Page 1
23.1
CHAPTER – 23
HEAT AND TEMPERATURE
EXERCISES
1. Ice point = 20° (L
0
) L
1
= 32°
Steam point = 80° (L
100
)
T = 100
L L
L L
0 100
0 1
?
?
?
= 100
20 80
20 32
?
?
?
= 20°C
2. P
tr
= 1.500 × 10
4
Pa
P = 2.050 × 10
4
Pa
We know, For constant volume gas Thermometer
T = 16 . 273
P
P
tr
? K = 16 . 273
10 500 . 1
10 050 . 2
4
4
?
?
?
= 373.31
3. Pressure Measured at M.P = 2.2 × Pressure at Triple Point
T = 16 . 273
P
P
tr
? = 16 . 273
P
P 2 . 2
tr
tr
?
?
= 600.952 K ? 601 K
4. P
tr
= 40 × 10
3
Pa, P = ?
T = 100°C = 373 K, T = 16 . 273
P
P
tr
? K
? P =
16 . 273
P T
tr
?
=
16 . 273
10 49 373
3
? ?
= 54620 Pa = 5.42 × 10
3
pa ˜ 55 K Pa
5. P
1
= 70 K Pa, P
2
= ?
T
1
= 273 K, T
2
= 373K
T = 16 . 273
P
P
tr
1
? ? 273 = 16 . 273
P
10 70
tr
3
?
?
? P
tr
273
10 16 . 273 70
3
? ?
T
2
= 16 . 273
P
P
tr
2
? ? 373 =
3
2
10 16 . 273 70
273 P
? ?
?
? P
2
=
273
10 70 373
3
? ?
= 95.6 K Pa
6. P
ice point
= P
0°
= 80 cm of Hg
P
steam point
= P
100°
90 cm of Hg
P
0
= 100 cm
t = ? ?
?
?
100
P P
P P
0 100
0
= 100
100 90
100 80
?
?
?
= 200°C
7. T ? =
0
T
V V
V
? ?
T
0
= 273,
V = 1800 CC, V ? = 200 CC
T ? = 273
1600
1800
? = 307.125 ? 307
8. R
t
= 86 ?; R
0°
= 80 ?; R
100°
= 90 ?
t = 100
R R
R R
0 100
0 t
?
?
?
= 100
80 90
80 86
?
?
?
= 60°C
9. R at ice point (R
0
) = 20 ?
R at steam point (R
100
) = 27.5 ?
R at Zinc point (R
420
) = 50 ?
R
?
= R
0
(1+ ? ? + ? ?
2
)
? R
100
= R
0
+ R
0
?? +R
0
??
2
?
0
0 100
R
R R ?
= ? ? + ? ?
2
Page 2
23.1
CHAPTER – 23
HEAT AND TEMPERATURE
EXERCISES
1. Ice point = 20° (L
0
) L
1
= 32°
Steam point = 80° (L
100
)
T = 100
L L
L L
0 100
0 1
?
?
?
= 100
20 80
20 32
?
?
?
= 20°C
2. P
tr
= 1.500 × 10
4
Pa
P = 2.050 × 10
4
Pa
We know, For constant volume gas Thermometer
T = 16 . 273
P
P
tr
? K = 16 . 273
10 500 . 1
10 050 . 2
4
4
?
?
?
= 373.31
3. Pressure Measured at M.P = 2.2 × Pressure at Triple Point
T = 16 . 273
P
P
tr
? = 16 . 273
P
P 2 . 2
tr
tr
?
?
= 600.952 K ? 601 K
4. P
tr
= 40 × 10
3
Pa, P = ?
T = 100°C = 373 K, T = 16 . 273
P
P
tr
? K
? P =
16 . 273
P T
tr
?
=
16 . 273
10 49 373
3
? ?
= 54620 Pa = 5.42 × 10
3
pa ˜ 55 K Pa
5. P
1
= 70 K Pa, P
2
= ?
T
1
= 273 K, T
2
= 373K
T = 16 . 273
P
P
tr
1
? ? 273 = 16 . 273
P
10 70
tr
3
?
?
? P
tr
273
10 16 . 273 70
3
? ?
T
2
= 16 . 273
P
P
tr
2
? ? 373 =
3
2
10 16 . 273 70
273 P
? ?
?
? P
2
=
273
10 70 373
3
? ?
= 95.6 K Pa
6. P
ice point
= P
0°
= 80 cm of Hg
P
steam point
= P
100°
90 cm of Hg
P
0
= 100 cm
t = ? ?
?
?
100
P P
P P
0 100
0
= 100
100 90
100 80
?
?
?
= 200°C
7. T ? =
0
T
V V
V
? ?
T
0
= 273,
V = 1800 CC, V ? = 200 CC
T ? = 273
1600
1800
? = 307.125 ? 307
8. R
t
= 86 ?; R
0°
= 80 ?; R
100°
= 90 ?
t = 100
R R
R R
0 100
0 t
?
?
?
= 100
80 90
80 86
?
?
?
= 60°C
9. R at ice point (R
0
) = 20 ?
R at steam point (R
100
) = 27.5 ?
R at Zinc point (R
420
) = 50 ?
R
?
= R
0
(1+ ? ? + ? ?
2
)
? R
100
= R
0
+ R
0
?? +R
0
??
2
?
0
0 100
R
R R ?
= ? ? + ? ?
2
23.Heat and Temperature
23.2
?
20
20 5 . 27 ?
= ? × 100 + ? × 10000
?
20
5 . 7
= 100 ? + 10000 ?
R
420
= R
0
(1+ ? ? + ? ?
2
) ?
0
0
R
R 50 ?
= ? ? + ??
2
?
20
20 50 ?
= 420 × ? + 176400 × ?? ? ??
2
3
??? 420 ? + 176400 ??
??
20
5 . 7
= 100 ? + 10000 ? ?? ? ? ??
