9. KTG & Thermodynamics part-2

7/29/2019 9. KTG & Thermodynamics part-2

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Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 186

EXERCISE # 33.4 Heat lost in time t = Pt = ML

t le; esa "ek {kfr = Pt = ML

L =M

Pt

3.17 v =92

r2 . (r0 r

w) g integer

= 180 m/sec.

h =310

49

81

8.92

32400

m

3.21 For minimum value of m, the final temperature of the mixture must be 0C.

m ds U;wure eku ds fy;s feJ.k dk vfUre rki 0C gksuk pkfg;sA

20 2

1 10 + 20 80 = m 540 + m.1. 100

m =640

1700=

32

85gm.

3.22 T = 2g

when the temperature is raised length changes to (1 + T)when the lift accelerates upwards

geff

= g + a

new period of pendulum,

tc rkieku c


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3.24 100 0

=

0

60 101011

100

= 11 103

= 1.1 102

3.26 As steam has comparatively large amount of heat to provide in the form of latent heat we check what amount of

heat is required by the water and ice to go up to 100C, that is

(miL + m

iS

wT) + m

w. S

w. T

= [(200 80) + (200 1 100)] + (200 1 45)

= 45,000 cal.

That is given by m mass of steam, then

ms.L = 45,000

ms=

540

000,45=

6

500= 83.3 gm

therefore 83.3 gm steam converts into water of 100C.

Total water = 200 + 200 + 83.3 = 483.3 gm

steam left = 16.7 gm.pwafd Hkki ds ikl xqIr m"ek ds :i esa iznku djus ds fy, Hkkjh ek=kk esa ok"i gS] ge ;g Kkr djrs gSa fd cQZ rFkk ikuhdks 100C rd tkus gsrq fdruh m"ek pkfg;sA og gS -

(miL + m

iS

wT) + m

w. S

w. T

= [(200 80) + (200 1 100)] + (200 1 45)

= 45,000 cal.

;s m nzO;eku dh Hkki }kjk nh x;h gS] rcm

s.L = 45,000

ms=

540

000,45=

6

500= 83.3 gm

vr% 83.3 gm Hkki 100C ikuh esa cny tkrh gSA

dqy ikuh = 200 + 200 + 83.3 = 483.3 gmcph gqbZ Hkki = 16.7 gm.

3.27 Heat liberated when 300 g water at 25C goes to water at 0C :

Q = ms = (300) (1) (25)

= 7500 cal

From Q = mL, this much heat can melt mass of ice given by

m =L

Q=

80

7500

= 93.75 g

i.e., whole ice will not melt.Hence, the mixture will be at 0C

Mass of water in mixture = 300 + 93.75 = 393.75 g and

Mass of ice in mixture = 100 93.75 = 6.25 g

300 g ty dk rkieku 25C ls 0C rd ykus esa fudyh "ek :

4.1 Since s<

bthere will be less contraction in steel as compared to brass.

pwafd s<

b gS rks LVhy esa dkals dh rqyuk esa de ladqpu gksxkA

Q = ms Dq

= (300) (1) (25)

= 7500 cal

Q = mL ls , bruh "ek] cQZ dh ek=kk tks fi?kyk ldrh gS oks gksxh


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6.1 Gas thermometers are more sensitive than liquid thermometers because coefficient of thermal expansion of

gases are more then liquids.

xSl rkiekih] nzo rkiekih ls T;knk lqxzkgh gksrk gS D;ksafd xSlksa dk rkih; lkj xq.kkad nzoksa ls T;knk gksrk gSA

6.2 Coefficient of thermal expansion of gases are more than liquids

xSlksa ds rkih; izlkj xq.kkad dk eku nzoks a ls T;knk gksrk gSA

6.3 Conductivity of metal is more so it expands more as compared to liquid initially.

/kkrq dh pkydrk T;knk gS vr% ;s nzO; ds rwyuk esa T;knk izlkfjr gksrk gSA

EXERCISE # 41*. Let

0be the initial length of each strip before heating.

