23.HEAT AND TEMPERATURE by Aditya Bhardwaj

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

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

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

Ptr = 40 × 10 Pa, P = ?

T = 100°C = 373 K,

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 = ? T2 = 373K T1 = 273 K, P=

T= T2 =

P1  273.16 Ptr P2  273.16 Ptr

 273 =  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  100 =  100 = 200°C t= 90  100 P100  P0

V T0 T0 = 273, V  V V = 1800 CC, V = 200 CC 1800  273 = 307.125  307 T = 1600 Rt = 86; R0° = 80; R100° = 90 R t  R0 86  80 t=  100 =  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   R0 R 2  100 =  +  R0 23.1

 Ptr

70  273.16  10 3 273

 P2 =

373  70  10 3 = 95.6 K Pa 273

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 –5 –4 L1 = L0 (1 + t) = 10(1 + 10 × 35) = 10 + 35 × 10 = 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  = 11 × 10 /°C L0 = 12 cm, 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 –5 al = 2.3 × 10 /°C –2 –5 2 d2 = d1 (1 + t) = 2 × 10 (1 + 2.3 × 10 10 ) = 0.02 + 0.000046 = 0.020046 m = 2.0046 cm –5 Lst = LAl at 20°C Al = 2.3 × 10 /°C –5 st = 1.1 × 10 /°C So, Lost (1 – st × 20) = LoAl (1 – AI × 20) 

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

23.Heat and Temperature –3

16. T1 = 20°C, L = 0.055mm = 0.55 × 10 m –6 st = 11 × 10 /°C t2 = ? We know, L = L0T In our case, –3 –6 0.055 × 10 = 1 × 1.1 I 10 × (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 ƒ4°C = 1 g/m 17. ƒ0°C=0.098 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 LAl LFe –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. …(1) LFe = LFe(1 + Fe × T) 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+=

19. g1 = 9.8 m/s , T1 = 2

T2 = 2

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 dAl = 2.000 cm dSt = 2.005 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

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) –6 –6  l(1 – 24 × 10 × 40) = 20 (1 – 9 × 10 × 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 2 Area of cross section of can = 125 m 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 3 T = 10°C, VHG – Vg = 1.6 cm –6 –6 g = 6.5 × 10 /°C, Hg = ?, g= 3 × 6.5 × 10 /°C …(1) VHg = vHG(1 + HgT) Vg = vg(1 + gT) …(2) VHg – Vg = VHg –Vg + VHgHg T – Vgg T –6  1.6 = 1000 × Hg × 10 – 1000 × 6.5 × 3 × 10 × 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

23.Heat and Temperature  Vƒ g = Vƒb g ƒb ƒ    1     1   b 

880 900 = 1  1.2  10 3  1  1.5  10 3  –3 –3  880 + 880 × 1.5 × 10 () = 900 + 900 × 1.2 × 10 () –3 –3  (880 × 1.5 × 10 – 900 × 1.2 × 10 ) () = 20 –3  (1320 – 1080) × 10 () = 20   = 83.3°C ≈ 83°C L = 100°C A longitudinal strain develops if and only if, there is an opposition to the expansion. Since there is no opposition in this case, hence the longitudinal stain here = Zero. 1 = 20°C, 2 = 50°C –5 steel = 1.2 × 10 /°C Longitudinal stain = ? L L Stain = = =  L L –5 –4 = 1.2 × 10 × (50 – 20) = 3.6 × 10 2 –6 2 A = 0.5mm = 0.5 × 10 m T1 = 20°C, T2 = 0°C –5 11 2 s = 1.2 × 10 /°C, Y = 2 × 2 × 10 N/m Decrease in length due to compression = L …(1) Stress F L FL Y=  =  L = …(2) Strain A L AY Tension is developed due to (1) & (2) Equating them, FL  F = AY L = AY -5 –5 11 = 1.2 × 10 × (20 – 0) × 0.5 × 10 2 × 10 = 24 N 1 = 20°C, 2 = 100°C 2 –6 2 A = 2mm = 2 × 10 m –6 11 2 steel = 12 × 10 /°C, Ysteel = 2 × 10 N/m Force exerted on the clamps = ? 

26.

27.

28.

29.

F    A  = Y  F = Y  L  L = YLA = YA L Strain L 11 –6 –6 = 2 × 10 × 2 × 10 × 12 × 10 × 80 = 384 N 30. Let the final length of the system at system of temp. 0°C = ℓ Initial length of the system = ℓ0 When temp. changes by .  Strain of the system =  1  0 

total stress of system But the total strain of the system = total young' s mod ulusof of system Now, total stress = Stress due to two steel rod + Stress due to Aluminium = ss + s ds  + al at  = 2% s  + 2 Aℓ  Now young’ modulus of system = s + s + al = 2s + al

23.5



Steel Aluminium Steel

23.Heat and Temperature  Strain of system = 

2 s  s    s  al  2  s   al

  0 2 s  s    s  al  = 0 2  s   al

1   al  al  2 s  s    ℓ = ℓ0    al  2 s   31. The ball tries to expand its volume. But it is kept in the same volume. So it is kept at a constant volume. So the stress arises P V =BP= B = B × V  V     v  11

–6

= B × 3 = 1.6 × 10 × 10 × 3 × 12 × 10 × (120 – 20) = 57.6 × 19  5.8 × 10 pa.  32. Given 0 = Moment of Inertia at 0°C  = Coefficient of linear expansion To prove,  = 0 = (1 + 2) Let the temp. change to  from 0°C T =  Let ‘R’ be the radius of Gyration, 2 0 = MR where M is the mass. Now, R = R (1 + ), 2 2 2 2 Now,  = MR = MR (1 + )  = MR (1 + 2) 2 2 [By binomial expansion or neglecting   which given a very small value.] (proved) So,  = 0 (1 + 2) 33. Let the initial m.. at 0°C be 0

 K  = 0 (1 + 2) T = 2

(from above question)

At 5°C,

T1 = 2

 0 (1  2) = 2 K

At 45°C,

T2 = 2

 (1  90 )  0 (1  2 45) = 2 0 K K

T2 = T1

1  90 = 1  10

1  90  2.4  10 5 1  10  2.4  10

5

 0 (1  25)  (1  10 ) = 2 0 K K

1.00216 1.00024

 T –2 % change =  2  1  100 = 0.0959% = 9.6 × 10 % T   1 T2 = 50°C, T = 30°C 34. T1 = 20°C, 5  = 1.2 × 10 /°C  remains constant V V (II)  = (I)  = R R –5 Now, R = R(1 + ) = R + R × 1.2 × 10 × 30 = 1.00036R From (I) and (II) V V V  = R R 1.00036R V = 1.00036 V (1.00036 V  V ) –2 % change = × 100 = 0.00036 × 100 = 3.6 × 10 V      23.6