ROUTERA


Thermal And Chemical Effects Of Electric Current

Class 12th Concepts Of Physics Part 2 HC Verma Solution



Short Answer
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:

We know Joule’s heating effect –


Joule’s heating effect is the process by which the passage of an electric current through a conductor produces heat


When a constant potential difference is applied across a bulb, due to Joule’s heating effect, the temperature of the bulb increases.


We know in case of metals, the temperature co-efficient is positive, which means, with increase in temperature, the Resistance of the metal increases.


In case of metals, the resistance and temperature is related as -



Where,


R= resistance at some temperature T


R0 = resistance at zero temperature


and = Temperature co-efficient.


From ohm’s law


Voltage,


Where


I is the current


R is the resistance


Thus, from the above formula, we get that with increase in resistance, current decreases. if the voltage remains constant


Now, the heat generated by the resistance is constantly radiated through the connecting wires into the surroundings.


Thus, the value of its temperature is maintained constant and so its resistance.


As a result, the current through the bulb filament becomes constant.



Question 2.

Two unequal resistances R1 and R2 are connected across two identical batteries of emf ϵ and internal resistance r. Can the thermal energies developed in R1 and R2 be equal in a given time. If yes, what will be the condition?




Answer:

Given ∈ as emfs of the circuits with internal resistances r


Let the currents passing through the


resistance R1 and R2 be i1 and i2 , respectively for time t.


According to Kirchhoff’s Voltage Law, sum of


all voltages around any closed loop in a


circuit must equal zero.



ie,


Looking into the circuit 1-we get-



(1)


Similarly, the current in the other circuit,


(2)


The heat lost through a resistor r is given by


(3)


Now the total thermal energies through the resistances from Joule’s heating effect are given


by -


=


From(1), (2) and (3)


⇒( = (


⇒( = (


⇒( = (






Hence, the condition when the thermal energies developed in


R1and R2 be equal in a given time is when the internal resistance r–




Question 3.

When a current passes through a resistor, its temperature

increases. Is it an adiabatic process?


Answer:

We know that adiabatic process occurs without the transfer of heat or mass of substances between a thermodynamic system and its surroundings.


Now, in this case, No.


There is no exchange of heat with surrounding in case of adiabatic process.


In case of resistance,


Heat is developed by the virtue of Joule heating effect in the capacitor, which state that the heat developed in the resistance is given by


H =i2 RT


Where,


I is the current in the circuit


R is the resistacne of thetemperature


T is the time perod for which current is allowed to transfer in the circuit.


The heat developed by the resistor led to increase in the operating temperature of the resistor due to which the resistance of the resistor increase with the increase of the resistance given by the formua below:



Where,


R= resistance at some temperature T


R0 = resistance at zero temperature


and = Temperature co-efficient.


Thus, the heat generated is exchanged between the resistor and surrounding.


Thus, the above process cannot be an adiabatic process.



Question 4.

Apply the first law of thermodynamics to a resistor carrying a current i. Identify which of the quantities ΔQ, ΔU and ΔW are zero, which are positive and which are negative.


Answer:

First law of thermodynamics states –


the total energy of an isolated system is constant, energy can neither created nor destroyed, but can be transformed from one form to another



where


ΔU is the change in the internal energy,


Q is the heat added to the system,


and ∆W is the work done by the system.


Now, here 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 in the form of heat, thus ∆Q is positive.


As the temperature of the resistor rises, ΔU is positive.



Question 5.

Do all the thermocouples have a neutral temperature?


Answer:

Thermocouples are of two wire legs made from different metals. These wires legs are welded together at one end, creating a junction.


This junction is where the temperature is measured.


When the change in temperature occurs at junction, a voltage is created.


Now neutral temperature for the thermocouple.
is the temperature of the hot junction at which the thermo-emf in a thermocouple becomes maximum


For a thermocouple with constants a and b having the same sign, the neutral temperature will be less than the cold junction temperature of thermocouple, as



Hence there will be no neutral or inversion temperature, since the hot junction temperature cannot be less than the temperature of cold junction of thermocouple.