2
3
??? 420 ? + 176400 ??
10. L
1
= ?, L
0
= 10 m, ? = 1 × 10
–5
/°C, t= 35
L
1
= L
0
(1 + ?t) = 10(1 + 10
–5
× 35) = 10 + 35 × 10
–4
= 10.0035m
11. t
1
= 20°C, t
2
= 10°C, L
1
= 1cm = 0.01 m, L
2
=?
?
steel
= 1.1 × 10
–5
/°C
L
2
= L
1
(1 + ?
steel
?T) = 0.01(1 + 101 × 10
–5
× 10) = 0.01 + 0.01 × 1.1 × 10
–4
= 10
4
× 10
–6
+ 1.1 × 10
–6
= 10
–6
(10000 + 1.1) = 10001.1
=1.00011 × 10
–2
m = 1.00011 cm
12. L
0
= 12 cm, ? = 11 × 10
–5
/°C
tw = 18°C ts = 48°C
Lw = L
0
(1 + ?tw) = 12 (1 + 11 × 10
–5
× 18) = 12.002376 m
Ls = L
0
(1 + ?ts) = 12 (1 + 11 × 10
–5
× 48) = 12.006336 m ?
?L ????12.006336 – 12.002376 = 0.00396 m ? 0.4cm ?
13. d
1
= 2 cm = 2 × 10
–2
t
1
= 0°C, t
2
= 100°C
?
al
= 2.3 × 10
–5
/°C
d
2
= d
1
(1 + ??t) = 2 × 10
–2
(1 + 2.3 × 10
–5
10
2
)
= 0.02 + 0.000046 = 0.020046 m = 2.0046 cm
14. L
st
= L
Al
at 20°C ?
Al
= 2.3 × 10
–5
/°C
So, Lo
st
(1 – ?
st
× 20) = Lo
Al
(1 – ?
AI
× 20) ?
st
= 1.1 × 10
–5
/°C
(a) ?
Al
st
Lo
Lo
=
) 20 1 (
) 20 1 (
st
Al
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
(b) ?
Al 40
st 40
Lo
Lo
=
) 40 1 (
) 40 1 (
st
AI
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
=
273
10 10 3 . 2 1
Lo
Lo
5
st
Al
? ? ?
?
?
=
00044 . 1
00092 . 1 99977 . 0 ?
= 1.0002496 ˜1.00025
St 100
Al 100
Lo
Lo
=
) 100 1 (
) 100 1 (
st
Al
? ? ?
? ? ?
=
00011 . 1
00092 . 1 99977 . 0 ?
= 1.00096
15. (a) Length at 16°C = L
L = ? T
1
=16°C, T
2
= 46°C
? = 1.1 × 10
–5
/°C
?L = L ??? = L × 1.1 × 10
–5
× 30
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
2
L
?
?
?
?
?
?
?
? ? ?
= 1.1 × 10
–5
× 30 × 100% = 0.033%
(b) T
2
= 6°C
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
L
L
?
?
?
?
?
?
?
? ? ?
= – 1.1 × 10
–5
× 10 × 100 = – 0.011%
Page 3
23.1
CHAPTER – 23
HEAT AND TEMPERATURE
EXERCISES
1. Ice point = 20° (L
0
) L
1
= 32°
Steam point = 80° (L
100
)
T = 100
L L
L L
0 100
0 1
?
?
?
= 100
20 80
20 32
?
?
?
= 20°C
2. P
tr
= 1.500 × 10
4
Pa
P = 2.050 × 10
4
Pa
We know, For constant volume gas Thermometer
T = 16 . 273
P
P
tr
? K = 16 . 273
10 500 . 1
10 050 . 2
4
4
?
?
?
= 373.31
3. Pressure Measured at M.P = 2.2 × Pressure at Triple Point
T = 16 . 273
P
P
tr
? = 16 . 273
P
P 2 . 2
tr
tr
?
?
= 600.952 K ? 601 K
4. P
tr
= 40 × 10
3
Pa, P = ?
T = 100°C = 373 K, T = 16 . 273
P
P
tr
? K
? P =
16 . 273
P T
tr
?
=
16 . 273
10 49 373
3
? ?
= 54620 Pa = 5.42 × 10
3
pa ˜ 55 K Pa
5. P
1
= 70 K Pa, P
2
= ?
T
1
= 273 K, T
2
= 373K
T = 16 . 273
P
P
tr
1
? ? 273 = 16 . 273
P
10 70
tr
3
?
?
? P
tr
273
10 16 . 273 70
3
? ?
T
2
= 16 . 273
P
P
tr
2
? ? 373 =
3
2
10 16 . 273 70
273 P
? ?
?
? P
2
=
273
10 70 373
3
? ?
= 95.6 K Pa
6. P
ice point
= P
0°
= 80 cm of Hg
P
steam point
= P
100°
90 cm of Hg
P
0
= 100 cm
t = ? ?
?
?
100
P P
P P
0 100
0
= 100
100 90
100 80
?
?
?
= 200°C
7. T ? =
0
T
V V
V
? ?
T
0
= 273,
V = 1800 CC, V ? = 200 CC
T ? = 273
1600
1800
? = 307.125 ? 307
8. R
t
= 86 ?; R
0°
= 80 ?; R
100°
= 90 ?
t = 100
R R
R R
0 100
0 t
?
?
?
= 100
80 90
80 86
?
?
?
= 60°C
9. R at ice point (R
0
) = 20 ?
R at steam point (R
100
) = 27.5 ?
R at Zinc point (R
420
) = 50 ?
R
?
= R
0
(1+ ? ? + ? ?
2
)
? R
100
= R
0
+ R
0
?? +R
0
??
2
?
0
0 100
R
R R ?
= ? ? + ? ?
2
23.Heat and Temperature
23.2
?
20
20 5 . 27 ?
= ? × 100 + ? × 10000
?
20
5 . 7
= 100 ? + 10000 ?
