Thermal and Chemical Effect

Chapter # 33

Thermal and Chemical Effects of Electric Current

SOLVED EXAMPLES

Example 33.1 Find the heat developed in each of the three resistors shown in figure in 1 minute.

6

3 9V Sol.

1

The equivalent resistance of 6 and 3 resistance is (6 )  (3 ) = 2 . 6  3  This connected in series with the 1 resistor. The equivalent resistance of the circuit is R = 2 + 1 = 3. The current through the battery is 9V = 3 A. 3 The current through the 1 resistor is , therefore, 3 A. The heat developed in this resistor is H = i 2 Rt = ( 3A ) 2 × ( 1 ) × ( 60 s ) = 540 J. The current through the 6 resistor is

i=

( 3A ) ×

3 = 1 A. 6  3

The heat developed in it = ( 1A ) 2 × ( 6 ) × ( 60 s ) = 360 J. The current through the 3 resistor is 3 A – 1 A = 2 A. The heat developed in it = ( 2A )2 ( 3 ) × ( 60 s ) = 720 J. Example 33.2 The cold junction of a thermocouple is maintained at 100 C. No thermo–emf is developed when the hot junction is maintained at 5300 C. Find the neutral temperature. Sol. Clearly, 5300 C is the inversion temperature i of the couple. If n be the netural temperature and c be the temperature of the cold junction, i – n = n – c n =

or ,

i - c 530 0 C - 10 0 C = = 2700 C. 2 2

Example 33.3 Uses

Copper 2.76 0.012 Iron 16.6 – 0.030 Using table, find a and b coefficients for a copper–iron thermocouple. Sol.

aCu , Fe = aCu , Pb – aFe , Pb = 2.76 V / 0C – 16.6 V / 0C = – 13.8 V / 0C. bCu , Fe = bCu , Pb – bFe , Pb = 0.012 V / (0C)2 + 0.030 V / (0C)2 = 0.042 V / (0C)2 .

QUESTIONS

FOR

SHORT

ANSWER

1.

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

2.

Two unequal resistances R1 and R2 are connected across two identical batteries of emf  and internal resistance r (figure 33-Q1). Can the thermal energies developed in R1 and R2 be equal in a given time. If yes, what will be condition? manishkumarphysics.in

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Chapter # 33


3.

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

4.

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

5.

Do all the thermocouples have a neutral temperature?

6.

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

7.

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

8.

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

9.

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

Objective - I 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 ?

fuEu esa ls dkSulk oØ fdlh fuf'pr le; esa fo|qr /kkjk ds Qyu ds :i esa fdlh izfrjks/k esa mRiUu Å"ek ÅtkZ O;Dr djrk gS -

(A*) 1 2.

(C) 3

(D) 4

A constant current i is passed through a resistor. Taking the temperature coefficient of resistance into account, indicate which of the plots shown in fig. best represents the rate of production of thermal energy in the resistor. fdlh izfrjks/k ls fu;r /kkjk i izokfgr gksrh gSA rki izfrjks/k xq.kkad dks izHkkoh ekurs gq, fuEu fp=k esa dkSulk oØ fdlh izfrjks/k esa m"ek ÅtkZ mRiUu gksus dh nj dks vf/kd lgh iznf'kZr djrk gS -

(A) 1 3.

(B) 2

(B) 2

(C) 3

(D*) 4

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 worng (C) b is correct but a is wrong (D) Both a and b are wrong manishkumarphysics.in

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Chapter # 33


rki&oS|qr ;qXe ds fy;s fuEu dFkuksa ij fopkj dhft;s (a) mnklhu rki B.Mh laf/kds rki ij fuHkZj ugha djrk gSA (b) A lR; gS rFkk B vlR; gSA (A) a o b nksuksa lR; gSA (B*) a lR; gS rFkk b vlR; gSA (C) b lR; gS rFkk a vlR; gSA (D) a o b nksuksa vlR; gSA 4.

