrease the resistance of an electrolyte as the temperature increases?

#### 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:

#### Question 1:

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?

Figure

#### Answer:

Plot (a) is the correct option.

When current passes through a resistor, the heat produced,

*H* = *I*^{2}*Rt*,

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:

#### Question 2:

A constant current *i* is passed through a resistor. Taking the temperature coefficient of resistance into account, indicate which of the plots shown in the figure best represents the rate of production of thermal energy in the resistor.

Figure

#### 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* =* **i*^{2}*Rt*,

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:

$\frac{dU}{dt}=\frac{d}{dt}\left({i}^{2}Rt\right)={i}^{2}R$,

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, $\frac{dU}{dt}$ also increases linearly, which is best represented by plot *d*.

#### Page No 218:

#### Question 3:

Consider the following statements regarding a thermocouple.

(A) The neutral temperature does not depend on the temperature of the cold junction.

(B) The inversion temperature does not depend on the temperature of the cold junction.

(a) A and B are correct.

(b) A is correct but B is wrong.

(c) *B* is correct but A is wrong.

(d) A and B are wrong.

#### 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:

#### Question 4:

The heat developed in a system is proportional to the current through it.

(a) It cannot be Thomson heat.

(b) It cannot be Peltier heat.

(c) It cannot be Joule heat.

(d) It can be any of the three heats mentioned above.

#### 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:

#### Question 5:

Consider the following statements.

(A) Free-electron density is different in different metals.

(B) Free-electron density in a metal depends on temperature.

Seebeck Effect is caused

(a) due to both A and B

(b) due to A but not due to B

(c) due to B but not due to A

(d) neither due to A nor due to B

#### 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:

#### Question 6:

Consider the statements A and B in the previous question. Peltier Effect is caused

(a) due to both A and B

(b) due to A but not due to B

(c) due to B but not due to A

(d) neither due to A nor due to B

#### 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:

#### Question 7:

Consider the statements A and B in question 5. Thomson Effect is caused

(a) due to both A and B

(b) due to A but not due to B

(c) due to B but not due to A

(d) neither due to A nor due to B

#### 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:

#### Question 8:

Faraday constant

(a) depends on the amount of the electrolyte

(b) depends on the current in the electrolyte

(c) is a universal constant

(d) depends on the amount of charge passed through the electrolyte

#### Answer:

(c) is a universal constant

Faraday^{,}s constant is a universal constant. Its value is 9.6845×10^{7} 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:

#### Question 1:

Two resistors of equal resistances are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature changes. In a given time interval,

(a) equal amounts of thermal energy must be produced in the resistors

(b) unequal amounts of thermal energy may be produced

(c) the temperature must rise equally in the resistors

(d) the temperature may rise equally in the resistors

#### 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* = *i*^{2}*Rt*.

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:

#### Question 2:

A copper strip *AB* and an iron strip *AC* are joined at *A*. The junction *A* is maintained at 0°C and the free ends *B* and *C* are maintained at 100°C. There is a potential difference between

(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*

#### 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:

#### Question 3:

The constants *a* and *b* for the pair silver-lead are 2.50 μV°C^{−1} and 0.012μV°C^{−2}, respectively. For a silver-lead thermocouple with colder junction at 0°C,

(a) there will be no neutral temperature

(b) there will be no inversion temperature

(c) there will not be any thermo-emf even if the junctions are kept at different temperatures

(d) there will be no current in the thermocouple even if the junctions are kept at different temperatures

#### 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, ${\theta}_{n}=-\frac{a}{b},$ 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:

#### Question 4:

An electrolysis experiment is stopped and the battery terminals are reversed.

(a) The electrolysis will stop.

(b) The rate of liberation of material at the electrodes will increase.

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

(d) Heat will be produced at a greater rate.