R
420
= R
0
(1+ ? ? + ? ?
2
) ?
0
0
R
R 50 ?
= ? ? + ??
2
?
20
20 50 ?
= 420 × ? + 176400 × ?? ? ??
2
3
??? 420 ? + 176400 ??
??
20
5 . 7
= 100 ? + 10000 ? ?? ? ? ??
2
3
??? 420 ? + 176400 ??
10. L
1
= ?, L
0
= 10 m, ? = 1 × 10
–5
/°C, t= 35
L
1
= L
0
(1 + ?t) = 10(1 + 10
–5
× 35) = 10 + 35 × 10
–4
= 10.0035m
11. t
1
= 20°C, t
2
= 10°C, L
1
= 1cm = 0.01 m, L
2
=?
?
steel
= 1.1 × 10
–5
/°C
L
2
= L
1
(1 + ?
steel
?T) = 0.01(1 + 101 × 10
–5
× 10) = 0.01 + 0.01 × 1.1 × 10
–4
= 10
4
× 10
–6
+ 1.1 × 10
–6
= 10
–6
(10000 + 1.1) = 10001.1
=1.00011 × 10
–2
m = 1.00011 cm
12. L
0
= 12 cm, ? = 11 × 10
–5
/°C
tw = 18°C ts = 48°C
Lw = L
0
(1 + ?tw) = 12 (1 + 11 × 10
–5
× 18) = 12.002376 m
Ls = L
0
(1 + ?ts) = 12 (1 + 11 × 10
–5
× 48) = 12.006336 m ?
?L ????12.006336 – 12.002376 = 0.00396 m ? 0.4cm ?
13. d
1
= 2 cm = 2 × 10
–2
t
1
= 0°C, t
2
= 100°C
?
al
= 2.3 × 10
–5
/°C
d
2
= d
1
(1 + ??t) = 2 × 10
–2
(1 + 2.3 × 10
–5
10
2
)
= 0.02 + 0.000046 = 0.020046 m = 2.0046 cm
14. L
st
= L
Al
at 20°C ?
Al
= 2.3 × 10
–5
/°C
So, Lo
st
(1 – ?
st
× 20) = Lo
Al
(1 – ?
AI
× 20) ?
st
= 1.1 × 10
–5
/°C
(a) ?
Al
st
Lo
Lo
=
) 20 1 (
) 20 1 (
st
Al
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
(b) ?
Al 40
st 40
Lo
Lo
=
) 40 1 (
) 40 1 (
st
AI
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
=
273
10 10 3 . 2 1
Lo
Lo
5
st
Al
? ? ?
?
?
=
00044 . 1
00092 . 1 99977 . 0 ?
= 1.0002496 ˜1.00025
St 100
Al 100
Lo
Lo
=
) 100 1 (
) 100 1 (
st
Al
? ? ?
? ? ?
=
00011 . 1
00092 . 1 99977 . 0 ?
= 1.00096
15. (a) Length at 16°C = L
L = ? T
1
=16°C, T
2
= 46°C
? = 1.1 × 10
–5
/°C
?L = L ??? = L × 1.1 × 10
–5
× 30
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
2
L
?
?
?
?
?
?
?
? ? ?
= 1.1 × 10
–5
× 30 × 100% = 0.033%
(b) T
2
= 6°C
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
L
L
?
?
?
?
?
?
?
? ? ?
= – 1.1 × 10
–5
× 10 × 100 = – 0.011%
23.Heat and Temperature
23.3
16. T
1
= 20°C, ?L = 0.055mm = 0.55 × 10
–3
m
t
2
= ? ?
st
= 11 × 10
–6
/°C
We know,
?L = L
0
??T
In our case,
0.055 × 10
–3
= 1 × 1.1 I 10
–6
× (T
1
+T
2
)
0.055 = 11 × 10
–3
× 20 ± 11 × 10
–3
× T
2
T
2
= 20 + 5 = 25°C or 20 – 5 = 15°C
The expt. Can be performed from 15 to 25°C
17. ƒ
0°C
=0.098 g/m
3
, ƒ
4°C
= 1 g/m
3
ƒ
0°C
=
T 1
ƒ
C 4
? ? ?
?
? 0.998 =
4 1
1
? ? ?
? 1 + 4 ? =
998 . 0
1
? 4 + ? = 1
998 . 0
1
? ? ? = 0.0005 ˜ 5 × 10
–4
As density decreases ? = –5 × 10
-4
18. Iron rod Aluminium rod
L
Fe
L
Al
?
Fe
= 12 × 10
–8
/°C ?
Al
= 23 × 10
–8
/°C
Since the difference in length is independent of temp. Hence the different always remains constant.
L ?
Fe
= L
Fe
(1 + ?
Fe
× ?T) …(1)
L ?
Al
= L
Al
(1 + ?
Al
× ?T) …(2)
L ?
Fe
– L ?
Al
= L
Fe
– L
Al
+ L
Fe
× ?
Fe
× ?T – L
Al
× ?
Al
× ?T
Al
Fe
L
L
=
Fe
Al
?
?
=
12
23
= 23 : 12
19. g
1
= 9.8 m/s
2
, g
2
= 9.788 m/s
2
T
1
=
1
1
g
l
2 ? T
2
=
2
2
g
l
2 ? =
g
) T 1 ( l
2
1
? ?
?
?
Steel
= 12 × 10
–6
/°C
T
1
= 20°C T
2
= ?
T
1
= T
2
?
1
1
g
l
2 ? =
2
1
g
) T 1 ( l
2
? ?
? ?
1
1
g
l
=
2
1
g
) T 1 ( l ? ?
?
8 . 9
1
=
788 . 9
T 10 12 1
6
? ? ? ?
?
?
8 . 9
788 . 9
= 1+ 12 × 10
–6
× ?T
? 1
8 . 9
788 . 9
? = 12 × 10
–6
?T ? ?T =
6
10 12
00122 . 0
?