The heat developed in a system is propertional 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 fdlh fudk; esa mRiUu m"ek blls izokfgr /kkjk ds lekuqikrh gS (A) ;g FkkWelu m"ek ugha gks ldrh gSA (B) ;g isYVh;j m"ek ugha gks ldrh gSA (C*) ;g twy m"ek ugha gks ldrh gSA (D) ;g mDr rhuksa m"ekvksa esa ls dksbZ Hkh ugha gks ldrh gSA

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. 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 fuEufyf[kr nks dFkuksa ij fopkj dhft;s (a) fHkUu&fHkUu /kkrqvksa esa eqDr bysDVªkuW ?kuRo fHkUu&fHkUu gksrk gSA (b) /kkrq esa eqDr bysDVªkWu ?kuRo rki ij fuHkZj djrk gSA lhcsd izHkko mRiUu gksrk (A*) a o b nksuksa ds dkj.k (B) a ds dkj.k] fdUrq b ds dkj.k ugha (C) b ds dkj.k] fdUrq a ds dkj.k ugha (D) u rks a ds dkj.k u gh b ds dkj.k

gS %

6.

Consider the statement 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 fiNys iz'u esa fn;s x;s dFkuksa A rFkk B ij fopkj dhft;sA isYVh;j izHkko mRiUu gksrk gS (A) a o b nksuksa ds dkj.k (B*) a ds dkj.k] fdUrq B ds dkj.k ugha (C) b ds dkj.k] fdUrq a ds dkj.k ugha (D) u rks a ds dkj.k u gh b ds dkj.k

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 iz'u la[;k 5 esa fn;s x;s dFkuksa a o b ij fopkj dhft;sA Fkkelu izHkko mRiUu gksrk (A) a o b nksuksa ds dkj.k (B) a ds dkj.k] fdUrq b ds dkj.k ugha (C*) u rks a ds dkj.k u gh b ds dkj.k (D) u rks a ds dkj.k u gh b ds dkj.k

8.

gS -

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. QSjkMs fu;rkad : (A) fo|qr vi?kV~; dh ek=kk ij fuHkZj djrk gSA (B) fo|qr vi?kV~; ls izokfgr /kkjk ij fuHkZj djrk gSA (C*) ,d lkoZf=kd fu;rkad gSA (D) fo|qr vi?kV~; ls izokfgr vkos'k dh ek=kk ij fuHkZj djrk gSA

Objective - II

1.

Two resistors having equal resistance are joined in series and current is passed through the combination. Negect any variation is 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 tepmerature may rise equally in the resistors

nks ,d leku izfrjks/k Js.khØe esa tksM+s x;s gSa rFkk bl la;kstu ls /kkjk izokfgr dh tk jgh gSA rki ifjorZu ds dkj.k manishkumarphysics.in

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Chapter # 33


izfrjks/k esa ifjorZu ux.; eku yhft;sA fdlh fuf'pr le;karjky esa (A*) izfrjks/k ij leku ek=kk esa m"ek ÅtkZ mRiUu gksxhA (B) izfrjks/kksa ij vleku ek=kk esa m"ek ÅtkZ mRiUu gksxhA (C) izfrjks/kksa esa fuf'pr :i ls leku rki o`f) gksxhA (D*) izfrjks/kksa esa rki o`f) leku gks ldrh gSA 2.

A copper strip AB and an iron strip AC are joined at A. The junction A is maintained at 0oC and the free ends B and C are maintained at 100oC. There is potential different 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 rkacs dh iV~Vh AB rFkk yksgs dh iV~Vh AC, A ij tqM+h gqbZ gSA laf/k A dk rki 0oC rFkk eqDr fljksa B o C dk rki 100oC fLFkj j[kk x;k gSA foHkokUrj gksxk (A*) rkacs dh iV~Vh ds nksuksa fljksa ds chp (B*) laf/k ij rkacs ds fljs rFkk yksgs ds fljs ds chp (C*) yksgs dh iV~Vh ds nksuksa fljksa ds chp (D*) eqDr fljksa B o C ds chp

3.

