Thermal & Chemical Effects of Current & Thermoelectricity

HEATING EFFECT OF CURRENT, JOULE'S LAW

(i) Whenever work is transformed into heat or heat into work, the quantity of work done is mechanically equivalent to the quantity of heat.

(ii) Thus Work $\propto$Heat.

or Work = a constant x heat

or W = JH

Where J is known as Joule's constant or mechanical equivalent of heat. It is defined as that mechanical work which produces unit calorie of heat.

(iii) Expression for heat produced in a conductor due to current flow through it-

(a) Let the potential difference between the points A and B of a conductor is V, on account of which a current i flows through it.

(b) As potential difference is the work done per unit test charge

$\therefore$ W = QV But Q = it

$\therefore$ W = Vit

But heat developed H = $\frac{W}{J}=\frac{Vit}{J}=\frac{i^{2}RT}{J}$

Where R is the resistance of wire.

(c) If i i is in ampere, V is in volt and R is in ohm, then

$W=VitJoule$

= Vit x 10⁷ ergs.

$\therefore$ Heat developed H = $\frac{Vit}{J}$ calories

$=\frac{Vit}{4.2}=0.24Vitcals.$

= 0.24 i²Rt cals.

JOULE'S LAWS OF HEATING EFFECTS OF CURRENT

(i) The heat developed in a conductor is given by

H = i²Rt Joule = 0.24 i²Rt cals

(ii) The amount of heat developed in a conductor, in a given time, is directly proportional to the square of the current. i.e. $H\propto i^2$, when R and t are constants.

(iii) The amount of heat developed in a conductor by a given current in a given time is directly proportional to the resistance of the conductor.

i.e. $H\propto R$, when i and t are constants.

(iv) The amount of heat developed in a given conductor due to a given current is proportional to the time of flow of the current.

i.e. $H\propto t$, when i and R are constants.

ELECTRICAL ENERGY OR WORK

If Q units of charge be carried between two points differing in potential by V, then electrical work done is

W = Q x V Joule

= Q x V x 10⁷ ergs

POWER

(i) The rate at which work is done is defined as power

$\therefore Power=\frac{Q\times V}{t}=Vi$

= i²R

$=\frac{V^2}{R}$

(ii) Thus electric power = potential difference × current i.e. 1 watt = 1 Volt x 1 Ampere

= 1 Joule/sec

= 10⁷ ergs/sec

(iii) The practical unit of power is kilowatt

1 kilowatt = 1000 watt

(iv) Horse-power = 550 ft-lbs per sec.

= 550 × 12 × 2.54 ×453.6 gm-cm/sec

= 746 × 10⁷ ergs/sec = 746 watt.

Illustration 1: A coil of resistance 2$\Omega$ is immersed in 1 kgm of water and is connected to the terminals of the battery of internal resistance 4 $\Omega$ and emf 6 volt for 3 minutes. The increase in temperature of water is :

(A) 0.93°C (B) 0.085°C

(C) 1.92°C (D) 4.31°C

Solution: Heat developed in coil = Heat absorbed by water

$i^2\frac{Rt}{J}=ms\Delta t$

$\therefore \Delta t=\frac{i^{2}Rt}{Jms}={\left(\frac{E}{R+r}\right)}^2\frac{Rt}{Jms}$

$={\left(\frac{6}{4+2}\right)}^2\times \frac{2\times 180}{4.2\times 1\times {10}^3}\approx 0.085{}^{\circ }C$

Illustration 2: A Daniel cell has an emf of 1.08 volt and internal resistance 0.5$\Omega$. It is successively connected to two wires whose resistances are 2$\Omega$ and 3$\Omega$. The ratio of the amounts of heat developed in two wires will approximately be :

(A) 4 : 3 (B) 3 : 4

(C) 2 : 3 (D) 3 : 2

Solution: Heat developed in a wire

H = i²Rt

or $\frac{E^{2}Rt}{{(R+r)}^2}$

$\therefore H\propto \frac{R}{{(R+r)}^2}$

$\therefore \frac{H_1}{H_2}=\left(\frac{R_1}{R_2}\right)\times {\left(\frac{R_2+r}{R_1+r}\right)}^2$