?
?
? T
2
– 20 = – 101.6 ? T
2
= – 101.6 + 20 = – 81.6 ˜ – 82°C ?
20. Given
d
St
= 2.005 cm, d
Al
= 2.000 cm
?
S
= 11 × 10
–6
/°C ?
Al
= 23 × 10
–6
/°C
d ?s = 2.005 (1+ ?
s
?T) (where ?T is change in temp.)
? d ?s = 2.005 + 2.005 × 11 × 10
–6
?T
d ?
Al
= 2(1+ ?
Al
?T) = 2 + 2 × 23 × 10
–6
?T
The two will slip i.e the steel ball with fall when both the
diameters become equal.
So,
? 2.005 + 2.005 × 11 × 10
–6
?T = 2 + 2 × 23 × 10
–6
?T
? (46 – 22.055)10
-6
× ?T = 0.005
? ?T =
945 . 23
10 005 . 0
6
?
= 208.81
Aluminium
Steel
Page 4
23.1
CHAPTER – 23
HEAT AND TEMPERATURE
EXERCISES
1. Ice point = 20° (L
0
) L
1
= 32°
Steam point = 80° (L
100
)
T = 100
L L
L L
0 100
0 1
?
?
?
= 100
20 80
20 32
?
?
?
= 20°C
2. P
tr
= 1.500 × 10
4
Pa
P = 2.050 × 10
4
Pa
We know, For constant volume gas Thermometer
T = 16 . 273
P
P
tr
? K = 16 . 273
10 500 . 1
10 050 . 2
4
4
?
?
?
= 373.31
3. Pressure Measured at M.P = 2.2 × Pressure at Triple Point
T = 16 . 273
P
P
tr
? = 16 . 273
P
P 2 . 2
tr
tr
?
?
= 600.952 K ? 601 K
4. P
tr
= 40 × 10
3
Pa, P = ?
T = 100°C = 373 K, T = 16 . 273
P
P
tr
? K
? P =
16 . 273
P T
tr
?
=
16 . 273
10 49 373
3
? ?
= 54620 Pa = 5.42 × 10
3
pa ˜ 55 K Pa
5. P
1
= 70 K Pa, P
2
= ?
T
1
= 273 K, T
2
= 373K
T = 16 . 273
P
P
tr
1
? ? 273 = 16 . 273
P
10 70
tr
3
?
?
? P
tr
273
10 16 . 273 70
3
? ?
T
2
= 16 . 273
P
P
tr
2
? ? 373 =
3
2
10 16 . 273 70
273 P
? ?
?
? P
2
=
273
10 70 373
3
? ?
= 95.6 K Pa
6. P
ice point
= P
0°
= 80 cm of Hg
P
steam point
= P
100°
90 cm of Hg
P
0
= 100 cm
t = ? ?
?
?
100
P P
P P
0 100
0
= 100
100 90
100 80
?
?
?
= 200°C
7. T ? =
0
T
V V
V
? ?
T
0
= 273,
V = 1800 CC, V ? = 200 CC
T ? = 273
1600
1800
? = 307.125 ? 307
8. R
t
= 86 ?; R
0°
= 80 ?; R
100°
= 90 ?
t = 100
R R
R R
0 100
0 t
?
?
?
= 100
80 90
80 86
?
?
?
= 60°C
9. R at ice point (R
0
) = 20 ?
R at steam point (R
100
) = 27.5 ?
R at Zinc point (R
420
) = 50 ?
R
?
= R
0
(1+ ? ? + ? ?
2
)
? R
100
= R
0
+ R
0
?? +R
0
??
2
?
0
0 100
R
R R ?
= ? ? + ? ?
2
23.Heat and Temperature
23.2
?
20
20 5 . 27 ?
= ? × 100 + ? × 10000
?
20
5 . 7
= 100 ? + 10000 ?
R
420
= R
0
(1+ ? ? + ? ?
2
) ?
0
0
R
R 50 ?
= ? ? + ??
2
?
20
20 50 ?
= 420 × ? + 176400 × ?? ? ??
2
3
??? 420 ? + 176400 ??
??
20
5 . 7
= 100 ? + 10000 ? ?? ? ? ??
2
3
??? 420 ? + 176400 ??
10. L
1
= ?, L
0
= 10 m, ? = 1 × 10
–5
/°C, t= 35
L
1
= L
0
(1 + ?t) = 10(1 + 10
–5
× 35) = 10 + 35 × 10
–4
= 10.0035m
11. t
1
= 20°C, t
2
= 10°C, L
1
= 1cm = 0.01 m, L
2
=?
?
steel
= 1.1 × 10
–5
/°C
L
2
= L
1
(1 + ?
steel
?T) = 0.01(1 + 101 × 10
–5
× 10) = 0.01 + 0.01 × 1.1 × 10
–4
= 10
4
× 10
–6
+ 1.1 × 10
–6
= 10
–6
(10000 + 1.1) = 10001.1
=1.00011 × 10
–2
m = 1.00011 cm
12. L
0
= 12 cm, ? = 11 × 10
–5
/°C
tw = 18°C ts = 48°C
Lw = L
0
(1 + ?tw) = 12 (1 + 11 × 10
–5
× 18) = 12.002376 m
Ls = L
0
(1 + ?ts) = 12 (1 + 11 × 10
–5
× 48) = 12.006336 m ?
?L ????12.006336 – 12.002376 = 0.00396 m ? 0.4cm ?
13. d
1
= 2 cm = 2 × 10
–2
t
1
= 0°C, t
2
= 100°C
?
al
= 2.3 × 10
–5
/°C
d
2
= d
1
(1 + ??t) = 2 × 10
–2
(1 + 2.3 × 10
–5
10
2
)
= 0.02 + 0.000046 = 0.020046 m = 2.0046 cm
14. L
st
= L
Al
at 20°C ?