Question 6.

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


Answer:

In thermocouple when, keeping the temperature of the cold junction constant, the temperature of the hot junction is gradually increased.


The thermodynamic emf rises to a maximum at a temperature (θn) called neutral temperature and then gradually decreases and eventually becomes zero at a particular temperature (θi) called temperature of inversion.


If, the inversion temperature and neutral temperature both measured in degree Celsius, then we can say "inversion temperature is always double the neutral temperature"


But if measured in other units such as Kelvin, inversion temperature may not be double the neutral temperature



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 cannot be the arithmetic mean of the inversion temperature and the temperature of the cold junction always.


Neutral temperature is defined as the temperature of the hot junction of a thermocouple attains its maximum value, when the clod junction is maintained at a constant temperature of 0oC.


While the inversion temperature is the critical temperature at which a non-ideal gas experiences a decrease in temperature will experience a temperature decrease and above which will experience a temperature increase.


Neutral temperature does not change for a metal whether the temperature of cold junction and inversion temperature change by any value.


This is valid only Celsius is the unit for both the temperatures.



Question 8.

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


Answer:

An electrolytic cell is an electrochemical cell that drives a non-spontaneous redox reaction through the application of electrical energy through the electrodes.


Now, the electrodes in an electrolytic cell cannot have fixed polarity like that of a battery.



Lets take an example of electrolytic cell as show below:



The redox reaction for the above cell is given as below:


Zn(S) +Cu2+ (aq) → Zn3+ (aq) + Cu(s)


The cell potential is 1.1 eV


If external potential less than 1.1 eV is applied to the cell the electron flow from Zn rod to Cu rod hence current flows from Cu to Zn. Zinc rod will act as Cathode while the copper rod will act as Anode.


If the external potential is greater than 1.1V. Electrons flow from Cu to Zn and current flows from Zn to Cu. The Zinc rod act as anode while copper rod acts as cathode.


Thus, we an infer that the electrodes of the electrolytic cell cannot have fixed polarity.



Question 9.

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


Answer:

As temperature increases, volume of the electrolyte decreases due to which there is rapid decrease of the mean free path which decreases the viscosity of the liquids, so the resistance of the electrolyte will also decrease as with an increase in temperature


Thus, the resistance of the electrolyte will decrease.




Objective I
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?




Answer:

|

When current passes through a resistor, the heat produced by Joule’s heating effect is given by-




where,


I =current


R = resistance of the resistor


t = time for which current is flowing


The graph shows the variation of the heat generated by the resistor with respect to the current.


Form the above equation, we get



Which is same as the equation of parabola.


Since, heat produced for a given time in a resistor varies with the square of current flowing through it.


Thus, the curve a is the correct option.


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 figure base represents the rate of production of thermal energy in the resistor.




Answer:

When current passes through a resistor, the heat produced by Joule’s heating effect is given by-




where,


I = current


R = resistance of the resistor


t = time for which current is flowing


Since with increase in the temperature of the resistor, its


resistance is also increased.


The rate of production of thermal energy in the resistor will be-



where


i = current flowing through the resistor and


R = resistance of the resistor


From above equation, the rate of production of thermal energy in


the resistor is directly proportional to the resistance.


Since, due to the heat generated by the resistance the operating temperature of the resistance increases given by the formula below:


R=Ro(1+αT)


Where


Ro is the initial resistance of the resistor


α is the coefficient of thermal conductivity


T is the temperature


Thus, the resistance of the resistor start increasing with the increase in temperature.


From the graph,


also increases linearly with time starting from a point , which is best shown by plot d.


Thus, option D is the correct option.


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. Both A and B are correct.

B. A is correct but B is wrong.

C. B is correct but A is wrong.

D. Both A and B are wrong.


Answer:

|

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.


Thus, (b) is the correct option.


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:

When current passes through a resistor, the heat produced by Joule’s heating effect is given by-




where,


I =current


R = resistance of the resistor


t = time for which current is flowing


Joule heat is directly proportional to the square of the current passing through the resistor.


