Thermal Expansion, Calorimetry Practice Questions - DPP for NEET

 Page 1


1. (c)  Due to volume expansion of both mercury and flask,
the change in volume of mercury relative to flask is
given by 
0
[ ] [ 3]
L g mg
VVV D = g - g Dq = g - a Dq
= 50[180  × 10
–6
 – 3 × 9 × 10
–6
] (38 – 18)
= 0.153 cc
2. (a) g
real
 = g
app.
 + g
vessel
So (g
app.
 + g
vessel
)
glass
= (g
app.
 + g
vessel
)
steel
Þ 153 × 10
–6
 + (g
vessel
)
glass
= (144 × 10
–6
 + g
vessel
)
steel
Further , (g
vessel
)
steel 
= 3a = 3 × (12 × 10
–6
) = 36 × 10
–6
/°C
Þ 153 × 10
–6
 + (g
vessel
)
glass 
= 144 × 10
–6
 + 36 × 10
–6
Þ (g
vessel
)
glass 
= 3a = 27 × 10
–6
/°C
Þa = 9 × 10
–6
/°C
3. (c) Initial diameter of tyre = (100 – 6) mm = 994 mm,
So, initial radius of tyre R = 
994
497mm
2
=
and change in diameter DD = 6 mm, so
6
3mm
2
R D ==
After increasing temperature by Dq, tyre will fit onto
wheel
Increment in the length (circumference) of the iron tyre
3
LLL
g
D = ´ a ´ Dq = ´ ´ Dq [As
3
g
a´ ]
22
3
RR
g æö
pD = p Dq
ç÷
èø
Dq
Þ
 
5
3 33
3.6 10 497
R
R
-
D´
=
g
´´
500C ÞDq=°
4. (b)
0
LL D = aDq
Rod A : 0.075 = 20 × a
A
 × 100
Þ
6
75
10/
2
A
C
-
a= ´°
Rod B : 0.045 = 20 × a
B
 × 100
Þ
6
45
10/
2
B
C
-
a= ´°
For composite rod: x cm of A and (20 – x) cm of B we
have
A B A
a
B
a
20 cm
x (20 – ) x
0.060 = x a
A
 × 100 + (20 – x) a
B
 × 100
= 
66
75 45
10 100 (20 ) 10 100
22
xx
--
éù
´´ + -´ ´´
êú
ëû
On solving we get x = 10 cm.
5. (b) Due to volume expansion of both liquid and vessel,
the change in volume of liquid relative to container is
given by 
0
[]
Lg
VV D = g - g Dq
Given V
0
 = 1000 cc, a
g
 = 0.1× 10
–4
/°C
\
44
3 3 0.1 10 / 0.3 10 /
gg
CC
--
g=a=´ ´°= ´°
\
44
1000[1.82 10 0.3 10 ] 100 15.2 V cc
--
D= ´ - ´ ´=
6. (b) g
 r
 = g
 a
 + g
 v
 ; where g
 r
 = coefficient of real expansion,
g
 a
 = coefficient of apparent expansion and
g
 v
 = coefficient of expansion of vessel.
For copper 
r Cu
C3 C3A g =+a =+
For silver 
r Ag
S3 g = +a
Ag Ag
C S 3A
C3A S3
3
-+
Þ + = +a Þa=
7. (d)
0
(1) VV = + gDq
Þ
Change in volume
V – V
0
 = DV = A.Dl = V
0
gDq
Þ
65
0
4
. 10 18 10 (100 0)
0.004 10
V
l
A
--
-
Dq ´ ´ ´-
D==
´
= 45 × 10
–3
m = 4.5 cm
8. (b) Loss of weight at 27°C is
= 46 – 30 = 16 = V
1
 × 1.24 r
1
 × g ...(i)
Loss of weight at 42°C is
= 46 – 30.5 = 15.5 = V
2
 × 1.2 r
1
 × g ...(ii)
Now dividing (i) by (ii), we get
1
2
16 1.24
15.5 1.2
V
V
=´
But 
2
21
1
15.5 1.24
1 3 ( ) 1.001042
16 1.2
V
tt
V
´
=+a-==
´
Þ3a (42° – 27°) = 0.001042
Þ a = 2.316 × 10
–5
/°C
9. (b) Heat lost in t sec = mL or heat lost per sec = .
mL
t
 This
must be the heat supplied for keeping the substance in
molten state per sec.
\
mL Pt
P orL
tm
==
10. (b) Initially ice will absorb heat to raise it's temperature to
0°C then it's melting takes place
If m
1
 = Initial mass of ice, m
1
' = Mass of ice that melts
and m
W
 = Initial mass of water
Page 2