#### 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:

#### Question 5:

The electrochemical equivalent of a material depends on

(a) the nature of the material

(b) the current through the electrolyte containing the material

(c) the amount of charge passed through the electrolyte

(d) the amount of this material present in the electrolyte

#### 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:

#### 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?

#### 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* = *i*^{2}*Rt*

= 2 × 2 × 25 × 60

= 100 × 60 J = 6000 J

#### Page No 218:

#### Question 2:

A coil of resistance 100 Ω 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^{−1}, how long will it take to raise the temperature of the coil by 15°C?

#### 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.

$\Rightarrow H=c\u2206T\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{V}^{2}}{R}t=c\u2206T\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{36}{100}t=4\times 15\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{6000}{36}=166.7\mathrm{s}=2.8\mathrm{min}$

#### Page No 218:

#### 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?

#### 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

$P=\frac{{V}^{2}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{{V}^{2}}{P}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{{\left(250\right)}^{2}}{500}=125\mathrm{\Omega}$

Now, *P* = 1000 W

$\Rightarrow R=\frac{{V}^{2}}{P}=\frac{{\left(250\right)}^{2}}{1000}=62.5\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

#### Page No 219:

#### Question 4:

A heater coil is to be constructed with a nichrome wire (ρ = 1.0 × 10^{−6} Ωm) that 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?

#### 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

$P=\frac{{V}^{2}}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{{V}^{2}}{P}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{{\left(250\right)}^{2}}{500}=125\mathrm{\Omega}$

(b) We know:

$R=\mathrm{\rho}\frac{l}{A}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{RA}{\mathrm{\rho}}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\frac{125\times 0.5\times {10}^{-6}}{{10}^{-6}}=62.5\mathrm{m}$

(c) Let *n* be the number of turns in the coil. Then,

$l=2\mathrm{\pi}rn\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}n\mathit{=}\frac{\mathit{l}}{\mathit{2}\mathit{\pi}\mathit{r}}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}n\mathit{=}\frac{\mathit{62}\mathit{.}\mathit{5}}{\mathit{2}\mathit{\times}\mathit{3}\mathit{.}\mathit{14}\mathit{\times}\mathit{4}\mathit{\times}{\mathit{10}}^{\mathit{-}\mathit{3}}}\mathit{\approx}2500$

#### Page No 219:

#### Question 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 the connecting wires? Resistivity of copper = 1.7 × 10^{−8}^{ }Ωm

#### Answer:

Let *R* be the resistance of the bulb. If *P* is the power consumed by the bulb when operated at voltage *V,* then

$R=\frac{{V}^{2}}{P}=\frac{{\left(250\right)}^{2}}{100}=625\mathrm{\Omega}$

Resistance of the copper wire,

${R}_{\mathit{c}}=\mathrm{\rho}\frac{l}{\mathrm{A}}=\frac{1.7\times {10}^{-8}\times 10}{5\times {10}^{-6}}=0.034\mathrm{\Omega}$

The effective resistance,

${R}_{eff}=R+{R}_{c}=625.034\mathrm{\Omega}$

The current supplied by the power station,

$i=\frac{\mathrm{V}}{{R}_{eff}}=\left\{\frac{220}{625.034}\right\}\mathrm{A}$

The power supplied to one side of the connecting wire,

$P\text{'}={i}^{2}{R}_{c}\phantom{\rule{0ex}{0ex}}={\left(\frac{220}{625.034}\right)}^{2}\times 0.034$

The total power supplied on both sides,

$2P\text{'}={\left(\frac{220}{625.034}\right)}^{2}\times 0.034\times 2\phantom{\rule{0ex}{0ex}}=0.0084\mathrm{W}=8.4\mathrm{mW}$

#### Page No 219:

#### Question 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?

#### Answer:

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

$R=\frac{{V}^{2}}{P}=\frac{220\times 220}{60}=806.67\mathrm{\Omega}$

(a) Now the supply drops to *V' *= 180 V.