We know,


Peltier heat is defined as the change in the temperature at the junction due to the flow of current in a circuit consists of two different kinds of conductor which lead to a heating or cooling effect is observed at the junction between the two material.


Peltier heat is directly proportional to the current passing through the junction.


Thomson heat is defined as the evolution or absorption of heat when electric current passes through a circuit composed of a single material that has a temperature difference along its length


Also, in case of Thomson heat is also directly proportional to the current passing through the section of the wire.


Question 5.

Consider the following two statements.

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

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

See beck 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:

|

The Seebeck effect is a phenomenon, in which a temperature difference between two dissimilar electrical conductors or semiconductors produces a voltage difference between the two substances.


The cause of this voltage difference is the diffusion or migration of free electrons from a higher electron-density region to a lower electron-density region.


The free electron-density of the electrons varies from metal to metal and also electron-density changes with change in temperature.


Hence, both the statements are the causes of Seebeck Effect.


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:

|

Peltier effect is an effect whereby heat is given out or absorbed when an electric current passes across a junction between two materials.


These materials are generally metals.


Now, this is caused due to the difference in density of free electrons in different metals.


Thus, the option B is the correct option


Question 7.

Consider the statements A and B in question 5. Thomas’s effect is caused.

Options 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:

|

Thomson effect is the process where the evolution or absorption of heat takes place when electric current passes through a circuit composed of a single material that has a temperature difference along its length.


If a metallic conductor has non-uniform temperature distribution along its length, the density of the free electrons varies for different sections.


The electrons diffuses from the region with higher concentration to the region with lower concentration of free electrons.


Thus, the free-electron density in a material depends on temperature for the Thomson effect.


Hence, option C is the correct option.


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:

|

Faradays constant is universal constant.


Its value is equal to 9.6845×107 C/kg which is constant on entire universe.


It does not depend on the amount of the electrolyte, current in the electrolyte and on the amount of charge passed through the electrolyte.


Thus, option C is the correct option.



Objective Ii
Question 1.

Two resistors having 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:

|

When current passes through a resistor, the heat produced by Joule’s heating effect is given by-




where,


I =current


R = resistance of the resistor


t = time for which current is flowing


When the resistors are in series, the current through them will be same.


So, the amount of thermal energy evolved in the resistors is same for each of the resistor.


The rise or fall in the temperature of the resistance will depend on the shape and size of the resistor.


So, the rise in the temperature of the two resistances may become equal.


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:

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 as shown in fig.


Due to Seebeck effect, there will be generation of thermo-emf between the points that are at different temperatures.


Here, all junctions and ends are at different temperatures, i.e., 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, a potential difference exists among them.


Question 3.

The constants a and b for the pair silver-lead are 2.50 μV°C–1 and 0.012 μV°C–1 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:

|

Thermocouples are of two wire legs made from different metals. These wires legs are welded together at one end, creating a junction.


This junction is where the temperature is measured.


When the change in temperature occurs at junction, a voltage is created.


Now neutral temperature for the thermocouple.
is the temperature of the hot junction at which the thermo-emf in a thermocouple becomes maximum


For a thermocouple with constants a and b having the same sign, the neutral temperature will be less than the cold junction temperature of thermocouple, as



Hence, there will be no neutral or inversion temperature, since the hot junction temperature cannot be less than the temperature of cold junction of thermocouple.


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:

|

An electrolytic cell is an electrochemical cell that drives a non-spontaneous redox reaction through the application of electrical energy through the electrodes


Electrolytic cells have both the electrodes of the same material.


So, on reversing the terminals of the battery, the direction of the charge flow reverses, but the rate of the electrolysis will remain constant.


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:

|

The electrochemical equivalent of a substance is defined as -


the mass of the chemical element (in grams) transported by 1 coulomb of electric charge.


The electrochemical equivalent of a substance is calculated by taking the ratio of the relative atomic mass of the substance to its valency.


Thus, it is only dependent on the nature of the material.



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




where,


I =current


R = resistance of the resistor


t = time for which current is flowing


Putting the value in the above formula, we get



= 2 × 2 × 25 × 60= 100 × 60 J = 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.