1. (c)  Due to volume expansion of both mercury and flask,
the change in volume of mercury relative to flask is
given by 
0
[ ] [ 3]
L g mg
VVV D = g - g Dq = g - a Dq
= 50[180  × 10
–6
 – 3 × 9 × 10
–6
] (38 – 18)
= 0.153 cc
2. (a) g
real
 = g
app.
 + g
vessel
So (g
app.
 + g
vessel
)
glass
= (g
app.
 + g
vessel
)
steel
Þ 153 × 10
–6
 + (g
vessel
)
glass
= (144 × 10
–6
 + g
vessel
)
steel
Further , (g
vessel
)
steel 
= 3a = 3 × (12 × 10
–6
) = 36 × 10
–6
/°C
Þ 153 × 10
–6
 + (g
vessel
)
glass 
= 144 × 10
–6
 + 36 × 10
–6
Þ (g
vessel
)
glass 
= 3a = 27 × 10
–6
/°C
Þa = 9 × 10
–6
/°C
3. (c) Initial diameter of tyre = (100 – 6) mm = 994 mm,
So, initial radius of tyre R = 
994
497mm
2
=
and change in diameter DD = 6 mm, so
6
3mm
2
R D ==
After increasing temperature by Dq, tyre will fit onto
wheel
Increment in the length (circumference) of the iron tyre
3
LLL
g
D = ´ a ´ Dq = ´ ´ Dq [As
3
g
a´ ]
22
3
RR
g æö
pD = p Dq
ç÷
èø
Dq
Þ
 