Al
= 2.3 × 10
–5
/°C
So, Lo
st
(1 – ?
st
× 20) = Lo
Al
(1 – ?
AI
× 20) ?
st
= 1.1 × 10
–5
/°C
(a) ?
Al
st
Lo
Lo
=
) 20 1 (
) 20 1 (
st
Al
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
(b) ?
Al 40
st 40
Lo
Lo
=
) 40 1 (
) 40 1 (
st
AI
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
=
273
10 10 3 . 2 1
Lo
Lo
5
st
Al
? ? ?
?
?
=
00044 . 1
00092 . 1 99977 . 0 ?
= 1.0002496 ˜1.00025
St 100
Al 100
Lo
Lo
=
) 100 1 (
) 100 1 (
st
Al
? ? ?
? ? ?
=
00011 . 1
00092 . 1 99977 . 0 ?
= 1.00096
15. (a) Length at 16°C = L
L = ? T
1
=16°C, T
2
= 46°C
? = 1.1 × 10
–5
/°C
?L = L ??? = L × 1.1 × 10
–5
× 30
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
2
L
?
?
?
?
?
?
?
? ? ?
= 1.1 × 10
–5
× 30 × 100% = 0.033%
(b) T
2
= 6°C
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
L
L
?
?
?
?
?
?
?
? ? ?
= – 1.1 × 10
–5
× 10 × 100 = – 0.011%
23.Heat and Temperature
23.3
16. T
1
= 20°C, ?L = 0.055mm = 0.55 × 10
–3
m
t
2
= ? ?
st
= 11 × 10
–6
/°C
We know,
?L = L
0
??T
In our case,
0.055 × 10
–3
= 1 × 1.1 I 10
–6
× (T
1
+T
2
)
0.055 = 11 × 10
–3
× 20 ± 11 × 10
–3
× T
2
T
2
= 20 + 5 = 25°C or 20 – 5 = 15°C
The expt. Can be performed from 15 to 25°C
17. ƒ
0°C
=0.098 g/m
3
, ƒ
4°C
= 1 g/m
3
ƒ
0°C
=
T 1
ƒ
C 4
? ? ?
?
? 0.998 =
4 1
1
? ? ?
? 1 + 4 ? =
998 . 0
1
? 4 + ? = 1
998 . 0
1
? ? ? = 0.0005 ˜ 5 × 10
–4
As density decreases ? = –5 × 10
-4
18. Iron rod Aluminium rod
L
Fe
L
Al
?
Fe
= 12 × 10
–8
/°C ?
Al
= 23 × 10
–8
/°C
Since the difference in length is independent of temp. Hence the different always remains constant.
L ?
Fe
= L
Fe
(1 + ?
Fe
× ?T) …(1)
L ?
Al
= L
Al
(1 + ?
Al
× ?T) …(2)
L ?
Fe
– L ?
Al
= L
Fe
– L
Al
+ L
Fe
× ?
Fe
× ?T – L
Al
× ?
Al
× ?T
Al
Fe
L
L
=
Fe
Al
?
?
=
12
23
= 23 : 12
19. g
1
= 9.8 m/s
2
, g
2
= 9.788 m/s
2
T
1
=
1
1
g
l
2 ? T
2
=
2
2
g
l
2 ? =
g
) T 1 ( l
2
1
? ?
?
?
Steel
= 12 × 10
–6
/°C
T
1
= 20°C T
2
= ?
T
1
= T
2
?
1
1
g
l
2 ? =
2
1
g
) T 1 ( l
2
? ?
? ?
1
1
g
l
=
2
1
g
) T 1 ( l ? ?
?
8 . 9
1
=
788 . 9
T 10 12 1
6
? ? ? ?
?
?
8 . 9
788 . 9
= 1+ 12 × 10
–6
× ?T
? 1
8 . 9
788 . 9
? = 12 × 10
–6
?T ? ?T =
6
10 12
00122 . 0
?
?
?
? T
2
– 20 = – 101.6 ? T
2
= – 101.6 + 20 = – 81.6 ˜ – 82°C ?
20. Given
d
St
= 2.005 cm, d
Al
= 2.000 cm
?
S
= 11 × 10
–6
/°C ?
Al
= 23 × 10
–6
/°C
d ?s = 2.005 (1+ ?
s
?T) (where ?T is change in temp.)
? d ?s = 2.005 + 2.005 × 11 × 10
–6
?T
d ?
Al
= 2(1+ ?
Al
?T) = 2 + 2 × 23 × 10
–6
?T
The two will slip i.e the steel ball with fall when both the
diameters become equal.
So,
? 2.005 + 2.005 × 11 × 10
–6
?T = 2 + 2 × 23 × 10
–6
?T
? (46 – 22.055)10
-6
× ?T = 0.005
? ?T =
945 . 23
10 005 . 0
6
?
= 208.81
Aluminium
Steel
23.Heat and Temperature
23.4
Now ?T = T
2
–T
1
= T
2
–10°C [ ? T
1
= 10°C given]
?T
2
= ?T + T
1
= 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)
? l =
99904 . 0
99964 . 0 20 ?
= 20.012 cm
Let initial breadth of aluminium = b
b(1 – ?
Al
?T) = 30(1 – ?
0
??)
?b =
) 40 10 24 1 (
) 40 10 9 1 ( 30
6
6
? ? ?
? ? ? ?
?
?
=
99904 . 0
99964 . 0 30 ?
= 30.018 cm
22. V
g
= 1000 CC, T
1
= 20°C
V
Hg
= ? ?
Hg
= 1.8 × 10
–4
/°C
?
g
= 9 × 10
–6
/°C
?T remains constant
Volume of remaining space = V ?
g
– V ?
Hg
Now
V ?
g
= V
g
(1 + ?
g
?T) …(1) ?
V ?
Hg
= V
Hg
(1 + ?
Hg
?T) …(2) ?