Answer:

Given-



Resistance of the coil, R = 100 Ω,
Emf of the battery, V = 6 V,
Change in temperature, ∆T = 15°C



Using Joule’s heating effect heat produced across the coil,



This heat generated is used to increase the temperature of the coil.









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 the resistance of the coil be R


Given


voltage =250 V


power =500 W



The power P consumed by a coil of resistance R when connected across a supply V is given by Joule’s heating effect -






Now when, P = 1000 W






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?


Answer:

(a) Let R be the resistance of the coil.


Given, resistivity, ρ = 1.0 × 10–8 Ω m


The power P consumed by a coil of resistance R when connected across a supply V is given by Joule’s heating effect-






(b) We know resistance R is given by-



where


ρ = resistivity of the material


l = length of the wire


A= area of cross section of the wire


Given, cross-sectional area of the wire is 0.5 mm2





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


Given, radius of each turn is 4.0 mm


Then,



Where,


l = length of the coil


r=radius of the coil






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.


Answer:

Given


voltage = 250V


power p=100W


Resistivity of copper = 1.7 × 10–8 Ω m.


area of cross section 5 mm2 = 5 ×10-6 m.


Let R be the resistance of the bulb.


If P is the power consumed by the bulb when operated at voltage V, then by Joule’s heating effect-




Putting the value in the above formula, we get




Resistance of the copper wire is given by,



where


l = length of the coil


ρ= resistivity of the coil


A= area of the coil




The effective resistance,




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




The power supplied to one side of the connecting wire is given using Joule’s heating effect,




The total power supplied on both sides,






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?


Answer:

If P is the power consumed by the bulb when operated at voltage V, then by Joule’s heating effect-






Now when the supply drops to V’ = 180 V



The power consumed,





(b) Now the supply changes to, V” = 240 V.


Therefore,






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?


Answer:

Given,


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 Joule’s heating effect-



Putting the values in the above formula, we get





(a) We know that for minimum power to be consumed, output voltage should be minimum.


The minimum output voltage will be,



The current through the bulb,


Where


V is the voltage across the bulb


R is the resistance of the bulb


Putting the values in the above formula, we get



=0.45 A


Now, the power consumed by the bulb,





(b) For maximum power to be consumed, output voltage should be maximum.


The maximum output voltage,






The current through the bulb,





Power consumed by the bulb in this case is,






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


The operating voltage is V and power consumed is P.



Therefore, the resistance of the bulb can be calculated as follows,




Putting the values in the above formula, we get




The power fluctuations is given as p = 150 W


So, the voltage fluctuation that the bulb can withstand is given by-


Resistance of the bulb,


Where


V is the velocity


P is the power fluctuations






Hence the 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?


Answer:

Given


The operating voltage V


Power consumed P,


Now the resistance of the immersion heater,





Mass of water m is ,


m =1100× 1000


= 10 Kg



We know specific heat of water, s = 4200 Jkg-1 K-1



Given rise in temperature, θ = 25°C



Heat required to raise the temperature of the given mass of water is given by –







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


Given the heat evolve is only 60%.


So,






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?


Answer:

Given-


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 = θ2θ1 = 75°C
Mass of water required to boil , m = 800 × 1 = 800 gm = 0.8 Kg



Now heat required for boiling water,


Where


m is the mass of the water


s is the specific heat of the water


θ is the change in temperature


Putting the value in the above formula, we get




We know, to convert watt-hour to watt-sec,



1000 watt – hour = 1000 × 3600 watt sec.



Hence cost of boiling 4 cups of water




(b)


Given,


Initial temperature, θ1 = 5°C
Final temperature, θ2 = 100°C
Change in temperature, θ = θ2θ1 = 95°C



heat required for boiling water,


Where


m is the mass of the water


s is the specific heat of the water


θ is the change in temperature


Putting the value in the above formula, we get


= 0.8 × 4200 × 95


= 319200


Again converting into watt-second,
Cost of boiling 4 cups of water


=11000×3600×319200


= Rs 0.09



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.


