33: Thermal And Chemical Effects Of Electric Current : Neet-xii-physics

Home Learn Fast select Course

Change Unit Change Subject Last Menu

NEET-XII-Physics

33: Thermal and Chemical Effects of Electric Current

with Solutions - page 2

Note: Please signup/signin free to get personalized experience.

10 minutes can boost your percentage by 10%

Note: Please signup/signin free to get personalized experience.

Qstn #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.
digAnsr: c
Ans : (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:

Qstn #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
digAnsr: a
Ans : (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:

Qstn #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
digAnsr: b
Ans : (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:

Qstn #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
digAnsr: c
Ans : (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:

Qstn #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
digAnsr: c
Ans : (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:

#
Section : iii

Qstn #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
digAnsr: a,d
Ans : (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:

Qstn #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
digAnsr: a,b,c,d
Ans : (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:

Qstn #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
digAnsr: a,b
Ans : (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:

Qstn #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.
digAnsr: c
Ans : (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:

Qstn #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
digAnsr: a
Ans : (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:

#
Section : iv

Qstn #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?
Ans : 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:

Qstn #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?
Ans : 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\,\mathrm{\,s\,}=2.8\,\mathrm{\,min\,}``
Page No 218:

Qstn #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?
Ans : 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{{\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 \,}``
`` ``
Page No 219: