Solution Manual for Concepts of Physics Volume 2 by HC Verma

CHAPTER – 23
HEAT AND TEMPERATURE
EXERCISES

1. Ice point = 20° (L0) L1 = 32°
Steam point = 80° (L100)
L1  L 0 32  20
T=  100 =  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
P 2.050  10 4
T=  273.16 K =  273.16 = 373.31
Ptr 1.500  10 4
3. Pressure Measured at M.P = 2.2 × Pressure at Triple Point
P 2.2  Ptr
T=  273.16 =  273.16 = 600.952 K  601 K
Ptr Ptr
3
4. Ptr = 40 × 10 Pa, P = ?
P
T = 100°C = 373 K, T=  273.16 K
Ptr
T  Ptr 373  49  10 3 3
P= = = 54620 Pa = 5.42 × 10 pa ≈ 55 K Pa
273.16 273.16
5. P1 = 70 K Pa, P2 = ?
T1 = 273 K, T2 = 373K
P1 70  10 3 70  273.16  10 3
T=  273.16  273 =  273.16  Ptr
Ptr Ptr 273
P2 P2  273 373  70  10 3
T2 =  273.16  373 = 3
 P2 = = 95.6 K Pa
Ptr 70  273.16  10 273
6. 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
V
7. T = T0 T0 = 273,
V  V
V = 1800 CC, V = 200 CC
1800
T =  273 = 307.125  307
1600
8. Rt = 86; R0° = 80; R100° = 90
R t  R0 86  80
t=  100 =  100 = 60°C
R100  R 0 90  80
9. 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

, 23.Heat and Temperature
27.5  20
 =  × 100 +  × 10000
20
7 .5
 = 100  + 10000 
20
2 50  R 0 2
R420 = R0 (1+  +  )  =  + 
R0
50  20 3
 = 420 ×  + 176400 ×     420  + 176400 
20 2
7 .5 3
 = 100  + 10000     420  + 176400 
20 2
–5
10. L1 = ?, L0 = 10 m,  = 1 × 10 /°C, t= 35
L1 = L0 (1 + t) = 10(1 + 10–5 × 35) = 10 + 35 × 10–4 = 10.0035m
11. 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
12. 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
13. 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
14. 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
Lo st (1   Al  20) 1  2.3  10 5  20 0.99954
(a)  = = = = 0.999
Lo Al (1   st  20) 1  1.1 10  5  20 0.99978
Lo 40st (1   AI  40) 1  2.3  10 5  20 0.99954
(b)  = = = = 0.999
Lo 40 Al (1   st  40) 1  1.1 10  5  20 0.99978
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, T2 = 46°C
–5
 = 1.1 × 10 /°C
–5
L = L = L × 1.1 × 10 × 30
 L   L 
% of error =   100 % =   100 % = 1.1 × 10–5 × 30 × 100% = 0.033%
 L   2 
(b) T2 = 6°C
 L   L  –5
% of error =   100 % =   100 % = – 1.1 × 10 × 10 × 100 = – 0.011%
 L   L 

23.2

, 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
4+=  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
2 2
19. g1 = 9.8 m/s , g2 = 9.788 m/s
l1 l2 l1(1  T )
T1 = 2 T2 = 2 = 2
g1 g2 g
–6
Steel = 12 × 10 /°C
T1 = 20°C T2 = ?
T1 = T2
l1 l1(1  T ) l1 l (1  T )
 2 = 2  = 1
g1 g2 g1 g2
1 1  12  10 6  T 9.788 –6
 =  = 1+ 12 × 10 × T
9 .8 9.788 9 .8
9.788 –6 0.00122
  1 = 12 × 10 T  T =
9 .8 12  10  6
 T2 – 20 = – 101.6  T2 = – 101.6 + 20 = – 81.6 ≈ – 82°C 
20. Given
dSt = 2.005 cm, dAl = 2.000 cm
–6 –6
S = 11 × 10 /°C Al = 23 × 10 /°C Steel
ds = 2.005 (1+ s T) (where T is change in temp.)
–6
 ds = 2.005 + 2.005 × 11 × 10 T Aluminium
–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
0.005  10 6
 T = = 208.81
23.945
23.3

, 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)
30  (1  9  10 6  40) 30  0.99964
b = 6
= = 30.018 cm
(1  24  10  40) 0.99904
22. Vg = 1000 CC, T1 = 20°C
–4
VHg = ? 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 1000 1.8  10 4
 =  =
VHg g VHg 9  10  6
9  10 3
 VHG = = 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
 Hg = = 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

23.4