$=\frac{2}{3}\times {\left(\frac{3.5}{2.5}\right)}^2$ $=\frac{2}{3}\times \frac{49}{25}\approx 4:3$

(v) Kilo-watt-hour or Board of Trade (B.O.T.) unit-

(a) Energy consumed in a given time is the product of power and time

(b) When power of one watt is consumed for an interval of one hour, then energy consumed

= 1 watt x 1 hour

= 1 watt - hour

= 0.001 kilo-watt-hour

(c) Board of Trade (B.O.T.) unit of electric power is kilo-watt-hour by which the consumption of electric energy is measured and charged by power supply authorities.

1 kilo watt-hour = 10³ x 60 × 60 watt-sec

= 36 x 10⁵ Joule

(d) Rule for calculation of cost-

(i) Number of B.O.T. units =

$\frac{Sumofwattagesofallsystems\times time(inhour)}{1000}$

(ii) Total cost =

Number of B.O.T. units x rate of charge per unit

(iii) Total cost =

$\frac{(\sum P)or(\sum Vi)or(\sum i^{2}R)or\left(\sum \frac{V^2}{R}\right)\times timeinhour\times rateperunit}{1000}$

Illustration 3: A dwelling house is installed with 15 lamps, each of resistance 10³$\Omega$ and 4ceiling fans each driven by 1/8th horse-power motor. If the lamps and fans are run on an average for 6 hours daily, then the number of B.O.T. units consumed by lamps in a month of 31 days will be:

(A) 135 (B) 150

(C) 165 (D) 180

Solution: Number of B.O.T. units consumed by 15 lamps

$=\frac{V^2\times 15\times 6\times 31}{R\times 1000}$

$=\frac{220\times 220\times 15\times 6\times 31}{{10}^3\times 1000}\approx 135$

Chemical effects of current

Definitions of various terms

(i) Electrolysis: The process of splitting up or decomposing a liquid by passing an electric current through it, is defined as electrolysis.

(ii) Electrolyte: The compound, whether fused or in solution, which undergoes decomposition by an electric current, is defined as electrolyte.

(iii) Anions and Kations: The decomposed substance appears in the form of ions. The ions appearing at the anode are known as anions and those at the cathode are known as cations.

(iv) Ionisation: The phenomenon of separation of a molecule into oppositely charged ions is known as ionisation.

(v) Equivalent weight- The ratio of the atomic weight of an element to its valency is defined as its equivalent weight.

(vi) Atomic weight: The ratio of the average mass of an atom of an element of the mass of an atom of hydrogen, taken as 1.008, is defined as the atomic weight of that element.

(vii) Valency: The valency of an element is the number of atoms of hydrogen or chlorine which combines with or is displaced by one atom of the element.

(viii) Molecular weight: The molecular weight of a substance is the ratio of the mass of one molecule of the substance to the mass of an atom of oxygen which is taken as 16.

(ix) Gram-equivalent: It is the weight in grams of an element which will combine with or replace 1 gm of hydrogen.

Gram-equivalent = $\frac{Gram-atom}{valency}$

(x) Gram-molecule: The molecular weight of any substance. expressed in grams is defined as gram-molecule of that substance.

(xi) Avogadro number: The number of molecules of a substance in its one-gram-molecule is known as Avogadro number.

(xii) Normal solution: A solution containing one gram-equivalent of the solute per litre is called a normal solution.

Faraday's laws of Electrolysis

(i) First law: The towal mass of ions liberated at an electrode, during electrolysis, is proportional to the quantity of electricity which passes through the electrolyte.

i.e. $m\propto QButQ=it$

$\therefore m\propto it$

Hence the first law may also be stated as follows

The mass of ions liberated at an electrode during electrolysis is proportional to

(a) the current following through the electrolyte, and

(b) the time for which the current flows

(ii) Second law: If same quantity of electricity is passed through different electrolytes, the masses of the substances (ions) deposited at the respective cathodes are directly proportional to their chemical equivalents (equivalent weights).

i.e. $m\propto E$ (chemical equivalent)

(iii) Electro-chemical equivalent (E.C.E.):

(a) The electro-chemical equivalent of an element is its mass in grams deposited on the electrode by the passage of 1 coulomb of charge through it i.e. by the passage of 1 ampere current for 1 second.