Length after heating will be

ekuk xeZ djus ls igys izR;sd iV~Vh dh yEckbZ 0gSA

xeZ djus ds ckn yEckbZ gksxh -lB

= l0

( 1 + BT) = (R + d ) and rFkk

lC

= l0

( 1 + CT) = R

T1

T1

R

dR

C

B

BC

R

d

T)(1R

d1 CB [ From binomial expansion ] [f}in izes; ls]

R =T)(

d

CB

or;kT

dR

and rFkk

CB1

2. The temperature of ice will f irst increase from 10C to 0C. Heat supplied in this process will be :

Q1

= mSi(10) m = mass of ice

Si= specific heat of ice

Then ice starts melting. Temperature during melt ing will remain constant (0C)

Heat supplied in the process will be

Q2

= mL L = Latent heat of melting

Now the temperature of water will increase from 0C to 100C. Heat supplied will be :

Q3

= mSw

(100)

Sw

= Specific heat of water

Finally water at 100C will be converted into steam t 100C and during this process temperature again

remains constant. Temperature versus heat supplied graph will be as follows :

Temp

Heat supplied

cQZ dk rkieku igys 10C ls 0C rd c


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bl izf;k esa nh xbZ "ek gksxh -Q

2= mL L = xyu dh xqIr "ek

vc ikuh dk rkieku 0C ls 100C rd c


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Now

Heat gained by the ice cube = Heat lost by the container.

i.e., (mass of ice) (latent heat of fusion of ice) + (mass of ice) (specific heat of water) (300 K 273K)

= m f

i

T

T

S.dT..

Substituting the values, we have

(0.1)(8 104) + (0.1) (103) (27) = m 300

500

)BTA( dT or 10700 = m

300

500

2

2

BTAT

After substituting the values of A and B and the proper limits, we get

m = 0.495 Kg. Ans.

ekuk m crZu dk nzO;eku gScrZu dk izkjfEHkd rkieku T

i= (227 + 273) = 500 K

rFkk crZu dk vfUre rkieku Tf= (27 + 273) = 300 K

vccQZ ds ?ku }kjk izkIr dh xbZ "ek = crZu }kjk [kksbZ xbZ "ek

i.e., (cQZ dk nzO;eku) (xyu dh xqIr "ek) + (cQZ dk nzO;eku) (ty dh fof'k"V "ek) (300 K 273K) = m f

i

T

T

S.dT..

ekuksa dks izfrLFkkfir djus ij gesa izkIr gksrk gS

(0.1)(8 104) + (0.1) (103) (27) = m 300

500

)BTA( dT

;k 10700 = m

300

500

2

2

BTAT

A o B ds ekuksa dks izfrLFkkfir djus ij rFkk mi;qZDr lhekvksa dk mi;ksx djus ijm = 0.495 Kg. Ans.

5. Given fn;k gS 1

= 2

or;k 1

at =

2

st

a

s

2

1

or;ksa

s

21

1

6. Heat released by 5 kg of water when its temperature falls from 20C to 0C is,

Q1= ms = (5) (103) (20 0) = 105 cal

when 2 kg ice at 20C comes to a temperature of 0C, it takes an energy

Q2= ms = (2) (500) (20) = 0.2 105 cal

The remaining heat

Q = Q1 Q

2= 0.8 105 cal will melt mass m of the ice, where

m = 3

5

1080

108.0

= 1 kg

So, the temperature of the mixture will be 0C,

mass of water in it is 5 + 1 = 6 kg and mass of ice is 2 1 = 1 kg.5 kg ikuh }kjk eqDr m"ek tc bldk rkieku 20C ls 0C rd fxj tkrk gS ,


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Q1= ms = (5) (103) (20 0) = 105 cal

tc 2 kg cQZ 20C ls 0C ds rkieku rd vkrh gS, rks ;s tkZ ysrh gSAQ

2= ms = (2) (500) (20) = 0.2 105 cal

cph gqbZ m"ek Q = Q1

Q2

= 0.8 105 cal nzO;eku m dk cQZ fi?ky tk;sxk

m = 3

5

1080

108.0

= 1 kg

vr% feJ.k dk rkieku 0C gksxkAblesa ty dk nzO;eku gS 5 + 1 = 6 kg rFkk cQZ dk nzO;eku 2 1 = 1 kg.

7. When the temperature is increased, volume of the cube will increase while density of liquid will decrease. The

depth upto which the cube is submerged in the liquid remains the same, hence the upthrust will not change.

F = F

Vi

Lg = V

i

Lg (V

i= volume immersed)

(Ahi) (

L) (g) = (1 + 2

sT) (Ah

i) g

T1 L

L

Solving this equation, we get L= 2

s.

tc rki c


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10. S = 2100 J kg1 C1

L = 3.36 105 J kg1

420 = m S + (1) 103 L

420 = m S (5) + 3.36 10

2

420 336 = m(2100) 5

m =125

1 1000 = 8 gm.

11.A

mg= Y m =

g

AY = 4kg.

