UNIT # 05

THERMAL PHYSICS
EXERCISE –I 8. Let at temperature , volume increases by 2%
1. Let  = temperature on X–scale corresponding to then according to question (ekuk rki  ij vk;ru o`f¼
50°C on Y–scale 2%gSrksiz'ukuqlkj)
ekuk  = Y iSekus ij 50°C ds laxr X iSekus ij rki gS 100 = 98 [1+3.3 × 10 (–4)]
  = 60.4 + 4 = 64.4 °C
375°C -30°C
FL
9. L=Lthermal– Lcontact force= 0  1L= AY (rod – 1)
 1
X 50°C
Y
1 2

–125°C –70°C
FL
X  ( 125) 50  ( 70) 2 L = AY (rod – 2)  Y11 = Y22
   X = 1375 2
375  ( 125) 30  ( 70)
10. Pressure at the bottom in both arms will be equal
2. For centigrade and Fahrenheit scale (nksuksaHkqtkvksaesalrgijnkclekugksxk)
(lsUVhxzsMoQkjsugkbViSekusdsfy;s)
 0   0  l1  l 2
F  32 C0 100  1  t  .l1   1  t  l 2    l t  l t
 C  (140  32) = 60°C 1 2 2 1 1 2
212  32 100  0 180

11. Strain (foÏfr)

3. Slope of line AB (js[kk AB dk <ky)
l
C 100  0 100 5 t=   = –12 × 10–6 × (75 – 25) = – 6 × 10–4
=    l
F 212  32 180 9

12. Coefficient of linear expansion of brass is greater

4. If we take two fixed points as tripe point of water
than that of steel.
and 0 K. Then (;fn ge nks fLFkj fcUnq ty dk f=d fcUnq
(ihrydkjs[kh;izlkjxq.kkadLVhydhrqyukesavf/kdgksxk)
,oa 0 K ysosa rc)
TX  0 TY  0

13. l  l 0 1  T   l  l 0  l 0T
450 TX = 200 TY 9TX = 4TY
200 450
l 0 A T 104  100 4 A 2
X  LFP   l  T  106  100  6    3
5. = constant (for all temperature scales) 0 B B
UFP  LFP
(fu;r (lHkh rki iSekus ds fy;s)
14. Clearance = R'–R but 2R' = 2R (1+ sT)
where (tgka)
 R' – R = RsT
LFP  lower fixed point (fuEu fLFkj fcUnq) = (6400) (1.2 × 10–5) (30) = 2.3 km
UFP  Upper fixed point (mPp fLFkj fcUnq)
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15. x= lA  lB  l A 1   A T  l B 1   B T   l A  l B

X   5  C0 60  5 C
  =  C = 65°C
95   5  100  0 95  5 100  l A A  l B B

16. For rod A (NM+ A ds fy, ) l = l0 A(100)

6  10 5
6. L = 6 × 10–5 = L  = =5°C l
1  12  106 For rod B (NM+ B ds fy, ) = 2l0B(100)
2
7. Expansions of a metal is same as photographic For rod C (NM+ C ds fy, )
enlargement.  d1 will increase by 0.3% 2l = xA(100) + (3l0 – x)B(100)
,d/kkrqdkfoLrj.kQksVksdscM+sdjusdhrjggksrkgS 5 4
 d1, 0.3% ls c<+sxk  x  l & 3l0–x = l 0
3 0 3

1
17. l   Expansion in dx = 0 = surrounding's temperature (ifjos'k dk rki)

80  60  80  60 
    1 x  dx T  100 1.76  105  1.2  106 x  dx
2
  k  30  ...(i)
0
0 t  2 
2
 6  x 
2
 60  50  60  50 
= 100 1.76  10  x  1.2  10    
5
and  k  30  ...(ii)
 2 0 t  2 
=3.76 mm  t = 48 sec
T2