5
3 33
3.6 10 497
R
R
-
D´
=
g
´´
500C ÞDq=°
4. (b)
0
LL D = aDq
Rod A : 0.075 = 20 × a
A
 × 100
Þ
6
75
10/
2
A
C
-
a= ´°
Rod B : 0.045 = 20 × a
B
 × 100
Þ
6
45
10/
2
B
C
-
a= ´°
For composite rod: x cm of A and (20 – x) cm of B we
have
A B A
a
B
a
20 cm
x (20 – ) x
0.060 = x a
A
 × 100 + (20 – x) a
B
 × 100
= 
66
75 45
10 100 (20 ) 10 100
22
xx
--
éù
´´ + -´ ´´
êú
ëû
On solving we get x = 10 cm.
5. (b) Due to volume expansion of both liquid and vessel,
the change in volume of liquid relative to container is
given by 
0
[]
Lg
VV D = g - g Dq
Given V
0
 = 1000 cc, a
g
 = 0.1× 10
–4
/°C
\
44
3 3 0.1 10 / 0.3 10 /
gg
CC
--
g=a=´ ´°= ´°
\
44
1000[1.82 10 0.3 10 ] 100 15.2 V cc
--
D= ´ - ´ ´=
6. (b) g
 r
 = g
 a
 + g
 v
 ; where g
 r
 = coefficient of real expansion,
g
 a
 = coefficient of apparent expansion and
g
 v
 = coefficient of expansion of vessel.
For copper 
r Cu
C3 C3A g =+a =+
For silver 
r Ag
S3 g = +a
Ag Ag
C S 3A
C3A S3
3
-+
Þ + = +a Þa=
7. (d)
0
(1) VV = + gDq
Þ
Change in volume
V – V
0
 = DV = A.Dl = V
0
gDq
Þ
65
0
4
. 10 18 10 (100 0)
0.004 10
V
l
A
--
-
Dq ´ ´ ´-
D==
´
= 45 × 10
–3
m = 4.5 cm
8. (b) Loss of weight at 27°C is
= 46 – 30 = 16 = V
1
 × 1.24 r
1
 × g ...(i)
Loss of weight at 42°C is
= 46 – 30.5 = 15.5 = V
2
 × 1.2 r
1
 × g ...(ii)
Now dividing (i) by (ii), we get
1
2
16 1.24
15.5 1.2
V
V
=´
But 
2
21
1
15.5 1.24
1 3 ( ) 1.001042
16 1.2
V
tt
V
´
=+a-==
´
Þ3a (42° – 27°) = 0.001042
Þ a = 2.316 × 10
–5
/°C
9. (b) Heat lost in t sec = mL or heat lost per sec = .
mL
t
 This
must be the heat supplied for keeping the substance in
molten state per sec.
\
mL Pt
P orL
tm
==
10. (b) Initially ice will absorb heat to raise it's temperature to
0°C then it's melting takes place
If m
1
 = Initial mass of ice, m
1
' = Mass of ice that melts
and m
W
 = Initial mass of water
64
DPP/ P 22
By Law of mixture
Heat gained by ice = Heat lost by water Þ m
1
 × (20) +
m
1
' × L = m
W
c
W 
[20]
Þ 2 × 0.5 (20) + m
1
' × 80 = 5 × 1 × 20
Þ m
1
'  = 1 kg
So final mass of water = Initial mass of water + Mass
of ice that melts = 5 + 1 = 6 kg.
11. (a) If mass of the bullet is m gm,
then total heat required for bullet to just melt down
Q
1
 = mcDq + mL = m × 0.03 (327 – 27) + m ×6
= 15m cal = (15m × 4.2)J
Now when bullet is stopped by the obstacle, the loss in
its mechanical energy = 
32
1
( 10)
2
m vJ
-
´
(As m gm = m × 10
–3
kg)
As 25% of this energy is absorbed by the obstacle,
23 23
2
7513
10 10
100 2 8
Q mv mv J
--
= ´ ´ =´
Now the bullet will melt if 
21
QQ ³
i.e.
23
min
3
10 15 4.2 410 m/s
8
mv mv
-
´ ³ ´ Þ=
 12. (c) Heat gain = heat lost
3
(16 12) (19 16)
4
A
AB
B
C
CC
C
- = - Þ=
and
5
(23 19) (28 23)
4
B
BC
C
C
CC
C
- = - Þ=
Þ
15
16
A
C
C
C
=
....(i)
If q is the temperature when A and C are mixed then,
28
( 12) (28 )
12
A
AC
C
C
CC
C
-q
q- = -qÞ=
q-
...(ii)
On solving equation (i) and (ii) q = 20.2°C
13. (a) Same amount of heat is supplied to copper and water
so m
c
c
c
Dq
c
 = m
W
c
W
Dq
W
Þ
3
3
() 50 10 420 10
5
10 10 4200
ccc
W
WW
mc
C
mc
-
-
Dq ´ ´´
Dq = = =°
´´
14. (b) Heat lost by hot water = Heat gained by cold water in
beaker + Heat absorbed by beaker
Þ
440 (92 – q) = 200 × (q – 20) + 20 × (q – 20)
Þ
q = 68°C
15. (a)
16. (b) Firstly the temperature of bullet rises up to melting
point, then it melts. Hence according to W = JQ.
Þ
 
