SOLUTIONS MANUAL for Concepts of Physics (Volume 2) by Harish C Verma All Chapters 23-47

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CHAPTER – 23 HEAT AND TEMPERATURE EXERCISES 1.

Ice point = 20° (L0) L1 = 32° Steam point = 80° (L100) L1  L 0 32  20  100 = T=  100 = 20°C L100  L 0 80  20

2.

Ptr = 1.500 × 104 Pa 4 P = 2.050 × 10 Pa We know, For constant volume gas Thermometer T=

P 2.050  10 4  273.16 = 373.31  273.16 K = Ptr 1.500  10 4

3.

Pressure Measured at M.P = 2.2 × Pressure at Triple Point 2.2  Ptr P  273.16 =  273.16 = 600.952 K  601 K T= Ptr Ptr

4.

3 Ptr = 40 × 10 Pa, P = ?

T = 100°C = 373 K,

T=

P  273.16 K Ptr

T  Ptr 373  49  10 3 3 = = 54620 Pa = 5.42 × 10 pa ≈ 55 K Pa 273.16 273.16 P1 = 70 K Pa, P2 = ?

T1 = 273 K, T2 = 373K P=

5.

6.

7.

8.

9.

T=

P1  273.16 Ptr

 273 =

T2 =

P2  273.16 Ptr

 373 =

70  10 3  273.16 Ptr

P2  273

70  273.16  10 Pice point = P0° = 80 cm of Hg Psteam point = P100° 90 cm of Hg P0 = 100 cm P  P0 80  100 t=  100 =  100 = 200°C P100  P0 90  100

3

 Ptr

70  273.16  10 3 273

 P2 =

373  70  10 3 = 95.6 K Pa 273

V T0 T0 = 273, V  V V = 1800 CC, V = 200 CC 1800 T =  273 = 307.125  307 1600 Rt = 86; R0° = 80; R100° = 90 R t  R0 86  80  100 = t=  100 = 60°C R100  R 0 90  80 T =

R at ice point (R0) = 20 R at steam point (R100) = 27.5 R at Zinc point (R420) = 50 2 R = R0 (1+  +  ) 2  R100 = R0 + R0  +R0  R  R0 2  100 =  +  R0 23.1

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23.Heat and Temperature

27.5  20 =  × 100 +  × 10000 20 7 .5  = 100  + 10000  20 50  R 0 2 2 R420 = R0 (1+  +  )  =  +  R0 

50  20 3 = 420 ×  + 176400 ×     420  + 176400  20 2 7 .5 3  = 100  + 10000     420  + 176400  20 2 –5 L1 = ?, L0 = 10 m,  = 1 × 10 /°C, t= 35 L1 = L0 (1 + t) = 10(1 + 10–5 × 35) = 10 + 35 × 10–4 = 10.0035m t1 = 20°C, t2 = 10°C, L1 = 1cm = 0.01 m, L2 =?

–5 steel = 1.1 × 10 /°C –5 –4 L2 = L1 (1 + steelT) = 0.01(1 + 101 × 10 × 10) = 0.01 + 0.01 × 1.1 × 10 4 –6 –6 –6 = 10 × 10 + 1.1 × 10 = 10 (10000 + 1.1) = 10001.1 –2 =1.00011 × 10 m = 1.00011 cm –5 L0 = 12 cm,  = 11 × 10 /°C tw = 18°C ts = 48°C –5 Lw = L0(1 + tw) = 12 (1 + 11 × 10 × 18) = 12.002376 m –5 Ls = L0 (1 + ts) = 12 (1 + 11 × 10 × 48) = 12.006336 m L12.006336 – 12.002376 = 0.00396 m  0.4cm –2 d1 = 2 cm = 2 × 10 t1 = 0°C, t2 = 100°C al = 2.3 × 10–5 /°C d2 = d1 (1 + t) = 2 × 10–2 (1 + 2.3 × 10–5 102) = 0.02 + 0.000046 = 0.020046 m = 2.0046 cm –5 Lst = LAl at 20°C Al = 2.3 × 10 /°C –5 So, Lost (1 – st × 20) = LoAl (1 – AI × 20) st = 1.1 × 10 /°C 

10.

11.

12.

13.

14.

(a) 

Lo st (1   Al  20) 1  2.3  10 5  20 0.99954 = = = = 0.999 Lo Al (1   st  20) 0.99978 1  1.1 10  5  20

(b) 

Lo 40st (1   AI  40) 0.99954 1  2.3  10 5  20 = = = = 0.999 0.99978 Lo 40 Al (1   st  40) 1  1.1 10  5  20

=

Lo Al 1  2.3  10 5  10 0.99977  1.00092 = = 1.0002496 ≈1.00025  Lo st 273 1.00044

Lo100 Al (1   Al  100) 0.99977  1.00092 = = = 1.00096 Lo100St (1   st  100 ) 1.00011 15. (a) Length at 16°C = L L=?