18. Q dl = l0dT  l   dl    aT  bT  l 24. Q t  (x22 – x12)

2
0 dT
T1 For x1 = 0, x2=1 cm
7  (12–02)
a 2 b 3 3   3 2 7b 3  For x1 =1cm, x2 = 2cm
= l 0   T2  T1    T2  T1   =  aT1 
2
T l
2 3  2 3 1  0 7 1
t  (22 – 12)    t = 21 hrs
t 3
19. For simple pendulum (ljy yksyd ds fy;s) T = 2

l  T 1 l 1 25. From Stefan's law (LVhQu fu;e ls):  Q = AT
T4
 T  l1/ 2    
g T 2 l 2 1cal 4.2J
   5.67  108  1  T 4  T= 100 K
Assuming clock gives correct time at temperature s  m2 cal
ekuk yhft;s fd ?kM+h fn;sx;s rki ij lgh le;n'kkZrh gS Power
26. = Intensity (rhozrk)
6 1 6 Area
0  =   0  20 &
24  3600 2 24  3600 Power absorbed by the foil = Intensity at foil × Area
offoiliUuh}kjkvo'kksf"kr'kfDriUuhijrhozrk×iUuhdk
1
=   40  0   0  30C {ks=Qy)
2
   1.4  10 5 C 1 P0  A 0 T 4 
 P  A   4d2  A
4d2

20. A 0 (2T)4  A
20°C Now P'=  4P
4 (2d)2

 Q  Q
27 Q    
40°C t  A  t  B

K A A 100  70 K A  70  35 
 = B
 1  20 C  30 70
l0 (1+20 C)= l1 (1+20S)  l 1   l0
 1  20 S  KB KA 1
 KA =  K =
2 B 2
 1   2  31
21.  r = 1 +3 1 = 2 +3 2   2 =
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3 x dx
dx dx
28. dRH = KA  KA (1  x)
22. Let  = junction temperature (ekuk  = laf/k dk rki) 0
0
Net heat current at junction is zero
(laf/kijdqyÅ"ek/kkjk'kwU;gksrhgS)
l
1  ln(1  x) 
l
dx
RH =  dR H     
200 0
KA 0 (1  x) KA 0  0
3k(100–)+k(0–)+2k(50–) =0    °C
3
1
= KA  ln(1  l 0 )
0
23. Newton's law of cooling (U;wVu ds 'khryu fu;e ls)
OR
 Check dimensionally (foeh; fo'ys"k.k ls tkaps)
= k[ – 0]
t
2
 29. According to Wien's law ohu ds fu;ekuqlkj 36. Heat lost = Heat gained
m 1/T  mT. (Å"ekgkfu=Å"eko`f¼)
As the temperature of body increases, frequency msteam × 540=1100× 1 × (80–15)+20×1× (80 – 15)
corresponding to maximum energy in radiation (m)  mass of steam condensed = 0.13 kg
increases. Also area under the curve (la?kfurHkkidknzO;eku)
(pwafdoLrqdkrkic<+rkgS]fofdj.k(m)esavf/kdreÅtkZ
37. Water has high specific heat and due to this it
dslaxrvko`fÙkc<+rhgSAoØdsvUrxZr{ks=Qy) absorber more heat in rise of temperature.
(ikuhdhfof'k"VÅ"ekmPpgksrhgSrFkkblhdsdkj.krkiesa
 E  d  T 4
o`f¼gksusij;gvf/kdÅ"ekvo'kksf"krdjrhgS)
Q
30. = AT
T4 = same but Tred < Tgreen 38. When water is cooled to form ice, the energy is
t
released as heat so mass of water decreases.
as red Tred = greenTgreen (see VIBGYOR)
(tctyBaMkcQZcukrkgSrksÅtkZÅ"ekds:iesaeqDrgksrh
 Area of red star is greater
gS]vr%ikuhdknzO;eku?kVrkgS)
(yky rkjs dk {ks=Qy vf/kd gksxk)
39. If intermolecular forces vanish water behaves as gas.
31. Rate of cooling of water = Rate of cooling of alcohol
(;fn vUrjkf.od cy 'kwU; gS rks ikuh xSl dh rjg O;ogkj
(ikuh ds 'khryu dh nj = ,Ydksgy ds 'khryu dh nj)
djrk gS )
(250  10)  1  (5) (200s  10)  5
  4.5  103
130 67 Number of moles of water =  250
18
 Specific heat of alcohol (,YdksgkWy dh fof'k"V Å"ek) (ikuhdseksyksadhla[;k)
s = 0.62 Total volume of water at STP
(STP ij ikuh dk dqy vk;ru)
32. Amount of energy utilised in climbing = 22.4 × 250 × 10–3 m3 = 5.6 m3
(p<+usesami;ksxdhxbZÅtkZdhek=k)
mgh= 0.28 × 10 × 4.2 40. Heat removed in cooling water from 250C to 00C
(250C ls 00C rd B.Ms ikuh esa fu"dkflr Å"ek)
0.28  10  4.2
h= = 1.96 × 10–2 m = 1.96 cm = 100 × 1 × 25=2500 cal
60  10
Heat removed in converting water into ice at 00C
(0°C ij ikuh dks cQZ esa cnyus esa fu"dkflr Å"ek)
33. Entire KE gets converted into heat.
= 100 × 80 = 8000 cal
(lEiw.kZxfrtÅtkZ]Å"ekesaifjofrZrgkstkrhgS) Heat removed in cooling ice from
KE = ms  10 × 10 × 10 = 2 × 4200 ×  (B.MscQZesafu"dkflrÅ"ek)
 = 0.12°C 00 to –150C = 100 × 0.5 × 10 = 500 cal
Total heat removed in1hr 50min
34. M= mass of hallstone falling (1 hr 50 min esa fu"dkflr dqy Å"ek)
(M=fxjjgsvksysdknzO;eku) = 2500 + 8000 + 500=11000 cal
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m = mass of hallstone melting Heat removed per minute (izfr feuV fu"dkflr Å"ek)
(m=fi?kyjgsvksysdknzO;eku) 11000
As Mgh = mL. = =100 cal/min
110
m gh 10  103 1 P P P
So   3
 41.
M L 33  10 33