Subtracting (2) from (1)
V ?
g
– V ?
Hg
= V
g
– V
Hg
+ V
g
?
g
?T – V
Hg
?
Hg
?T
?
Hg
g
V
V
=
g
Hg
?
?
?
Hg
V
1000
=
6
4
10 9
10 8 . 1
?
?
?
?
? V
HG
=
4
3
10 8 . 1
10 9
?
?
?
?
= 500 CC. ?
23. Volume of water = 500cm
3
Area of cross section of can = 125 m
2
Final Volume of water
= 500(1 + ???) = 500[1 + 3.2 × 10
–4
× (80 – 10)] = 511.2 cm
3
The aluminium vessel expands in its length only so area expansion of base cab be neglected.
Increase in volume of water = 11.2 cm
3
Considering a cylinder of volume = 11.2 cm
3
Height of water increased =
125
2 . 11
= 0.089 cm
24. V
0
= 10 × 10× 10 = 1000 CC
?T = 10°C, V ?
HG
– V ?
g
= 1.6 cm
3
?
g
= 6.5 × 10
–6
/°C, ?
Hg
= ?, ?
g
= 3 × 6.5 × 10
–6
/°C
V ?
Hg
= v
HG
(1 + ?
Hg
?T) …(1)
V ?
g
= v
g
(1 + ?
g
?T) …(2)
V ?
Hg
– V ?
g
= V
Hg
–V
g
+ V
Hg
?
Hg
?T – V
g
?
g
?T
? 1.6 = 1000 × ?
Hg
× 10 – 1000 × 6.5 × 3 × 10
–6
× 10
? ?
Hg
=
10000
10 3 3 . 6 6 . 1
2 ?
? ? ?
= 1.789 × 10
–4
? 1.8 × 10
–4
/°C ?
25. ƒ
?
= 880 Kg/m
3
, ƒ
b
= 900 Kg/m
3
T
1
= 0°C, ?
?
= 1.2 × 10
–3
/°C,
?
b
= 1.5 × 10
–3
/°C
The sphere begins t sink when,
(mg)
sphere
= displaced water
Page 5
23.1
CHAPTER – 23
HEAT AND TEMPERATURE
EXERCISES
1. Ice point = 20° (L
0
) L
1
= 32°
Steam point = 80° (L
100
)
T = 100
L L
L L
0 100
0 1
?
?
?
= 100
20 80
20 32
?
?
?
= 20°C
2. P
tr
= 1.500 × 10
4
Pa
P = 2.050 × 10
4
Pa
We know, For constant volume gas Thermometer
T = 16 . 273
P
P
tr
? K = 16 . 273
10 500 . 1
10 050 . 2
4
4
?
?
?
= 373.31
3. Pressure Measured at M.P = 2.2 × Pressure at Triple Point
T = 16 . 273
P
P
tr
? = 16 . 273
P
P 2 . 2
tr
tr
?
?
= 600.952 K ? 601 K
4. P
tr
= 40 × 10
3
Pa, P = ?
T = 100°C = 373 K, T = 16 . 273
P
P
tr
? K
? P =
16 . 273
P T
tr
?
=
16 . 273
10 49 373
3
? ?
= 54620 Pa = 5.42 × 10
3
pa ˜ 55 K Pa
5. P
1
= 70 K Pa, P
2
= ?
T
1
= 273 K, T
2
= 373K
T = 16 . 273
P
P
tr
1
? ? 273 = 16 . 273
P
10 70
tr
3
?
?
? P
tr
273
10 16 . 273 70
3
? ?
T
2
= 16 . 273
P
P
tr
2
? ? 373 =
3
2
10 16 . 273 70
273 P
? ?
?
? P
2
=
273
10 70 373
3
? ?
= 95.6 K Pa
6. P
ice point
= P
0°
= 80 cm of Hg
P
steam point
= P
100°
90 cm of Hg
P
0
= 100 cm
t = ? ?
?
?
100
P P
P P
0 100
0
= 100
100 90
100 80
?
?
?
= 200°C
7. T ? =
0
T
V V
V
? ?
T
0
= 273,
V = 1800 CC, V ? = 200 CC
T ? = 273
1600
1800
? = 307.125 ? 307
8. R
t
= 86 ?; R
0°
= 80 ?; R
100°
= 90 ?
t = 100
R R
R R
0 100
0 t
?
?
?
= 100
80 90
80 86
?
?
?
= 60°C
9. R at ice point (R
0
) = 20 ?
R at steam point (R
100
) = 27.5 ?
R at Zinc point (R
420
) = 50 ?
R
?
= R
0
(1+ ? ? + ? ?
2
)
? R
100
= R
0
+ R
0
?? +R
0
??
2
?
0
0 100
R
R R ?
= ? ? + ? ?
2
23.Heat and Temperature
23.2
?
20
20 5 . 27 ?
= ? × 100 + ? × 10000
?
20
5 . 7
= 100 ? + 10000 ?
R
420
= R
0
(1+ ? ? + ? ?
2
) ?
0
0
R
R 50 ?
= ? ? + ??
2
?
20
20 50 ?
= 420 × ? + 176400 × ?? ? ??
2
3
??? 420 ? + 176400 ??
??
20
5 . 7
= 100 ? + 10000 ? ?? ? ? ??
2
3
??? 420 ? + 176400 ??
10. L
1
= ?, L
0
= 10 m, ? = 1 × 10
–5
/°C, t= 35
L
1
= L
0
(1 + ?t) = 10(1 + 10
–5
× 35) = 10 + 35 × 10
–4
= 10.0035m
11. t
1
= 20°C, t
2
= 10°C, L
1
= 1cm = 0.01 m, L
2
=?