Answer:

Case-I


Given -


When the supply voltage is 220 V.
Power consumed by the bulb = 100 W
Excess power = 100 − 40 = 60 W



Hence, power converted to light = 60% of 60 W


= 36 W


Case-II


Given


When the supply voltage is 200 V.



Power consumed,


Where


P is the power consumed by the bulb


V is the voltage


Putting the values in the above formula, we get




Excess power


= 82.64 − 40 = 42.64 W



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


Percentage drop in the light intensity is-





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?




Answer:

Given-


Resistance - 2.0 ΩΩ


heat capacity of the calorimeter together with water = 2000 J K–1



Looking into the circuit the effective resistance of the circuit,


2.0 Ω in parallel with 6.0 Ω and this combination in series with 1.0 Ω




Now, current i through the circuit, from ohm’s law-






Let take i’ as the current through the 6 Ω resistor.


Then, applying KCL, the algebraic sum of current entering and leaving a node is zero and from ohm’s law, we can write -









(a) Heat generated in the 2 Ω resistor, using Joule’s Heating effect




from (1) and substituting values




Given that the heat capacity of the calorimeter together with water is 2000 J K−1


Which means , 2000 J of heat raise the temp by 1 K.



Then, 5832 J of heat raises the temperature of water by




(b) When the 6 Ω resistor burn out, the effective resistance of the


circuit will become –






Current through the circuit,





Heat generated in the 2 Ω resistor, using Joule’s Heating effect





Given that the heat capacity of the calorimeter together with water is 2000 J K−1


Which means , 2000 J of heat raise the temp by 1 K.



So, 7200J will raise the temperature by




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.


Answer:

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.


We, know emf E is given by-




Putting the value in the above equation, we get









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:

Given –


Difference in temperature, θ = 40°C



Emf, E developed across junction is given by -



From table –










Substituting in eq. (1),







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 (33.1).


Answer:

Neutral temperature is given by -



Using table data –






Similarly






Thus, the neutral temperature becomes,






Since, temperatures is in Celsius, 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.


Answer:

(a) monovalent material –


We know


1 F (farady) is 1 mol of electrons



⇒1 F = one electron (1.602 x 10-19) ×Avogadro's number (6.022 x 1023)


=96,500 C of charge



We know that amount of charge required by 1 equivalent mass of the substance = 96500 C



For a monovalent material, equivalent mass is its molecular mass



⇒ Amount of charge required by 6.023 × 1023 atoms which is the Avogadro’s Number and is equal to 96500 C of charge



Therefore ,amount of charge required by 1 atom =



(b) For a divalent material-



As we know equivalent mass(1U) is 12 times its molecular mass(Z)


We know


1 F (faraday) is 1 mol of electrons



⇒1 F = one electron (1.602 x 10-19) ×Avogadro's number (6.022 x 1023)


= 96,500 C of charge



⇒ Amount of charge required is 12 times of 6.023 × 1023 =


= 96500 C of charge



⇒ Amount of charge required by 1 atom for a divalent material is –




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.


Answer:

Given


Equivalent mass of silver EAg = 107.9 g


Current passed = 0.005 A


We know


1 F (faraday) is 1 mol of electrons



⇒1 F = one electron (1.602 x 10-19) ×Avogadro's number (6.022 x 1023)


= 96,500 C of charge



⇒ Amount of charge required is 12 times of 6.023 × 1023 =


= 96500 C of charge



The Electrochemical equivalent of silver,



Where


E is the equivalent mass of silver


Agf is the amount of charge




Using the formula, from faraday’s law of electrolysis –



where,


m = the mass of the substance deposited on the electrode


z=electrochemical equivalence


i=current passing


t=time taken


we get:







So, 2.01 g of silver is liberated on electrode.



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



Electrochemical equivalent of silver, Z = 1.12 × 10−6 kg C−1


Using the formula, from faraday’s law of electrolysis –



where,


m = the mass of the substance deposited on the electrode


z=electrochemical equivalence


i=current passing


t=time taken


substituting the values






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.