The constance a and b for the pair silver-lead are 2.50 V/oC and 0.012 V/o(C)2 respectively. For a silver-lead thermocouple with colder junction at 0oC, (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 no current in the thermocouple even if the junction are kept at different temperature. pkanh rFkk lhls ds ;qXe ds fy;s fu;rkad a rFkk b ds eku Øe'k% 2.50 V/oC rFkk 0.012 V/o(C)2 gSA pkanh&lhlk rki&oS|rq ;qXe ftldh B.Mh laf/k dk rki 0oC gS] ds fy;s (A*) dksbZ mnklhu rki ugha gksxkA (B*) dksbZ O;qRØe.k rki ugha gksxkA (C) laf/k;ksa dks fHkUu&fHkUu rkiksa ij j[kus ij Hkh rki fo-ok-cy mRiUu ugha gksxkA (D) laf/k;ksa dks fHkUu&fHkUu rkiksa ij j[kus ij Hkh rki oS|r q ;qXe ls dksbZ /kkjk izokfgr ugha gksxhA

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 increased. (C*) The rate of liberation of material will remain the same (D) Heat will be produce at a greater rate ,d fo|qr&vi?kVu iz;ksx dks jksddj cSVjh ds VfeZuy iyV fn;s x;s gSa (A) fo|qr&vi?kVu :d tk;sxkA (B) bysDVªkM s ksa ij inkFkZ vo{ksfir gksus dh nj esa o`f) gks tk;sxhA (C*) inkFkZ vo{ksfir gksus dh nj ogh jgsxhA (D) m"ek mRiUu gksus dh nj esa o`f) gksxhA

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. fdlh inkFkZ dk oS|rq &jklk;fud rqY;kad fuHkZj djrk gS (A*) inkFkZ dh izÑfr ij (B) inkFkZ dks j[kus okys fo|qr vi?kV~; dh izÑfr ij (C) fo|qr vi?kV~; ls izokfgr vkos'k ij (D) fo|qr vi?kV~; esa mifLFkr inkFkZ dh ek=kk ij

WORKED OUT EXAMPLES 1. Sol.

A current of 30 A is registered when the terminals of a drya cell of emf 1.5 volts are connectd through an ammeter. Neglecting the meter resistance, find the amount of heat produced in the battery in 10 seconds. The current in the circuit will be i=

 r manishkumarphysics.in

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Chapter # 33


1.5 V giving r = 0.05 . r The amount of heat produced in the battery = i2 rt = (30 A)2 × (0.05 ) × 10 s = 450 J. or,

2.

Sol.

30 A =

A room heater is rated 500 W, 220 V. (a) Find the resistance of its coil. (b) If the supply votage drops to 200 V, what will be the power consumed? (c) If an electric bulb rated 100 W, 220 V is connected in series with this heater, what will be the power consumed by the heater and by the bulb when the supply is at 220 V? (a) The power consumed by a coil of resistance R when connected across a supply V is

V2 . R The resistance of the heater coil is, therefore P=

R=

(220 V )2 = 96.8 . 500 W

(b) The reasistance of the 100 W, 220 V, the power consumed will be

(220 V )2 V2 = = 413 W.. 96.8  R (c) The resistance of the 100 W, 220 V bulb is P=

(220 V )2 R= = 484 . 100 W If this is connected in series with the heater of 96.8 , the current i will be i=

220 V 2 = 0.379 A. 484   96.8 

Thus, the poer consumed by the heater = i2 × 96.8  = 0.144 × 96.8 W = 13.9 W = i2 × 484  = 69.7 W. 3. Sol.

A battery of emf  and internal resistance r is used in a circuit with a variable external resistance R. Find the value of R for which the power consumed in R is maximum. The current in the resitance R is

 r R The power consumed in R is i

P = i2R =

 2R (r  R ) 2

2 dP 2  (r  R)  2R(r  R)   . It is maximum when  (r  R) dR  

It is zero when (r + R)2 = 2R(r + R) or, R = r. 4.

The junctions of a a Ni–Cu thermocopule are maintained at 0ºC. Calcualte the Seebeck emf produced in the loop aNI, Cu = 16.3 × 10–8 V/ºC and bNi, Cu = –0.042 × 10–6 V/(ºC)–2.

Solution :

 Ni,Cu  aNi,Cu  

= (16.3 × 10–6 × 100)V +

1 bNi,Cu  2 2 1 (–0.042 × 10–6 × 10–4) N 2

= 1.42 × 10–3 V.

manishkumarphysics.in

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Chapter # 33 Thermal and Chemical Effects of Electric Current 5. Find the neutral and inversion temperatures for Ni – Cu thermodynamics with the cold junction at 0ºC. Use data from previous example. Solution : The neutral temperature is n = – =

a b

16.3  10 6

ºC = 388ºC. 0.042  10 6 The inversion temprature is double the neutral temperature, i.e., 776ºC. 6.