So, the power consumed,

$P\text{'}=\frac{V{\mathit{\text{'}}}^{2}}{R}=\frac{{\left(180\right)}^{2}}{806.67}=40\mathrm{W}$

(b) Now the supply increases to *V"* = 240 V. Therefore,

$P"=\frac{{{V}^{\mathit{\text{\'}}\mathit{\text{\'}}}}^{2}}{R}=71\mathrm{W}$

#### Page No 219:

#### Question 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?

#### 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

$R=\frac{{V}^{2}}{P}=\frac{(220{)}^{2}}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{48400}{100}=484\mathrm{\Omega}$

(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,

$i\text{'}=\frac{V\text{'}}{R}=\frac{217.8}{484}=0.45\mathrm{A}$

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,

$i"=\frac{V"}{R}=\frac{222.2}{484}=0.459\mathrm{A}$

Power consumed by the bulb,

*P" =* *i*" × *V*"

= 0.459 × 222.2 = 102 W

#### Page No 219:

#### 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?

#### Answer:

Given that the operating voltage is *V* and power consumed is *P*.

Therefore, the resistance of the bulb,

$R=\frac{{V}^{2}}{P}=\frac{(220\times 220)}{100}=484\mathrm{\Omega}\phantom{\rule{0ex}{0ex}}$

The power fluctuation, *p* = 150 W. So, the voltage fluctuation that the bulb can withstand,

$v=\sqrt{pR}=\sqrt{150\times 484}\phantom{\rule{0ex}{0ex}}=269.4\mathrm{V}=270\mathrm{V}$

The bulb will withstand up to 270 V.

#### Page No 219:

#### Question 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?

#### Answer:

Given the operating voltage *V* and power consumed *P*, the resistance of the immersion heater,

$R=\frac{{V}^{2}}{P}=\frac{{\left(220\right)}^{2}}{1000}=48.4\mathrm{\Omega}$

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,

$\left(\frac{{V}^{2}}{R}\right)$ × *t* × 60% = 1050000 J

$\Rightarrow \frac{(220{)}^{2}}{48.4}\times t\times \frac{60}{100}=1050000$

⇒ *t* = 29.17 minutes

#### Page No 219:

#### 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 Re 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?

#### Answer:

Time taken to boil 4 cups of water, *t* = 2 minutes

Volume of water boiled = 4 × 200 cc = 800 cc

Initial temperature, *θ*_{1} = 25°C

Final temperature, *θ*_{2} = 100°C

Change in temperature, *θ* = *θ*_{2} − *θ*_{1} = 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, *θ*_{1} = 5°C

Final temperature, *θ*_{2} = 100°C

Change in temperature, *θ* = *θ*_{2} − *θ*_{1} = 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:

#### Question 11:

The coil of an electric bulb takes 40 watts to start glowing. If more than 40 W are 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.

#### 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,

$p=\frac{36-25.584}{36}\times 100\phantom{\rule{0ex}{0ex}}\Rightarrow p=28.93\approx 29\%$

#### Page No 219:

#### Question 12:

The 2.0 Ω resistor shown in the 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?

Figure

#### Answer:

The effective resistance of the circuit,

${\mathrm{R}}_{eff}=\left(\frac{6\times 2}{6+2}\right)+1=\frac{5}{2}\mathrm{A}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Current *i* through the circuit,

$i=\frac{V}{{R}_{eff}}=\frac{6}{5/2}=\frac{12}{5}\mathrm{A}$

Let *i'* be the current through the 6 Ω resistor. Then,

*i*' × 6 = (*i* − *i'*) × 2

$\Rightarrow 6i\text{'}=\left(\frac{12}{5}\right)\times 2-2i\text{'}\phantom{\rule{0ex}{0ex}}\Rightarrow 8i\text{'}=\frac{24}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow i\text{'}=\frac{24}{(5\times 8)}=\frac{3}{5}\mathrm{A}\phantom{\rule{0ex}{0ex}}\Rightarrow i-i\text{'}=\frac{12}{5}-\frac{3}{5}=\frac{9}{5}\mathrm{A}$

(a) Heat generated in the 2 Ω resistor,

*H* =* *(*i - i'*)^{2}*Rt*

$\Rightarrow H=\left(\frac{9}{5}\right)\times \left(\frac{9}{5}\right)\times 2\times 15\times 60=5832\mathrm{J}$

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:

#### 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 bismuth-silver, *a* = − 46 × 10^{−6} V°C^{−1} and *b* = −0.48 × 10^{−6} V°C^{−2}.

