HC Verma Solutions: Chapter 23 - Heat & Temperature

# HC Verma Solutions: Chapter 23 - Heat & Temperature | Physics Class 11 - NEET PDF Download

``` 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
```

## Physics Class 11

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## FAQs on HC Verma Solutions: Chapter 23 - Heat & Temperature - Physics Class 11 - NEET

 1. What is heat and temperature?
Ans. Heat is the transfer of thermal energy between two objects or systems due to a temperature difference. Temperature, on the other hand, is a measure of the average kinetic energy of the particles in a substance.
 2. How is heat transferred between objects?
Ans. Heat can be transferred through three main mechanisms: conduction, convection, and radiation. Conduction occurs when heat is transferred through direct contact between two objects. Convection involves the transfer of heat through the movement of fluids or gases. Radiation is the transfer of heat through electromagnetic waves.
 3. How does temperature affect the behavior of objects?
Ans. Temperature affects the behavior of objects in various ways. As temperature increases, the kinetic energy of the particles in an object also increases, causing the object to expand. On the other hand, a decrease in temperature leads to a contraction of the object. Temperature also determines the direction of heat flow, with heat transferring from objects at higher temperatures to those at lower temperatures.
 4. What is specific heat capacity?
Ans. Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is a property that varies for different substances and is a measure of how effectively a substance can store heat energy.
 5. How can temperature be measured?
Ans. Temperature can be measured using various instruments such as thermometers. Common types of thermometers include mercury thermometers, alcohol thermometers, and digital thermometers. These instruments make use of the expansion or contraction of a liquid or metal to indicate the temperature.

## Physics Class 11

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