U =3

2

10AY2

LF J

12. = WYA

The graph is straight line passing through origin the slope of which isYA

.

vr% xzkQ ,d ljy js[kk gS] tks n'kkZ;s vuqlkj ewyfcUnq ls xqtjrh gS rFkk ftldk


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14. n = mk

=m

/yA =

m

yA

1.01

)109.4()10n( 79

= 140 n = 4.

15. In equilibrium,

lkE;koLFkk esamg = qE

In absence of electric field,

fo|qr {ks=k dh vuqifLFkfr esa

mg = 6rv

qE = 6qrv

m =3

4Rr3d. =

g

qE

dv6

qE

3

4 3

=g

qE

After substituting value we get,

eku frLFkkfir djus ij

q = 8 1019 C Ans.

16.L

L T

L = L T

mg = K L

m =YA

Lg L

11 6 510 10 10 10m

10

40 C 30 C

m = 3.14 3

PART - II

1. As we know that thermal capacity of a substance is defined as the amount of heat required to raise its tempera-

ture by 1C.

tSlk fd ge tkurs gS inkFkZ dh "ek /kkfjrk bl izdkj ifjHkkf"kr gS fd 1C rkieku c


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3. T = 2g

l

2

1d

2

1

T

dT

l

l 21 = 10.5%

4. Let time taken in boiling the water by the heater is t sec. Then

ekuk fd ghVj }kjk ikuh dks xeZ djus esa t lsd.M yxrk gSA rcQ = msT

2.4

836t = 1 1000 (40 10)

2.4

836t = 1000 30

t =836

2.4301000 = 150 sec

5. U = M s = 0.1 4184 20 = 8368 = 8.4 kJ

6. Elastic energy stored in the wire is rkj esa lafpr izR;kLFk fLFkfrt tkZ

U =2

1stress strain volume

U =2

1izfrcy fod fr vk;ru

=2

1

LA

F l AL

=2

1Fl

= 2

1

200 1 103

= 0.1J

7. Retarding force acting on a ball fall ing into a viscous fluid

';ku nzo esa fxj jgh xsan ij dk;Zjr eUnu cyF = 6Rvwhere R = radius of ball,

tgk R = xsan dh f=kT;kv = velocity of ball, xsan dk osxand rFkk = coefficient of viscosity ';kurk xq.kkad F R and rFkk F v

8. Youngs modulus =Strain

Strees

;ax xq.kkad = fod frizfrcy

Strain fodfr =Y

S

2

1

Volume

wireinstoredEnergy Stress Strain

2

1

vk;ru

tkZlaxzfgrrkj esa izfrcy fod fr

= 2

1S Y2

S

Y

S 2


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9. Terminal speed of spherical body in a viscous liquid is given by

';ku nzo esa xksykdkj oLrq dh lhekUr pky fuEu izdkj nh tkrh gSA

vT=

9

g)(r2 2

where = density of substance of body,tgk = oLrq ds inkFkZ dk ?kuRo

= density of liquid. nzo dk ?kuRoFrom given data fn;s x;s vkadMksa ls

lGold

lAg

T

T

)Gold(v

)Ag(v

vT(Ag) = 2.0

18

92.0

5.15.19

5.15.10

= 0.1 m/s

10. Let us consider the length of wire as L and cross-sectional area A, the material of wire has Young's modulus asY.

ekuk rkj dh yEckbZ LrFkk vuqizLFk dkV {ks=kQy A gS rFkk rkj ds inkFkZ dk ;ax xq.kkad Y gSA

Then for 1st case Y =L/

A/W

rc izFke fLFkfr ds fy, Y = L/A/W

For 2nd case, Y =L/'2

A/W

f}rh; fLFkfr ds fy, Y =L/'2

A/W

l' = l /2So, total elongation of both sides = 2' =

vr% nksuksa vksj dqy foLrkj = 2' =

11. The forces acting on the ball are gravity force, buoyancy force and viscous force. When ball acquires terminalspeed, it is in dynamic equilibrium, let terminal speed of ball is v

T. So,

xsan esa dk;Zjr cy xq:Ro cy, mRIykou cy rFkk ';ku cy gS A tc xsan lhekUr pky izkIr dj ysrh gS rc ;g xfrd lkE;oLFkkesa gksrh gSA ekuk xsan dh lhekUr pky V

TgSA vr%

V2g + kv

T2 = V

1g v

T=

k

g)(V 21

12. F =

xAY

and rFkk F2=

)3/(

x)A3(Y

= 9 F

9. KTG & Thermodynamics part-2

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