An electric current of 0.4 A is passed through a silver voltameter for half an hour. Find the amount of silver deposited on the cathode. ECE of silver = 1.12 × 10–6 kg/C. Solution : Using the formula m = Zit, the mass of silver deposted = (1.12 × 10–6 kg/C) (0.4 A) (30 × 60 s) = 8.06 × 10–4 kg = 0.806 g. 7.

A silver and a copper voltameter are connected in sereis with a 12.0 V battery of negligible resistance. It is found that 0.806 g of silver is deposited in half an hour. Find (a) the mass of the copper deposited and (b) the energy supplied by the battery. EOE of silver = 1.12 × 10–6 kg/C and that of copper = 6.6 × 10–7 kg/C. Solution : (a) For silver voltameter, the formula m = Zit gives 0.806 g = (1.12 × 10–6 kg/C) i(30 × 60 s) or, i = 0.4 A. As the two voltmeters are connected in seris, the same current passes through the copper voltameter. The mass of copper deposited is m = Zit = (6.6 × 10–7 kg/C) i(30 × 60 s) or, i = 0.4 A.

m1 z1 This could also be obtained by using m  Z for series circuit. 2 2 (b) Energy supplied by the battery = Vit = (12 V) (0.4 A) (30 × 60 s) = 8.64 kJ. 8.

A current of 1 A is passed through a dilute solution of sulphuric acid for some time to liberate 1 g of oxygen. How long was the current passed ? Faraday constant = 96500 C/mole. Solution : The relative atomic mass of oxygen = 16 and its valency = 2 so that the chemical equivalent E =

16 = 8. 2

Chemical equivalent of hydrogen = 1

E oxygen Ehydrogen or,

mhydrogen =

moxygen 8

We have, 1 g of oxygen =

E oxygen 8 = E = 1 hydrogen =

1g = 0.125 g. 8

1 gram-equivalent. 8

The charge needed to liberate

1 gram equivalent 8

=

1 faraday 8

96500 C = 1.12 × 104 C. 8 As the current is 1 A, the time taken is

=

1.2  10 4 C Q t= = 1A i = 1.2 × 104 s = 3 hours 20 minutes. manishkumarphysics.in

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Chapter # 33

1.

Ans. 2.

Ans. 3.

Ans. 4.

Ans. 5.

Ans. 6.

Ans. 7.


EXERCISE

An electric current of 2.0 A passes through a wire of resistance 25 . How much heat will be developed in 1 minute? 25  izfrjks/k okys rkj ls 2.0 A fo|qr /kkjk izokfgr dh tk jgh gSA ,d fefuV esa fdruh m"ek mRiUu gksxh\ 6.0 × 103 J A coil of resistance 100  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 long will it take to raise the temperature of the coil by 15ºC? 6.0 V fo-ok-cy okyh cSVjh ls 100  izfrjks/k dh dq.Myh tksM+h x;h gSA ;g eku yhft;s fd dq.Myh esa mRiUu lkjh m"ek dq.Myh dk rki c<+kus esa iz;D q r gksrh gSA ;fn dq.Myh dh m"ek /kkfjrk 4.0 J/K gS] blds rki esa 15ºC dh o`f) gksus esa fdruk le; yxsxk ? 2.8 minute The specificationon 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? ,d ghVj dq.Myh 250 V, 500 W dh gSA dq.Myh ds izfrjks/k dh x.kuk dhft;sA leku oksYVrk ij iz;qDr dh tkus okyh 1000 W dh dq.Myh dk izfrjks/k fdruk gksxk\ 125 , 62.5  A heater coil is to be constructed with a nichrome wire ( = 1.0 × 10–6 -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? ,d ghVj dq.Myh ukbØkse rkj ( = 1.0 × 10–6 -m) ls cuk;h x;h gS] bldks 250 V ds lzkrs ls tksM+us ij 500 W 'kfDr mRiUu gksrh gSA (a) dq.Myh dk izfrjks/k fdruk gS\ (b) ;fn rkj dk vuqiLz Fk dkV {ks=kQy 0.5 feeh2 gS] rks fdrus yEcs rkj dh vko';drk gksxh\ (c) ;fn izR;sd Qsjs dh f=kT;k 4.0 feeh- gS] dq.Myh esa Qsjksa dh dqy la[;k fdruh gS\ (a) 125  (b) 62.5 m (c) = 2500 turns A bulb with rathing 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 mm2. How much power will be consumed by the connecting wires? Resistivity of copper = 1.7 × 10–8 -m. 250 V, 100 W dk ,d cYc 5 mm2 vuqizLFk dkVj okys rkacs ds yEcs rkj dh lgk;rk ls 10 eh- nwj fLFkr lzkr s ls tksM+k –8 x;k gSA la;kstu rkjksa }kjk fdruh 'kfDr dke esa yh tk;sxhA rkacs dh izfrjks/kdrk = 1.7 × 10 -m. 8.4 mW 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? It the supply is suddenly increased to 240 V, what will be the power consumed? ,d fo|qr cYc dks 220 oksYV ds lzkrs ls tksMu+ s ij ;g 60 W 'kfDr xzg.k djrk gSA ;fn lzkrs oksYVrk 180 oksYV rd de gks tk;s] fdruh 'kfDr iz;D q r gksxh\ ;fn lzksr oksYVrk vpkud c<+dj 240 oksYV gks tk;s] rks fdruh 'kfDr [kpZ gksxh\ 40 W, 71 W 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 ? ,d ^^loksZ oksYVst LVscykbtj** dh fuxZr oksYVrk 220 V ± 1% gSA ;fn ,d 220 V, 100 W dk cYc blls tksM+k tk;s rks