(b) According to Faraday's first law-

$m\propto i.t$

or m = Z i.t

Where Z is the constant of proportionality known as the electro- chemical equivalent of the substance. It is numerically equal to the mass in grams of the element deposited when a unit current flows in unit time.

(c) According to Faraday's first law

m = ZQ

If same charge is passed two electrolytes, the

$\frac{m_1}{m_2}=\frac{Z_1}{Z_2}$

According to Faraday's second law

$\frac{m_1}{m_2}=\frac{E_1}{E_2}$

$\therefore \frac{m_1}{m_2}=\frac{Z_1}{Z_2}$

or Z² = $Z_1\times \frac{E_2}{E_1}$

(d) E.C.E. of any substance =

E.C.E. of hydrogen × chemical equivalent of the substance.

(e) If silver is taken as the standard substance for which E.C.E. is 0.001118 gm per coulomb, then E.C.E. of any substance = $0.001118\times \frac{Chemicalequivalentofsubs\tan ce}{chemicalequivalentofsilver}$

(iv) Faraday: (a) The quantity of electricity (i.e. charge) required to liberate a gram equivalent of a substance during electrolysis.

(b) As derived above

$\frac{E_1}{E_2}=\frac{Z_1}{Z_2}or\frac{E}{Z}=cons\tan t=F$

The constant F is known as one Faraday.

For example for copper E = 31.5 gm

and Z = 0.000329 gmC^–1

$\therefore F=\frac{31.5}{0.000329}=96500coulomb$

(c) According to Faraday's first law

m = Z.i.t. = ZQ

If p is the valency of the element, then electrons has to flow through the solution to deposit one atom.

$\therefore$ charge required to deposit 1 mol of the substance = Npe,

where N = Avogadro number and m = M

$\therefore$ M = ZNpe

$\therefore Z=\frac{M}{p}\frac{1}{Ne}=\frac{E}{Ne}=\frac{E}{F}$

Where F = Ne = Faraday constant

or F = 6.0229 × 10²³ × 1.602 × 10^–19 = 96487C $\approx$ 96500C

Illustration 4: In producing chlorine through electrolysis 100 KW power at 125 volt is being consumed. If the E.C.E. of chlorine is 0.367 ´ 10^–6 kg/coul, then the mass of chlorine liberated per minute will be:

(A) 16.3 ´ 10^–4 kg (B) 17.61 ´ 10^–3 kg

(C) 18.2 ´ 10^–3 kg (D) 10^–4 kg

Solution: Mass of chlorine liberated $=Zit=Z\frac{P}{V}t$

$=\frac{0.367\times {10}^{-6}\times {10}^5\times 60}{125}=17.61\times {10}^{-3}$

Some important points

(i) Capacity of a cell- It is expressed in kilowatt hours which involves both current and the voltage.

(ii) Energy capacity- It is expressed in kilowatt hours which involves both current and the voltage.

(iii) Efficiency- (a) It is the ratio of discharging capacity to the charging capacity.

(b) If discharge takes place slowly, the maximum available energy is 80% of that spent in charging.

(c) If the cell is short circuited, the discharge takes place suddenly and the cell is spoiled.

Applications of electrolysis

The phenomenon of electrolysis has been used as a valuable tool in

(i) identifying the components of certain liquids

(ii) making accurate measurement of current

(iii) calibrating an ammeter

(iv) determining the electro-chemical equivalents of elements

(v) electroplating

(vi) producing pure metals

(vii) electrolying

Thermoelectricity-Seebeck effect

(i) Thermocoule: If two wires of different metals are joined at their ends so as to from two junctions, then the resulting arrangement is called a thermocouple.

(ii) Seebeck effect: (a) When the two junctions of a thermocouple are maintained at different temperatures, then a current starts flowing through the wires. This is called Seebeck effect and the e.m.f. developed in the circuit is called thermo-emf.