35. HC = ms = ms (1)°C T T T

HK = ms  = ms (1)K = ms(1)°C density density
increases decreases
HF = ms  = ms (1°F) = ms(5/9)°C
 HC = HK > H F  RT 
For equation : Q P= M 
 w 

3
 At constant temperature   P n1C P1  n2 C P 2
For 1st graph  P At constant temperature. 48.   n Cv  n Cv
1 1 2 2

 RT  5R 7R
For 2nd graph : Q P=
 M w   = 1.5  CP1 =
2
; CP2 =
2

1 3R 5R
At constant P,   CV1 = ; CV2= ; then n1 = n2
T 2 2

dP
For 3rd graph L: = constant  P  T
dT 3KT mv 2rms
49. v rms  T Q T  mv 2rms
 density  = constant m 3K

42. If temperature is doubled, pressure will also be

50. Change in momentum (laosx esa ifjorZu)
RT = 2mv cos(45°)
doubled as P  M (;fn rki nqxquk gS rks) 45°
w
1 45°
=2 × 3.32 × 10 –27
× 10 ×
3
 100%increase  lsnkcHkhnqxqukgksxk 2
43. Volume can't be negative. = 4.7 × 10–24 kg ms –1.
(vk;ru ½.kkRed ughagksldrk)
At constant pressure (fu;r nkc ij) 51. Here V = aT + b where a,b > 0

V  T or V  (t + 273) nRT nR b b
So P =  but T  T so P2 > P1
aT  b a  b / T 2 1

3 th 
44. volume of air at 0°C occupies entire volume at  ,
4 52. PV= nRT  P = M RT
w
(0°Cij ok;q dk 3@4 vk;ru rki ij lEiw.kZ vk;ru ?ksjrk gS)
PM w (105 )(28  103 )
V1 V2 3 / 4V V   kg m–3= 1.25 g/litre
As T  T     = 171°C RT 8.3  273
1 2 273  60 273  
53. U1 = +ve;
45. Ideal gas equation (vkn'kZ xSl lehdj.k): PV = nRT U2 = 0
U3 = –ve
 P0   U1 > U2 > U3
So at V = V0; RT1 =   (V0) and at V = 2V0,
2 As volume increases, W = +ve.