blds }kjk mi;ksx dh xbZ vf/kdre ,oa U;wure 'kfDr fdruh gksxh\ Ans.

98 W, 102 W

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? ,d fo|qr cYc ij 220 V, 100 W vafdr gS] ;fn ;g 150 W ;k vf/kd ij iz;D q r gksxk rks ¶;wt gks tk;sxkA cYc oksYVrk

esa fdruk mrkj p<+ko lgu dj ldrk gSA Ans.

up to 270 V

9.

An immersion heater rated 1000 W, 220 V is used to beat 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 15ºC to 40ºC? 1000 W, 220 V dh be'kZu jkWM] 0.01 eh3 ikuh dks xeZ djus ds fy;s iz;D q r dh tk jgh gSA ;g ekurs gq, fd lzkrs oksYVrk 220 V gS rFkk nh xbZ 'kfDr dk 60% gh ty dks xeZ djus esa iz;D q r gksrk gS] ikuh dk rki 15ºC ls 40ºC rd c<+kus esa manishkumarphysics.in Page # 7

Chapter # 33 Ans. 10.

Ans. 11.


fdruk le; yxsxk? 29 minutes 20 feuV An electric kettle used to prepare tea, takes 2 minutes to boil 4 cups of water (1 up contains 200 cc of water) if the room temperature is 25ºC. (a) If the cost of power consumptionis Rs. 1.00 per unit = 1000 watt-hour), calculate the cost to boiling 4 cups of water. (b) What will be the corresponding cost if the room temperature drops to 5ºC? ,d fo|+rq dsVyh dk mi;ksx pk; cukus esa fd;k tk jgk gSA ;fn dejs dk rki 25ºC gS rks ;g 4 di ikuh ¼1 di esa 200V ?ku lseh ikuh½ dks 2 fefuV esa mcky nsrh gSA ;fn dejs dk rki (a) ;fn fo|qr 'kfDr [kpZ 1.00 :i;k izfr ;wfuV gks (1 ;wfuV = 1000 okWV&?kaVk), 4 de ikuh mckyus esa O;f;r jkf'k dh x.kuk dhft;sA (b) ;fn dejs dk rki 5ºC de gks tk;s rks O;; fdruk gksxk ? (a) 7 paisa iSlk (b) 9 paisa iSlk 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 intesity at a point if the supply voltage cahnges form 220 V to 200 V. ,d fo|qr cYc peduk izkjEHk djus ds fy;s 40 okWV ls vf/kd iznku dh tk;s] 60% vfrfjDr 'kfDr izdk'k esa rFkk 'ks"k m"ek esa ifjofrZr gks tkrh gSA cYc 200 V ij 100 okWV O;; djrk gSA ;fn lzksr oksYVrk 220 V ls 200 V jg tk;s rks izdk'k

dh rhozrk esa izfr'kr deh Kkr dhft;sA Ans.