(b) The Seebeck effect is perfectly reversible, i.e. if the hot and the cold junctions are interchanged, the direction of current is reversed.

(c) The thermo-emf is of the order of a few micro-volts per degree temperature difference.

thermoelectric series

(i) Seebeck arranged a number of metals in the form of a series called thermoelectic series. The arrangement of some of the metals forming the series are-

Sb, Fe, Zn, Ag, Au, Mo, Cr, Sn, Pb, Hg, Mn, Cu, Co, Ni, Bi.

(ii) In the above series, current flows through the cold junction from the metal which appear earlier to the metal which appears later.

(iii) Greater the separation of the two metals in the series, greater is the thermo emf generated.

Variation of thermo-emf with temperature-difference

(i) If the cold junction is kept at 0^o and the temperature of hot junction (t) is gradually increased, it is found that the thermo emf e first increases, attains a maximum value and then decreases to become zero again (Fig.)

(ii) If the temperature is further increased, the emf changes direction.

(iii) Netural temperature (t_n)- The temperature at which thermo-emf is maximum is defined as neutral temperature. t_n is fixed for a given thermo-couple.

(iv) Temperature of inversion (t_i)- The temperature at which thermo-emf changes its direction is called the temperature of inversion. t_i is an much above t_n as the tenoeratyre if cikd hybctuib us bekiw t_n.

(v) Mathematically

$e=\alpha t+\frac{1}{2}\beta t^2$

where a and b are constants for a given thermo- couple.

Illustration 5: The e.m.f. of a thermocouple, one junction of which is kept at 0°C, is given by e = at + bt²

The neutral temperature will be:

(A) $\frac{a}{b}$ (B) $-\frac{a}{b}$

(C) $\frac{a}{2b}$ (D) $-\frac{a}{2b}$

Solution: At neutral temperature e = maximum

$\therefore \frac{de}{dt}=0$

or $a+2b{t}_n=0$

$\therefore t_n=-\frac{a}{2b}$.

Illustration 6: In the above problem, the temperature of inversion will be :

(A) $\frac{a}{b}$ (B) $-\frac{a}{b}$

(C) $\frac{a}{2b}$ (D) $-\frac{a}{2b}$

Solution: $\because t_n=\frac{t_o+t_i}{2}$

$\therefore t_i=2t_n-t_o$

or $t_i=2\times -\frac{a}{2b}-0=-\frac{a}{b}$

Seebeck coefficient

The rate of change of thermo emf with temperature is called thermo-electric power or Seebeck coefficient (S).

$\therefore X=\frac{de}{dt}=\alpha +\beta t$

When t = tn, e is maximum

$\therefore \frac{de}{dt}=0$

$\therefore \alpha +\beta t_n=0or{t}_n=-\frac{\alpha }{\beta }$

Peltier effect

(i) This effect is the converse of Seebeck effect.

(ii) If a current is passed through a junction of two dissimilar metals, heat is eigher absorbed or evolved at the junction.

(iii) On reversing the direction of current, the heating effect is also reversed. If the seebeck current is in a certain direction through the hot junction, then an external current sent in the same direction through this junction produces a cooling at this junction and a heating at the other junction.

(iv) Peltier coefficient ($\pi$)- It is the amount of heat absorbed or evolved at a junction per second when a current of one ampere flows through it.

$\pi =T\frac{de}{dT}$

Thomson effect

(i) An emf is developed between two parts of a single metal if they are at different temperatures. This is called Thomson effect.

(ii) Thomson coefficient- If de is the potential difference between two points in a metal which have a temperature difference dt, then the ratio

$\sigma =\frac{de}{dT}=T\frac{d^{2}e}{d{T}^2}$

is defined as the thomson coefficient.

(iii) If one part of a metal is at a higher temperature than the other, the free electrons at hot part will have more kinetic energy and as a result these electrons will diffuse faster towards colder part than the electrons from cold to the hot part. This would result in the net transfer of electrons setting an electric current in the metal.

Applications of thermoelectric effects

(i) Measurement of temperature

(ii) Detection of heat radiations

(iii) Refrigeration

(iv) Power generation.