#### 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 + $\frac{1}{2}$**bθ*^{2}

⇒ *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:

#### 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 (33.1).

#### Answer:

Difference in temperature, *θ* = 40°C

Emf, *E _{cs}* =

*a*+ $\frac{1}{2}$

_{cs}θ*b*

_{cs}θ^{2}...(1)

*a*= [2.76 − (−43.7) μV

_{cs}= 46.46 μV/°C

*b*= [0.012 − (−0.47) μV/°C

_{cs}= 0.482 μV/°C

^{2}

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:

#### Question 15:

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

#### Answer:

Neutral temperature,

${\theta}_{n}=-\frac{a}{b}$

${a}_{CuFe}={a}_{CuPb}-{a}_{FePb}\phantom{\rule{0ex}{0ex}}=2.76-16.6=13.84\mathrm{\mu V}\xb0{\mathrm{C}}^{-1}$

${b}_{CuFe}={b}_{CuPb}-{b}_{FePb}\phantom{\rule{0ex}{0ex}}=0.012+0.030=0.042\mathrm{\mu V}\xb0{\mathrm{C}}^{-2}$

Thus, the neutral temperature,

${\theta}_{n}=\frac{-{a}_{CuFe}}{{b}_{CuFe}}=\frac{13.84}{0.042}=329.52\xb0\mathrm{C}=330\xb0\mathrm{C}$

The inversion temperature is double the neutral temperature, i.e. 659 $\xb0$C.

#### Page No 219:

#### 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.

#### 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 × 10^{23} atoms = 96500 C

∴ Amount of charge required by 1 atom = $\frac{96500}{6.023\times {10}^{23}}=1.6\times {10}^{-19}\mathrm{C}$

(b) For a divalent material,

equivalent mass =$\frac{1}{2}$molecular mass

⇒ Amount of charge required by $\frac{1}{2}$ × 6.023 × 10^{23} = 96500 C

∴ Amount of charge required by 1 atom = 1.6 × 2 × 10^{−19} = 3.2 × 10^{−19} C

#### Page No 219:

#### Question 17:

Find the amount of silver liberated at the cathode if 0.500 A of current is passed through an AgNO_{3} electrolyte for 1 hour. Atomic weight of silver is 107.9 g mol^{−1}.

#### Answer:

Equivalent mass of silver, *E _{Ag}* = 107.9 g (∵ A

*g*is monoatomic)

The ECE of silver,

${Z}_{\mathit{A}\mathit{g}}=\frac{{E}_{Ag}}{f}=\frac{107.9}{96500}=0.001118$

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:

#### 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}.

#### 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^{−6 }kg C^{−1}

Using the formula, *m* = *Zit*, we get:

$3\times {10}^{-3}=1.12\times {10}^{-6}\times i\times 180\phantom{\rule{0ex}{0ex}}$

$\Rightarrow i=\frac{3\times {10}^{-3}}{1.12\times {10}^{-6}\times 180}\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{1}{6.72}\times {10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow i=14.89\approx 15\mathrm{A}$

#### Page No 219:

#### Question 19:

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

#### Answer:

Let the required time be *t*.

Mass of 1 litre hydrogen,

$m=\frac{2}{22.4}\mathrm{g}\phantom{\rule{0ex}{0ex}}$

Using the formula, *m* = *Zit*, we get:

$\frac{2}{22.4}=\frac{1\times 5\times t}{96500}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{2\times 96500}{22.4\times 5}\phantom{\rule{0ex}{0ex}}\Rightarrow t=1723.21\mathrm{s}=28.7\mathrm{minutes}\approx 29\mathrm{minutes}$

#### Page No 219:

#### Question 20:

Two voltameters, one with a solution of silver salt and the other with 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 mol^{−1}.