2
1
.[ . . ] [ (475 25) ]
2
mv J m c mL J mS mL = Dq+ = -+
Þ
2
(475 25)
2
mv
mS mL
J
- +=
17. (b) Suppose m kg of ice melts then by using
() () Joules Joules
WH =
Þ
 Mgh = mL
Þ
 3.5 × 10 × 2000 = m × 3.5 × 10
5
Þ
 m = 0.2 kg = 200 gm
18. (d) Coefficient of volume expansion
4 12
() (10 9.7)
3 10
. .( ) 10 (100 0) T
-
r -r Dr-
g= = = =´
rD rDq ´-
Hence, coefficient of linear expansion
4
10/
3
C
-
g
a==°
19. (b) As we know
g 
real
 = g 
app.
 + g 
vessel
Þ g 
app.
 = g 
glycerine
 – g 
glass
= 0.000597 – 0.000027
= 0.00057/°C
20. (a)
0
1000
()
100
()
t
PP
tC
PP
-
= ´°
-
(60 50)
100 25
(90 50)
C
-
= ´ =°
-
21. (c) Since spec ific heat = 0.6 kc al/gm × °C
= 0.6 cal/gm × °C
From graph it is clear that in a minute, the temperature
is raised from 0°C to 50°C.
Þ Heat required for a minute
= 50 × 0.6 × 50 = 1500 cal.
Also from graph, boiling point of wax is 200°C.
22. (b) The horizontal parts of the curve, where the system
absorbs heat at constant temperature must depict
changes of state. Here the latent heats are proportional
to lengths of the horizontal parts. In the sloping parts,
specific heat capacity is inversely proportional to the
slopes.
23. (d) Let L
0
 be the initial length of each strip before heating.
Length after heating will be
q
Brass Strip
Copper Strip
R
d
( ) ( )
B 0B
L L1 a?T R d? = + =+
Page 3


1. (c)  Due to volume expansion of both mercury and flask,
the change in volume of mercury relative to flask is
given by 
0
[ ] [ 3]
L g mg
VVV D = g - g Dq = g - a Dq
= 50[180  × 10
–6
 – 3 × 9 × 10
–6
] (38 – 18)
= 0.153 cc
2. (a) g
real
 = g
app.
 + g
vessel
So (g
app.
 + g
vessel
)
glass
= (g
app.
 + g
vessel
)
steel
Þ 153 × 10
–6
 + (g
vessel
)
glass
= (144 × 10
–6
 + g
vessel
)
steel
Further , (g
vessel
)
steel 
= 3a = 3 × (12 × 10
–6
) = 36 × 10
–6
/°C
Þ 153 × 10
–6
 + (g
vessel
)
glass 
= 144 × 10
–6
 + 36 × 10
–6
Þ (g
vessel
)
glass 
= 3a = 27 × 10
–6
/°C
Þa = 9 × 10
–6
/°C
3. (c) Initial diameter of tyre = (100 – 6) mm = 994 mm,
So, initial radius of tyre R = 
994
497mm
2
=
and change in diameter DD = 6 mm, so
6
3mm
2
R D ==
After increasing temperature by Dq, tyre will fit onto
wheel
Increment in the length (circumference) of the iron tyre
3
LLL
g
D = ´ a ´ Dq = ´ ´ Dq [As
3
g
a´ ]
22
3
RR
g æö
pD = p Dq
ç÷
èø
Dq
Þ
 