 4P0  11P0 V0 54. Internal energy and volume depend upon states.
RT2 =  
 (2V0)  T2 – T1 =
5 10R (vkUrfjdÅtkZ ovk;ru voLFkkijfuHkZjdjrsgSa)

46. Number of moles remain constant 55. PT11 = constant & PV =nRT
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(eksyksadhla[;kfu;rgS)
V T V 12
 V  T12   12  v  
P1 V1 P2 V2 P1V1 P2V2 V T VT T
n1 + n2 = n1' + n2'  RT  RT  RT   RT 
1 2 1 2
56. U = 2P0V & W = P0V
PV PV (1.5P)V (1.5P)V So Q = W + U = 3P0V = 3P0V0
   
273 273 R  273 RT
 T = 273 × 3 K = ( 273 × 3 – 273) °C= 546°C 57. When water is heated from 0°C to 4°C, its volume
decreases. tc ikuh dks 0°C ls 4°C rd Å"ek nh tkrh gS
47. Total translational KE(dqy LFkkukUrfjr xfrt ÅtkZ) rksbldkvk;ru?kVrkgS
3 3  P V is negative (½.kkRed)
= nRT = PV Hence Cp – Cv < 0  Cp < Cv
2 2
4
 58. V  T4  V (PV)4
 P4V3 = constant  PV3/4 = constant  Q1   U W1 

C1  T   T T 
R R    1 (QW2 > W1)
 C =Cv + = 3R + = 3R + 4R = 7R C2  Q2   U  W2 
1 x 1  3/4  T   T T 

f1n1  f2 n2  f3 n3 (5n)(3)  (n)(5)  (5n)(6) 50

59. feq =   9. Q = U + W
n1  n2  n3 5n  n  5n 11
Q = +ve, as heat is absorbed from the atmosphere
ok;qe.My ls vo'kksf"kr Å"ek
60. U = a + bPV = a+bnRT
 U = bnRT = nCvT W=–ve as the volume decrease
 Cv = bR  Cp = bR + R (tc vk;ru ?kVrk gS)
 U = Q – W = +ve – (–ve) =+ve
C p bR  R b  1
   C  bR  b  Internal energy increases.
v (vkUrfjdÅtkZc<+rh gS)
EXERCISE –II
10. HA = (6 cal/s) × (6 – 2) s
1. All dimensions increase on heating. HB = (6 cal/s) × (6.5 – 4) s
(xjedjusijlkjhfoek;sac<+rhgSa) HA 4 8
 H  2.5  5
L2
1
2L 1 L 1 B
2. DC2 = L22 –  0 = 2L2 L2 –
4 4
12. For insulated chambers (Å"ekjks/kh izdks"Bksadsfy;s)
2L 1 (  1 L 1  )
 0 = 2L2(2L2) –  1 =42 n1 + n2 = n'1 + n2'
4
(final pressures become equal)
3. A part of liquid will evaporate immediately sucking
latent heat from the bulk of liquid. Hence a part of (vfUrenkccjkcjgksxk)
liquid will freeze.(nzo dk ,d Hkkx nzo ds vk;ru (ifjek.k) PV 2P.2V P 5P
  [3V]  P ' 
lsxqIrÅ"eklks[kdjmls'kh?kzrklsok"iesacnynsrkgS]vr% RT RT RT 3
nzO; dk ,d Hkkx te tk;sxk) For left chamber (cka;s izdks"B ds fy;s)

5P 3V
4. Qvap = Qfreezing PV = P'V'  V'  V'
3 5
m.(L) = M(L)  M = m
For right chamber (nka;s izdks"B ds fy;s)
L = latent heat of freezing (teus dh xqIr Å"ek)
m = mass of vapour (ok"i dk nzO;eku) 5P 12V
4PV= P'V' = V ' V ' 
M = mass of freezed tek nzO;eku 3 5

 Fraction of water which freezed P2

(ikuh dk ogHkkx tkstetk;sxk) 13. = constant (fu;r)

M M  RT
   P=p (Ideal gas equation) (vkn'kZ xSl lehdj.k)
m  M m  m 1  
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6. Mixture may attain intermediate temperature or P 2 P  RT   R

     PT   = constant
terminal temperatures of fusion or vapourisation.  
 M   M
(feJ.k dk rki e/;orhZ rki ds cjkcj ;k xyu ;k ok"iu ds  The graph of the above process on the P–T
vfUre rki ds cjkcj gks ldrk gS) diagram is hyperbola.
7. Water at 4°C has highest density (P-T vkjs[k ij mijksDr izØe dk oØ vfrijoyf;d gksxk)
(4°Cijikuhdk?kuRovR;f/kdgksrkgS) For the above process (mijksDr izØe ds fy;s)

 P2   P2  P2 P22 P
8. Q1 = U+ W1; Q2 = U + W2           P2  ...(i)
Ratio of specific heats 1 2  /2 2
(fof'k"VÅ"ekvksadkvuqikr) and