?
steel
= 1.1 × 10
–5
/°C
L
2
= L
1
(1 + ?
steel
?T) = 0.01(1 + 101 × 10
–5
× 10) = 0.01 + 0.01 × 1.1 × 10
–4
= 10
4
× 10
–6
+ 1.1 × 10
–6
= 10
–6
(10000 + 1.1) = 10001.1
=1.00011 × 10
–2
m = 1.00011 cm
12. L
0
= 12 cm, ? = 11 × 10
–5
/°C
tw = 18°C ts = 48°C
Lw = L
0
(1 + ?tw) = 12 (1 + 11 × 10
–5
× 18) = 12.002376 m
Ls = L
0
(1 + ?ts) = 12 (1 + 11 × 10
–5
× 48) = 12.006336 m ?
?L ????12.006336 – 12.002376 = 0.00396 m ? 0.4cm ?
13. d
1
= 2 cm = 2 × 10
–2
t
1
= 0°C, t
2
= 100°C
?
al
= 2.3 × 10
–5
/°C
d
2
= d
1
(1 + ??t) = 2 × 10
–2
(1 + 2.3 × 10
–5
10
2
)
= 0.02 + 0.000046 = 0.020046 m = 2.0046 cm
14. L
st
= L
Al
at 20°C ?
Al
= 2.3 × 10
–5
/°C
So, Lo
st
(1 – ?
st
× 20) = Lo
Al
(1 – ?
AI
× 20) ?
st
= 1.1 × 10
–5
/°C
(a) ?
Al
st
Lo
Lo
=
) 20 1 (
) 20 1 (
st
Al
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
(b) ?
Al 40
st 40
Lo
Lo
=
) 40 1 (
) 40 1 (
st
AI
? ? ?
? ? ?
=
20 10 1 . 1 1
20 10 3 . 2 1
5
5
? ? ?
? ? ?
?
?
=
99978 . 0
99954 . 0
= 0.999
=
273
10 10 3 . 2 1
Lo
Lo
5
st
Al
? ? ?
?
?
=
00044 . 1
00092 . 1 99977 . 0 ?
= 1.0002496 ˜1.00025
St 100
Al 100
Lo
Lo
=
) 100 1 (
) 100 1 (
st
Al
? ? ?
? ? ?
=
00011 . 1
00092 . 1 99977 . 0 ?
= 1.00096
15. (a) Length at 16°C = L
L = ? T
1
=16°C, T
2
= 46°C
? = 1.1 × 10
–5
/°C
?L = L ??? = L × 1.1 × 10
–5
× 30
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
2
L
?
?
?
?
?
?
?
? ? ?
= 1.1 × 10
–5
× 30 × 100% = 0.033%
(b) T
2
= 6°C
% of error = % 100
L
L
?
?
?
?
?
?
?
?
= % 100
L
L
?
?
?
?
?
?
?
? ? ?
= – 1.1 × 10
–5
× 10 × 100 = – 0.011%
23.Heat and Temperature
23.3
16. T
1
= 20°C, ?L = 0.055mm = 0.55 × 10
–3
m
t
2
= ? ?
st
= 11 × 10
–6
/°C
We know,
?L = L
0
??T
In our case,
0.055 × 10
–3
= 1 × 1.1 I 10
–6
× (T
1
+T
2
)
0.055 = 11 × 10
–3
× 20 ± 11 × 10
–3
× T
2
T
2
= 20 + 5 = 25°C or 20 – 5 = 15°C
The expt. Can be performed from 15 to 25°C
17. ƒ
0°C
=0.098 g/m
3
, ƒ
4°C
= 1 g/m
3
ƒ
0°C
=
T 1
ƒ
C 4
? ? ?
?
? 0.998 =
4 1
1
? ? ?
? 1 + 4 ? =
998 . 0
1
? 4 + ? = 1
998 . 0
1
? ? ? = 0.0005 ˜ 5 × 10
–4
As density decreases ? = –5 × 10
-4
18. Iron rod Aluminium rod
L
Fe
L
Al
?
Fe
= 12 × 10
–8
/°C ?
Al
= 23 × 10
–8
/°C
Since the difference in length is independent of temp. Hence the different always remains constant.
L ?
Fe
= L
Fe
(1 + ?
Fe
× ?T) …(1)
L ?
Al
= L
Al
(1 + ?
Al
× ?T) …(2)
L ?
Fe
– L ?
Al
= L
Fe
– L
Al
+ L
Fe
× ?
Fe
× ?T – L
Al
× ?
Al
× ?T
Al
Fe
L
L
=
Fe
Al
?
?
=
12
23
= 23 : 12
19. g
1
= 9.8 m/s
2
, g
2
= 9.788 m/s
2
T
1
=
1
1
g
l
2 ? T
2
=
2
2
g
l
2 ? =
g
) T 1 ( l
2
1
? ?
?
?
Steel
= 12 × 10
–6
/°C
T
1
= 20°C T
2
= ?
T
1
= T
2
?
1
1
g
l
2 ? =
2
1
g
) T 1 ( l
2
? ?
? ?
1
1
g
l
=
2
1
g
) T 1 ( l ? ?
?
8 . 9
1
=
788 . 9
T 10 12 1
6
? ? ? ?
?
?
8 . 9
788 . 9
= 1+ 12 × 10
–6
× ?T
? 1
8 . 9
788 . 9
? = 12 × 10
–6
?T ? ?T =
6
10 12
00122 . 0
?
?
?
? T
2
– 20 = – 101.6 ? T
2
= – 101.6 + 20 = – 81.6 ˜ – 82°C ?
20. Given
d
St
= 2.005 cm, d
Al
= 2.000 cm
?
S
= 11 × 10
–6
/°C ?
Al
= 23 × 10
–6
/°C
d ?s = 2.005 (1+ ?
s
?T) (where ?T is change in temp.)
? d ?s = 2.005 + 2.005 × 11 × 10
–6
?T
d ?
Al
= 2(1+ ?
Al
?T) = 2 + 2 × 23 × 10
–6
?T
The two will slip i.e the steel ball with fall when both the
diameters become equal.