5
3 33
3.6 10 497
R
R
-
D´
=
g
´´
500C ÞDq=°
4. (b)
0
LL D = aDq
Rod A : 0.075 = 20 × a
A
 × 100
Þ
6
75
10/
2
A
C
-
a= ´°
Rod B : 0.045 = 20 × a
B
 × 100
Þ
6
45
10/
2
B
C
-
a= ´°
For composite rod: x cm of A and (20 – x) cm of B we
have
A B A
a
B
a
20 cm
x (20 – ) x
0.060 = x a
A
 × 100 + (20 – x) a
B
 × 100
= 
66
75 45
10 100 (20 ) 10 100
22
xx
--
éù
´´ + -´ ´´
êú
ëû
On solving we get x = 10 cm.
5. (b) Due to volume expansion of both liquid and vessel,
the change in volume of liquid relative to container is
given by 
0
[]
Lg
VV D = g - g Dq
Given V
0
 = 1000 cc, a
g
 = 0.1× 10
–4
/°C
\
44
3 3 0.1 10 / 0.3 10 /
gg
CC
--
g=a=´ ´°= ´°
\
44
1000[1.82 10 0.3 10 ] 100 15.2 V cc
--
D= ´ - ´ ´=
6. (b) g
 r
 = g
 a
 + g
 v
 ; where g
 r
 = coefficient of real expansion,
g
 a
 = coefficient of apparent expansion and
g
 v
 = coefficient of expansion of vessel.
For copper 
r Cu
C3 C3A g =+a =+
For silver 
r Ag
S3 g = +a
Ag Ag
C S 3A
C3A S3
3
-+
Þ + = +a Þa=
7. (d)
0
(1) VV = + gDq
Þ
Change in volume
V – V
0
 = DV = A.Dl = V
0
gDq
Þ
65
0
4
. 10 18 10 (100 0)
0.004 10
V
l
A
--
-
Dq ´ ´ ´-
D==
´
= 45 × 10
–3
m = 4.5 cm
8. (b) Loss of weight at 27°C is
= 46 – 30 = 16 = V
1
 × 1.24 r
1
 × g ...(i)
Loss of weight at 42°C is
= 46 – 30.5 = 15.5 = V
2
 × 1.2 r
1
 × g ...(ii)
Now dividing (i) by (ii), we get
1
2
16 1.24
15.5 1.2
V
V
=´
But 
2
21
1
15.5 1.24
1 3 ( ) 1.001042
16 1.2
V
tt
V
´
=+a-==
´
Þ3a (42° – 27°) = 0.001042
Þ a = 2.316 × 10
–5
/°C
9. (b) Heat lost in t sec = mL or heat lost per sec = .
mL
t
 This
must be the heat supplied for keeping the substance in
molten state per sec.
\
mL Pt
P orL
tm
==
10. (b) Initially ice will absorb heat to raise it's temperature to
0°C then it's melting takes place
If m
1
 = Initial mass of ice, m
1
' = Mass of ice that melts
and m
W
 = Initial mass of water
64
DPP/ P 22
By Law of mixture
Heat gained by ice = Heat lost by water Þ m
1
 × (20) +
m
1
' × L = m
W
c
W 
[20]
Þ 2 × 0.5 (20) + m
1
' × 80 = 5 × 1 × 20
Þ m
1
'  = 1 kg
So final mass of water = Initial mass of water + Mass
of ice that melts = 5 + 1 = 6 kg.
11. (a) If mass of the bullet is m gm,
then total heat required for bullet to just melt down
Q
1
 = mcDq + mL = m × 0.03 (327 – 27) + m ×6
= 15m cal = (15m × 4.2)J
Now when bullet is stopped by the obstacle, the loss in
its mechanical energy = 
32
1
( 10)
2
m vJ
-
´
(As m gm = m × 10
–3
kg)
As 25% of this energy is absorbed by the obstacle,
23 23
2
7513
10 10
100 2 8
Q mv mv J
--
= ´ ´ =´
Now the bullet will melt if 
21
QQ ³
i.e.
23
min
3
10 15 4.2 410 m/s
8
mv mv
-
´ ³ ´ Þ=
 12. (c) Heat gain = heat lost
3
(16 12) (19 16)
4
A
AB
B
C
CC
C
- = - Þ=
and
5
(23 19) (28 23)
4
B
BC
C
C
CC
C
- = - Þ=
Þ
15
16
A
C
C
C
=
....(i)
If q is the temperature when A and C are mixed then,
28
( 12) (28 )
12
A
AC
C
C
CC
C
-q
q- = -qÞ=
q-
...(ii)
On solving equation (i) and (ii) q = 20.2°C
13. (a) Same amount of heat is supplied to copper and water
so m
c
c
c
Dq
c
 = m
W
c
W
Dq
W
Þ
3
3
() 50 10 420 10
5
10 10 4200
ccc
W
WW
mc
C
mc
-
-
Dq ´ ´´
Dq = = =°
´´
14. (b) Heat lost by hot water = Heat gained by cold water in
beaker + Heat absorbed by beaker
Þ
440 (92 – q) = 200 × (q – 20) + 20 × (q – 20)
Þ
q = 68°C
15. (a)
16. (b) Firstly the temperature of bullet rises up to melting
point, then it melts. Hence according to W = JQ.
Þ
 