Answer:

Given
Mass of 1 liter hydrogen,


m=222.4 g


Current = 5A


Let t be required time


Using the formula, from faraday’s law of electrolysis –



where,


m = the mass of the substance deposited on the electrode


z=electrochemical equivalence


i=current passing


t=time taken


substituting the values,






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.


Answer:

Given-
Mass of salt deposited, m = 1 g



Current, i = 2 A



Time, t = 1.5 hours = 5400 s


Atomic weight of silver is 107.9 g mol–1.


For the trivalent metal salt



Equivalent mass = 13 times its Atomic weight



The E.C.E of the salt


Z=Equivalent mass


96500=Atomic weight 3×96500


(a) Using the formula, from faraday’s law of electrolysis –



where,


m = the mass of the substance deposited on the electrode


z=electrochemical equivalence


i=current passing


t=time taken


Now, 1 gram is deposited and the element is trivalent metal






(b)Now the relation between equivalent mass and mass deposited on plates is given by –



Where


E1 is the equivalent mass of the trivalent metal


E2 is the equivalent mass of the silver


m1 is the mass of trivalent material deposited


m2 is the mass of the silver deposited


Atomic weight of silver is 107.9 g mol–1.


Substituting to calculate the mass deposited –





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.


Answer:

Given-
Current passing, 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 = 2 ×2 cm3 = 4 cm3


The specific gravity of silver = 10.5


Atomic weight = 107.9 g mol–1



Now, mass of silver deposited, m can be calculated



m = Volume × Specific gravity × 1000


putting the value in the above formula, we get


= 4 × 10-3 × 10.5 ×1000


= 42 kg



Using the formula, from faraday’s law of electrolysis –



where,


m = the mass of the substance deposited on the electrode


z=electrochemical equivalence


i=current passing


t=time taken


Substituting the values







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.




Answer:


Given:
Mass of silver deposited, m = 2.68 g



Time, t = 10 minutes = 600 s


Using the formula, from faraday’s law of electrolysis –



where,


m = the mass of the substance deposited on the electrode


z=electrochemical equivalence


i=current passing


t=time taken






Now, heat developed in the 20 Ω resistor from Joule’s heating effect,



Where


I is the current in the circuit


R is the resistance of the resistor


T is the time period for which the current is allowed to flow through the circuit


Putting the above values in the formula, we get






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 its 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.


Answer:

Let the current through the circuit i be.



Emf of battery, E = 12 V



Voltage drop across the voltmeter, V = 10 V



Internal resistance of the battery, r = 2 Ω



Applying Kirchhoff’s Law, the algebraic sum of voltage and voltage drop in the circuit is zero,


we get



Where, E = battery emf


V=voltage drop


r=internal resistance


I = current through the circuit





Using the formula, from faraday’s law of electrolysis –



where,


m = the mass of the substance deposited on the electrode


z=electrochemical equivalence


i=current passing


t=time taken





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.


Answer:

Given


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


Electrochemical equivalent of copper –3 × 10–7 kg C–1


specific heat capacity of water – 4200 J kg–1 K–1


Volume, V of copper deposited,


V = (2t)


Where


A is the area of the plates


t is the thickness of the plates


Putting the values in the above formula, we get









We know, mass of copper deposited,


m = Volume × Density





Using the Faraday's Laws


m = ZQ


where


m = mass of the substance


Q= charge


Z= Electrochemical equivalent


Putting the values in the above formula, we get
⇒ 18 × 10−5 = 3 × 10−7 × Q


Q = 6 × 102 C



Now,


The energy spent by the cell will be = Work done by the cell



W = V× Q


Where


V is the potential difference


Q is the charge


Putting the values in the above formula, we get



= 12 × 6 × 102



= 72 × 102 = 7.2 kJ



Let us take ∆θ as the rise in temperature of water.


If this amount of energy is used to heat 100 g of water, then-


Q = c×m ×∆θ


Where,


c= specific heat of water


Q= Heat


m = mass of water


∆θ = change in temperature


Substituting



⇒7.2 × 103 = 100 × 10−3 × 4200 × ∆θ



⇒ ∆θ = 17 K