So,
? 2.005 + 2.005 × 11 × 10
–6
?T = 2 + 2 × 23 × 10
–6
?T
? (46 – 22.055)10
-6
× ?T = 0.005
? ?T =
945 . 23
10 005 . 0
6
?
= 208.81
Aluminium
Steel
23.Heat and Temperature
23.4
Now ?T = T
2
–T
1
= T
2
–10°C [ ? T
1
= 10°C given]
?T
2
= ?T + T
1
= 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)
? l =
99904 . 0
99964 . 0 20 ?
= 20.012 cm
Let initial breadth of aluminium = b
b(1 – ?
Al
?T) = 30(1 – ?
0
??)
?b =
) 40 10 24 1 (
) 40 10 9 1 ( 30
6
6
? ? ?
? ? ? ?
?
?
=
99904 . 0
99964 . 0 30 ?
= 30.018 cm
22. V
g
= 1000 CC, T
1
= 20°C
V
Hg
= ? ?
Hg
= 1.8 × 10
–4
/°C
?
g
= 9 × 10
–6
/°C
?T remains constant
Volume of remaining space = V ?
g
– V ?
Hg
Now
V ?
g
= V
g
(1 + ?
g
?T) …(1) ?
V ?
Hg
= V
Hg
(1 + ?
Hg
?T) …(2) ?
Subtracting (2) from (1)
V ?
g
– V ?
Hg
= V
g
– V
Hg
+ V
g
?
g
?T – V
Hg
?
Hg
?T
?
Hg
g
V
V
=
g
Hg
?
?
?
Hg
V
1000
=
6
4
10 9
10 8 . 1
?
?
?
?
? V
HG
=
4
3
10 8 . 1
10 9
?
?
?
?
= 500 CC. ?
23. Volume of water = 500cm
3
Area of cross section of can = 125 m
2
Final Volume of water
= 500(1 + ???) = 500[1 + 3.2 × 10
–4
× (80 – 10)] = 511.2 cm
3
The aluminium vessel expands in its length only so area expansion of base cab be neglected.
Increase in volume of water = 11.2 cm
3
Considering a cylinder of volume = 11.2 cm
3
Height of water increased =
125
2 . 11
= 0.089 cm
24. V
0
= 10 × 10× 10 = 1000 CC
?T = 10°C, V ?
HG
– V ?
g
= 1.6 cm
3
?
g
= 6.5 × 10
–6
/°C, ?
Hg
= ?, ?
g
= 3 × 6.5 × 10
–6
/°C
V ?
Hg
= v
HG
(1 + ?
Hg
?T) …(1)
V ?
g
= v
g
(1 + ?
g
?T) …(2)
V ?
Hg
– V ?
g
= V
Hg
–V
g
+ V
Hg
?
Hg
?T – V
g
?
g
?T
? 1.6 = 1000 × ?
Hg
× 10 – 1000 × 6.5 × 3 × 10
–6
× 10
? ?
Hg
=
10000
10 3 3 . 6 6 . 1
2 ?
? ? ?
= 1.789 × 10
–4
? 1.8 × 10
–4
/°C ?
25. ƒ
?
= 880 Kg/m
3
, ƒ
b
= 900 Kg/m
3
T
1
= 0°C, ?
?
= 1.2 × 10
–3
/°C,
?
b
= 1.5 × 10
–3
/°C
The sphere begins t sink when,
(mg)
sphere
= displaced water
23.Heat and Temperature
23.5
? Vƒ ?
?
g = Vƒ ?
b
g
??
? ? ? ?
?
?
1
ƒ
???
? ? ? ?
b
b
1
ƒ
?
?
? ? ? ?
?3
10 2 . 1 1
880
=
? ? ? ?
?3
10 5 . 1 1
900
? 880 + 880 × 1.5 × 10
–3
( ??) = 900 + 900 × 1.2 × 10
–3
( ??)
? (880 × 1.5 × 10
–3
– 900 × 1.2 × 10
–3
) ( ??) = 20
? (1320 – 1080) × 10
–3
( ??) = 20
? ?? = 83.3°C ˜ 83°C
26. ?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.
27. ?
1
= 20°C, ?
2
= 50°C
?
steel
= 1.2 × 10
–5
/°C
Longitudinal stain = ?
Stain =
L
L ?
=
L
L ? ? ?
= ? ??
= 1.2 × 10
–5
× (50 – 20) = 3.6 × 10
–4
28. A = 0.5mm
2
= 0.5 × 10
–6
m
2
T
1
= 20°C, T
2
= 0°C
?
s
= 1.2 × 10
–5
/°C, Y = 2 × 2 × 10
11
N/m
2
Decrease in length due to compression = L ??? …(1)
Y =
Strain
Stress
=
L
L
A
F
?
? ? ?L =
AY
FL
…(2)
Tension is developed due to (1) & (2)
Equating them,
L ??? =
AY
FL
? F = ???AY
= 1.2 × 10
-5
× (20 – 0) × 0.5 × 10
–5
2 × 10
11
= 24 N
?
29. ?
1
= 20°C, ?
2
= 100°C
A = 2mm
2
= 2 × 10
–6
m
2
?
steel
= 12 × 10
–6
/°C, Y
steel
= 2 × 10
11
N/m
2
Force exerted on the clamps = ?
Strain
A
F
?
?
?
?
?
?
= Y ? F = L
L
L Y
?
? ?
=
L
A YL ? ? ?
= YA ???
= 2 × 10
11
× 2 × 10
–6
× 12 × 10
–6
× 80 = 384 N ?
30. Let the final length of the system at system of temp. 0°C = l
?
Initial length of the system = l
0
When temp. changes by ?.
Strain of the system =
?
?
?
?
?
0
1
But the total strain of the system =
system of ulusof mod s ' young total
system of stress total
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
Steel
Steel
Aluminium
????? ?
1 m
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