2
1
.[ . . ] [ (475 25) ]
2
mv J m c mL J mS mL = Dq+ = -+
Þ
2
(475 25)
2
mv
mS mL
J
- +=
17. (b) Suppose m kg of ice melts then by using
() () Joules Joules
WH =
Þ
 Mgh = mL
Þ
 3.5 × 10 × 2000 = m × 3.5 × 10
5
Þ
 m = 0.2 kg = 200 gm
18. (d) Coefficient of volume expansion
4 12
() (10 9.7)
3 10
. .( ) 10 (100 0) T
-
r -r Dr-
g= = = =´
rD rDq ´-
Hence, coefficient of linear expansion
4
10/
3
C
-
g
a==°
19. (b) As we know
g 
real
 = g 
app.
 + g 
vessel
Þ g 
app.
 = g 
glycerine
 – g 
glass
= 0.000597 – 0.000027
= 0.00057/°C
20. (a)
0
1000
()
100
()
t
PP
tC
PP
-
= ´°
-
(60 50)
100 25
(90 50)
C
-
= ´ =°
-
21. (c) Since spec ific heat = 0.6 kc al/gm × °C
= 0.6 cal/gm × °C
From graph it is clear that in a minute, the temperature
is raised from 0°C to 50°C.
Þ Heat required for a minute
= 50 × 0.6 × 50 = 1500 cal.
Also from graph, boiling point of wax is 200°C.
22. (b) The horizontal parts of the curve, where the system
absorbs heat at constant temperature must depict
changes of state. Here the latent heats are proportional
to lengths of the horizontal parts. In the sloping parts,
specific heat capacity is inversely proportional to the
slopes.
23. (d) Let L
0
 be the initial length of each strip before heating.
Length after heating will be
q
Brass Strip
Copper Strip
R
d
( ) ( )
B 0B
L L1 a?T R d? = + =+
DPP/ P 22
65
( )
C 0C
L L1 a ?T R? = +=
B
C
1 a ?T Rd
R1 a ?T
+ +
Þ=
+
( )
BC
d
11 a a ?T
R
Þ+ =+-
( )
BC
d
R
a a ?T
Þ=
-
( )
BC
11
R andR
?T aa
Þµµ
-
24. (a) A bimetallic strip on being heated bends in the form of
an arc with more expandable metal (A) outside (as
shown) correct.
A 
a
A
 
B 
a
B
 
a
A
 > a
B
 
A 
B 
a
A
 
a
B
 
25. (a) 26. (c) 27. (c)
r
1
v
1
A
1
 = r
2
v
2
A
2
m = 1500 kg/m
3
 × 0.1 m/s × 4 (cm)
2
msDT = 10000
1500 × 0.1 × 4 × 10
–4
 × 1500 × DT = 10000
10000 1000
909
TD== °C
1
2
3
1500
1350
1000 (1)
1110
9
T
-
r
r===
+ gDæö
+´´
ç÷
èø
kg/m
3
2 2 2 1 11
v A vA r =r
 Þ 1350 × v
2
 = 1500 × 0.1
v
2
 = 1/9 m/s
\ V olume rate of flow at the end of tube
= A
2
v
2 
 = 4 × 10
–4
1
9
´
= 
43
4
10m
9
-
´ = 
53
40
10m
9
-
´
V olume rate of flow at the entrance = A
1
v
1
= 0.1 × 4 × 10
–4 
= 4 × 10
–5
m
3
Hence,
 
difference of volume rate of flow at the two
ends
= 
5
40
4 10
9
-
æö
-´
ç÷
èø
  = 
53
4
10m
9
-
´
28. (d) Celsius scale was the first temperature scale and
Fahrenheit is the smallest unit measuring temperature.
29. (a) Linear expansion for brass (19 × 10
–4
) > linear expansion
for steel (11 × 10
–4
). On cooling the disk shrinks to a
greater extent than the hole and hence it will get loose.
30. (b) The latent heat of fusion of ice is amount of heat
required to convert unit mass of ice at 0°C into water
at 0°C. For fusion of ice
L = 80 cal/gm = 80000 cal/gm = 8000 × 4.2 j/kg
= 336000 J/kg.