T1 =16°C, –5  = 1.1 × 10 /°C –5 L = L = L × 1.1 × 10 × 30

T2 = 46°C

 L   L  % of error =   100 % =   100 % = 1.1 × 10–5 × 30 × 100% = 0.033%  L   2  (b) T2 = 6°C

 L  L   –5  100 % =   100 % = – 1.1 × 10 × 10 × 100 = – 0.011% % of error =   L  L   23.2

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23.Heat and Temperature –3

16. T1 = 20°C, L = 0.055mm = 0.55 × 10 m –6 t2 = ? st = 11 × 10 /°C We know, L = L0T In our case, 0.055 × 10–3 = 1 × 1.1 I 10–6 × (T1 +T2) –3 –3 0.055 = 11 × 10 × 20 ± 11 × 10 × T2 T2 = 20 + 5 = 25°C or 20 – 5 = 15°C The expt. Can be performed from 15 to 25°C 3 3 17. ƒ0°C=0.098 g/m , ƒ4°C = 1 g/m ƒ 4 C 1 1 ƒ0°C =  0.998 =  1 + 4 = 1  T 1   4 0.998

1  1   = 0.0005 ≈ 5 × 10–4 0.998 -4 As density decreases  = –5 × 10 18. Iron rod Aluminium rod LFe LAl –8 –8 Fe = 12 × 10 /°C Al = 23 × 10 /°C Since the difference in length is independent of temp. Hence the different always remains constant.

LFe = LFe(1 + Fe × T) …(1) LAl = LAl(1 + Al × T) …(2) LFe – LAl = LFe – LAl + LFe × Fe × T – LAl × Al × T L Fe  23 = Al = = 23 : 12 L Al  Fe 12 4+=

2

2

19. g1 = 9.8 m/s , T1 = 2

l1

T2 = 2

g1

Steel = 12 × 10 T1 = 20°C T1 = T2  2

g2 = 9.788 m/s

l1 g1

–6

= 2

l2 g2

= 2

l1(1  T ) g

/°C T2 = ?

l1(1  T ) g2

1  12  10 6  T 1 = 9 .8 9.788 9.788 –6   1 = 12 × 10 T 9 .8 



l1 l (1  T ) = 1 g1 g2

9.788 –6 = 1+ 12 × 10 × T 9 .8 0.00122  T = 12  10  6  T2 = – 101.6 + 20 = – 81.6 ≈ – 82°C  

 T2 – 20 = – 101.6 20. Given dSt = 2.005 cm, dAl = 2.000 cm –6 –6 S = 11 × 10 /°C Al = 23 × 10 /°C ds = 2.005 (1+ s T) (where T is change in temp.)

–6  ds = 2.005 + 2.005 × 11 × 10 T –6 dAl = 2(1+ Al T) = 2 + 2 × 23 × 10 T The two will slip i.e the steel ball with fall when both the diameters become equal.

So, –6 –6  2.005 + 2.005 × 11 × 10 T = 2 + 2 × 23 × 10 T -6  (46 – 22.055)10 × T = 0.005  T =

0.005  10 6 = 208.81 23.945 23.3

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Steel Aluminium

23.Heat and Temperature Now T = T2 –T1 = T2 –10°C [ T1 = 10°C given] T2 = T + T1 = 208.81 + 10 = 281.81 21. The final length of aluminium should be equal to final length of glass.

Let the initial length o faluminium = l l(1 – AlT) = 20(1 – 0)  l(1 – 24 × 10–6 × 40) = 20 (1 – 9 × 10–6 × 40)  l(1 – 0.00096) = 20 (1 – 0.00036) 20  0.99964 l= = 20.012 cm 0.99904 Let initial breadth of aluminium = b b(1 – AlT) = 30(1 – 0) b =

30  (1  9  10 6  40)

(1  24  10 22. Vg = 1000 CC, VHg = ?

6

 40)

=

30  0.99964 = 30.018 cm 0.99904

T1 = 20°C –4 Hg = 1.8 × 10 /°C –6 g = 9 × 10 /°C

T remains constant Volume of remaining space = Vg – VHg Now Vg = Vg(1 + gT) …(1) VHg = VHg(1 + HgT) …(2) Subtracting (2) from (1) Vg – VHg = Vg – VHg + VggT – VHgHgT Vg  Hg 1.8  10 4 1000  =  = VHg g VHg 9  10  6  VHG =

9  10 3

= 500 CC. 1.8  10  4 3 23. Volume of water = 500cm Area of cross section of can = 125 m2 Final Volume of water –4 3 = 500(1 + ) = 500[1 + 3.2 × 10 × (80 – 10)] = 511.2 cm The aluminium vessel expands in its length only so area expansion of base cab be neglected.

3 Increase in volume of water = 11.2 cm 3 Considering a cylinder of volume = 11.2 cm 11.2 Height of water increased = = 0.089 cm 125 24. V0 = 10 × 10× 10 = 1000 CC T = 10°C, VHG – Vg = 1.6 cm3 –6 –6 g = 6.5 × 10 /°C, Hg = ?, g= 3 × 6.5 × 10 /°C VHg = vHG(1 + HgT) …(1) Vg = vg(1 + gT) …(2) VHg – Vg = VHg –Vg + VHgHg T – Vgg T  1.6 = 1000 × Hg × 10 – 1000 × 6.5 × 3 × 10–6 × 10

1.6  6.3  3  10 2 –4 –4 = 1.789 × 10  1.8 × 10 /°C 10000 3 3 25. ƒ = 880 Kg/m , ƒb = 900 Kg/m –3 T1 = 0°C,  = 1.2 × 10 /°C, –3 b = 1.5 × 10 /°C The sphere begins t sink when, (mg)sphere = displaced water  Hg =

23.4

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