29%

12.

The 2.0  resistor shown in figure (33 E1) is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000 J/K. (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  resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes? fp=k esa iznf'kZr 2.0  izfrjks/k] ikuh ls Hkjs gq, dSyksjh ehVj esa Mqcks;k x;k gSA dSyksjh ehVj dh ikuh lfgr m"ek /kkfjrk 2000 J/K gSA (a) ;fn ifjiFk 15 fefuV rd fØ;k'khy jgrk gS] ikuh ds rki esa fdruh o`f) gksxh\ (b) ;fn 6.0  izfrjks/k tydj u"V gks tkrk gSA vxys 15 fefuV esa ikuh ds rki esa fdruh o`f) gksxh\

Ans.

(a) 2.9 ºC (b) 3.6ºC

13.

The temperatures of the junctions of a bismuth-silver thermocopule are maintained at 0ºC and 0.001ºC. Find the thermo-emf (Seeback emf) developed. For bismuthsilver, a = –46 × 10–6 V/degand b = – 0.48 × 10–6V/ deg2. fcLeFk pkanh rki&oS|rq ;qXe dh laf/k;ksa dk rki 0ºC rFkk 0.001ºC j[kk tkrk gSA mRiUu rki fo-ok-cy ¼lhcsd fo-ok-cy½ Kkr dhft;sA fcLeFk flYoj ds fy;s a = –46 × 10–6 oksYV/fMxzh rFkk b = – 0.48 × 10–6 oksYV/fMxzh2 – 4.6 × 10–6 V

Ans. 14.

Find the thermo-emf developed in a copper-silver thermocouple when the junctions are kept at 0ºC and 40ºC. Use the data in table (33.1). rkack] pkanh rki oS|rq ;qXe dh laf/k;ksa dk rki 0ºC rFkk 40ºC gS] blesa mRiUu rki fo-ok-cy Kkr dhft;sA ¼vkadM+s lkj.kh

ls izkIr dhft;s½ Ans.

1.04 × 10–6 V

15.

Find the neutral temperature and inversion temperature of copper-iron thermocople if the reference juction is kept at 0ºC. Use the dat in table (333.1). rkack yksgk rki oS|rq ;qXe dh lanHkZ laf/k dk rki 0ºC gS] bldk mnklhu rki rFkk O;qRØe.k rki Kkr dhft;sA ¼vkadM+ks lkj.kh

ls izkIr dhft;s½ Ans.

330ºC, 659 ºC

16.

Find the charge required to flow through an electrolyte to liberate one atom of (a) a monovalent material and (b) a divalent material. manishkumarphysics.in

Page # 8

Chapter # 33


,d ijek.kq eqDr djus ds fy;s fdlh fo|qr vi?kV~; ls izokfgr vkos'k Kkr dhft;sA (a) ,d la;kstd inkFkZ dk vkSj (b) f} la;kstd inkFkZ dk Ans.

(a) 1.6 × 10–19 C (b) 3.2 × 10–19 C

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/mole. ;fn AgNO3 fo|qr vi?kV~; ls 0.500 A /kkjk@?kaVs rd izokfgr dh tkrh gS] rks dSFkksM ij fu{ksfir pkanh dh ek=kk Kkr dhft;sA pkanh dk ijek.kq Hkkj 107.9 xzke@eksy gSA 2.01 g

Ans. 18.

Ans. 19.

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. ,d bysDVªkIs ysfVax ;wfuV esa 30 fefuV esa ihry dh IysV ij 3.0 xzke pkanh fu{ksfir gksrh gSA ;wfuV esa iz;D q r dh xbZ /kkjk Kkr dhft;sA pkanh dk fo|qr jklk;fud rqY;kad 1.12 × 10–6 fdxzk/dwykWe gSA 15 A Find the time required to liberate 1.0 litre of hydrogen at STP in an electrolytic cell by a current of 5.0 A. ekud rki o nkc (STP) ij fo|qr vi?kVuh lsy esa 5.0 A /kkjk izokfgr djds 1.0 yhVj gkbMªkstu eqDr djus ds fy,

vko';d le; Kkr dhft;sA Ans.