#### 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,

$Z=\frac{\mathrm{Equivalent}\mathrm{mass}}{96500}=\frac{\mathrm{Atomic}\mathrm{weight}}{3\times 96500}$

(a) Using the formula, *m* = *Zit,* we get:

$1\times {10}^{-3}=\frac{\mathrm{Atomic}\mathrm{weight}}{3\times 96500}\times 2\times 5400\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Atomic}\mathrm{weight}=\frac{3\times 96500\times {10}^{-3}}{2\times 5400}=26.8\times {10}^{-3}\mathrm{kg}/\mathrm{mole}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Atomic}\mathrm{weight}=26.8\mathrm{g}/\mathrm{mole}$

(b) Using the relation between equivalent mass and mass deposited on plates, we get:

$\frac{{E}_{\mathit{1}}}{{E}_{\mathit{2}}}=\frac{{m}_{1}}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{26.8}{3\times 107.9}=\frac{1}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{2}=12.1\mathrm{g}\phantom{\rule{0ex}{0ex}}$

#### Page No 219:

#### Question 21:

A brass plate of surface area 200 cm^{2} 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}.

#### Answer:

Given:

Current,* i* = 15 A

Surface area of the plate = 200 cm^{2},

Thickness of silver deposited= 0.1 mm = 0.01 cm

Volume of Ag deposited on one side = 200 × 0.01 cm^{3} = 2 cm^{3}

∴ Volume of Ag deposited on both side = 4 cm^{3}

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*

$\Rightarrow t=\frac{42\times 96500}{107.9\times 15}\mathrm{s}\phantom{\rule{0ex}{0ex}}\Rightarrow t=2504.17\mathrm{s}=42\mathrm{minutes}$

#### Page No 219:

#### Question 22:

The 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 20 Ω resistor during this period. Atomic weight of silver is 107.9 g mol^{−1}.

Figure

#### Answer:

Given:

Mass of silver deposited,* m *= 2.68 g

Time,* t *= 10 minutes = 600 s

Using the formula,* m* = *Zit*, we get:

$2.68\times {10}^{-3}=\frac{107.9\times {10}^{-3}}{96500}\times i\times 600\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{2.68\times 96500}{107.9\times 600}\phantom{\rule{0ex}{0ex}}\Rightarrow i=3.99=4\mathrm{A}$

Heat developed in the 20 Ω resistor,

$H={i}^{2}Rt\phantom{\rule{0ex}{0ex}}\Rightarrow H={\left(4\right)}^{2}\times 20\times 600\phantom{\rule{0ex}{0ex}}\Rightarrow H=192000\mathrm{J}=192\mathrm{kJ}$

#### Page No 219:

#### 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 voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9 g mol^{−1}.

#### 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:

$E=V+ir\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{E-V}{r}=\frac{12-10}{2}=1\mathrm{A}$

Using the formula *m* = *Zit*, we get:

$m=\frac{107.9}{96500}\times 1\times 0.5\times 3600=2.01\mathrm{g}$

#### Page No 219:

#### Question 24:

A plate of area 10 cm^{2} is to be electroplated with copper (density 9000 kg m^{−3}) to a thickness 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 × 10^{−7} kg C^{−1} and specific heat capacity of water = 4200 J kg^{−1}

#### Answer:

Surface area of the plate, *A *= 10 cm^{2} = 10 × 10^{−4} m^{2}

Thickness of copper deposited, *t *= 10 μm = 10^{−5}^{ }m

Density of copper = 9000 kg/m^{3}

Volume of copper deposited, *V* = *A*(2*t*)

*V *= 10 × 10^{−4} × 2 × 10 × 10^{−6}

= 2 × 10^{2} × 10^{−10}

= 2 × 10^{−8} m^{3}

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 × 10^{2} C

Energy spent by the cell = Work done by the cell

⇒*W* = *VQ*

= 12 × 6 × 10^{2}

= 72 × 10^{2} = 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 × 10^{3} = 100 × 10^{−3} × 4200 × ∆*θ*

⇒ ∆*θ* = 17 K

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