29 minutes

20.

Two voltameters, one having a solution of silver salt and the other of a trivalent-metal salt, are connceted in series and a current of 2A is maintained for 1.50 hours. It is found that 1.00 g of the trivalent-metal salt, are connected in series and a current of 2A is maintained-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/mole.

Js.khØe esa la;ksftr nks oksYVehVj gS] ,d pkanh ds yo.k dk foy;u rFkk nwljs esa f=kla;kstd /kkrq ds yo.k dk ?kksy gS] buesa 1.50 ?kaVs rd 2A izokfgr dh tkrh gSA ;g izfs {kr fd;k tkrk gS fd f=kla;kstd /kkrq dh 1.00 xzke ek=kk fu{ksfir gksrh gSA (A) f=kla;kstud /kkrq dk ijek.kq Hkkj fdruk gS\ (B) bl le;karj esa fdruh pkanh fu{ksfir gqbZ\ pkanh dk ijek.kq Hkkj 107.9 xzke@eksyA Ans.

(a) 26.8 g/mole (b) 12.1 g

21.

A brass plate having surface area 200 cm2 on one side is electroplated with 0.10 mm thick a 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. ,d ihry dh ifV~Vdk dk ,d vksj dk {ks=kQy 200 lseh2 gS] bl ij bysDVªkIs ysfVax }kjk 15A /kkjk izokfgr djds nksuksa vksj 0.10 feeh eksVh pkanh dh ijr p<+k;h tkrh gSA bl dk;Z esa yxk le; Kkr dhft;sA pkanh dk vkisf{kd ?kuRo 10.5 ,oa ijek.kq Hkkj 107.9 xzke@eksy gSA 42 minutes

Ans. 22.

Figure (33-E2) 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  resistor during this period. Atomic weight of silver is 107.9 g/mole. fp=k esa AgCl fo|qr vi?kV~; iznf'kZr fd;k x;k gS] ftlls /kkjk izokfgr gks jgh gSA ;g izfs {kr fd;k x;k gS fd dSFkksM ij 10 fefuV esa 2.68 xzke pkanh tek gksrh gSA bl dky esa 20  izfrjks/k ij mRiUu m"ek Kkr dhft;sA pkanh dk ijek.kq Hkkj 107.9 xzke@eksy

Ans.

190 kJ


Page # 9

Chapter # 33 Thermal and Chemical Effects of Electric Current 23. The potential difference across the terminals of a battery of emf 12 V and internal resistance 2  drops to 10 V when it is connected to a silver voltameter. Find the silver deposited at the cathod in half an hour. Atomic weight of silver is 107.9 g/mole. ,d cSVjh dk fo-ok-cy 12 V rFkk vkarfjd izfrjks/k 2  gS] tc bldks flYoj oksYVkehVj ls tksM+k tkrk gS rks bldh VfeZuy oksYVrk 10 V jg tkrh gSA dSFkksM ij vk/ks ?kaVs eas fu{ksfir pkanh dh ek=kk Kkr dhft;sA pkanh dk ijek.kq Hkkj 107.9 xzke@eksy Ans. 2g 24.

Ans.

A plate of area 10 cm2 is to be electroplated with copper (density 9000 kg/m3) 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 and specific heat capacity of water = 4200 J/kg-K. ,d IysV dk {ks=kQy 10 lseh2 gS] 12 V dk lsy iz;D q r djds bysDVªkIs ysfVax }kjk blds nksuksa vksj 10 ekbØks ehVj eksVh rkacs 3 ¼?kuRo 9000 fdxzk/eh ½ dh ijr p<+kuh gSA fu{ksiu dh ,d izfØ;k esa O;f;r ÅtkZ dh x.kuk dhft;sA ;fn bl ÅtkZ dk mi;ksx 100 xzke ikuh dks xeZ djus ds fy;s fd;k tk;s] rks ikuh ds rki esa o`f) dh x.kuk dhft;sA rkacs dk ECE = 3 × 10–7 fdxzk/dwykWe rFkk ikuh dh fof'k"V m"ek = 4200 twy/fdxzk-K 7.2 kJ, 17 K


Page # 10

Thermal and Chemical Effect

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