DC Pandey Solutions: Calorimetry and Heat Transfer

# DC Pandey Solutions: Calorimetry and Heat Transfer | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

19. Calorimetry and Heat
Tansfer
Introductory Exercise 19.1.
1.
As Heat gain = Heat loss
Q Q Q
1 2 3
+ =
Þ 140 15 80 ´ ´ + ´ 0.53 m
= ´ ´ 200 1 40
Þ m =
-
=
8000 1113
80
86 g is the mass of
ice melt
\  Mass of water
= + = 200 86 286 g g g
and mass of ice
= - = 140 86 54 g g g
while final temperature of mixture is
0°C.
2.
ms ms
A B
( ) ( ) 16 12 19 16 - = -
Þ          4 3 s s
A B
=
ms ms
B C
( ) ( ) 23 19 28 23 - = -
Þ           4 5 s s
B C
=
\     ms ms
A
( ) ( ) q q - = - 12
4
5
28
or
3
4
12
4
5
28 s s
B B
( ) ( ) q q - = -
or 15 12 16 28 ( ) ( ) q q - = -
or 31 448 180 q = +
Þ      q = ° 20 26 . C
3. mL ms = Dq
Þ 80 1 cal cal/ = °C (q - ° 0 C)
Þ q = ° 80 C
4. As Heat gain = Heat loss
Þ ( ) 100 529 80 - ´ = ´ m m
\  100 529 609 ´ = m
Þ m =
´
=
100 529
609
g 86.86 g of ice will
be formed.
5. P
d
dt
d
dt
ms
dm
dt
s = = =
q
q q ( ) D D
Þ
dm
dt
P
s
=
Dq

\
dm
dt
=
´
° ´ °
500 10
4200 10
6
J/s
J/ kg C C
= ´
5
4 2
10
4
.
kg/s = ´ 12 10
4
. kg/s
ice
–15°C
ice
0°C
water
0°C
Q
1
Q
2
140 g 140 g m g
water
0°C
200 g
Q
3 water
40°C
200 g
A
12°C
B
19°C
C
28°C
16°C 23°C
0°C
Page 2

19. Calorimetry and Heat
Tansfer
Introductory Exercise 19.1.
1.
As Heat gain = Heat loss
Q Q Q
1 2 3
+ =
Þ 140 15 80 ´ ´ + ´ 0.53 m
= ´ ´ 200 1 40
Þ m =
-
=
8000 1113
80
86 g is the mass of
ice melt
\  Mass of water
= + = 200 86 286 g g g
and mass of ice
= - = 140 86 54 g g g
while final temperature of mixture is
0°C.
2.
ms ms
A B
( ) ( ) 16 12 19 16 - = -
Þ          4 3 s s
A B
=
ms ms
B C
( ) ( ) 23 19 28 23 - = -
Þ           4 5 s s
B C
=
\     ms ms
A
( ) ( ) q q - = - 12
4
5
28
or
3
4
12
4
5
28 s s
B B
( ) ( ) q q - = -
or 15 12 16 28 ( ) ( ) q q - = -
or 31 448 180 q = +
Þ      q = ° 20 26 . C
3. mL ms = Dq
Þ 80 1 cal cal/ = °C (q - ° 0 C)
Þ q = ° 80 C
4. As Heat gain = Heat loss
Þ ( ) 100 529 80 - ´ = ´ m m
\  100 529 609 ´ = m
Þ m =
´
=
100 529
609
g 86.86 g of ice will
be formed.
5. P
d
dt
d
dt
ms
dm
dt
s = = =
q
q q ( ) D D
Þ
dm
dt
P
s
=
Dq

\
dm
dt
=
´
° ´ °
500 10
4200 10
6
J/s
J/ kg C C
= ´
5
4 2
10
4
.
kg/s = ´ 12 10
4
. kg/s
ice
–15°C
ice
0°C
water
0°C
Q
1
Q
2
140 g 140 g m g
water
0°C
200 g
Q
3 water
40°C
200 g
A
12°C
B
19°C
C
28°C
16°C 23°C
0°C
Introductory Excersise 19.2
1. Rest of the liquid will be heated due to
conduction and not convection.
2.
dQ
dt
k r d
dr
=
× - 4
2
p q ( )
\
dQ
dt
dr
r
k d × = -
2
4p q
or
dQ
dt
dr
r
k d
a
b
T
T
2
4
1
2
ò ò
= - p q
or
dQ
dt a b
k T T
1 1
4
2 1
-
æ
è
ç
ö
ø
÷
= - - p ( )
Þ
dQ
dt
k T T
a b
kab
T T
b a
=
-
-
=
-
-
4
1 1
4
1 2 1 2
p
p
( )
3.
d Q
d t
kA
t
=
Dq

Þ  k
dQ
dt
t
A
= ×
Dq
\   Unit of k
k
= = watt
m
m -
W/m - K
2
4.
K A
l
K A
l
1 1
1
2 2
2
D D q q
=
001 19
3 5
10
2
. ( )
.
( ) -
=
+ q q 0.08
or 2 19 28 10 ( ) ( ) - = + q q
or 38 280 30 - = q
or q = - = - °
242
30
8.07 C
dQ
dt
=
´ ´ +
´
-
0 . 0 1 8 .1
3.5
1 19
10
2
( )
= 7.74 W / m
2
5.
dQ
dt
dm
dt
L = × = ´ ´
0.44 kg
s
2.256
300
10
6
J/kg
= 3308 J/s . 8
= =
´ ´ -
´
-
kA
t
q q 50.2 0.15
1.2
( ) 100
10
2
= - 627.5 ( ) q 100
Þ q - = = 100
3308.8
627.5
5.27
Þ q = ° 105.27 C
6.
dQ
dt
kA
y
dm
dt
L =
- -
= ×
[ ( )] 0 q
= × = × r r
dV
dt
L A
dy
dt
L
Þ
kA
y
AL
dy
dt
q
r =
Þ
dy
dt
k
L y
=
q
r
(Proved)
7.
dQ
dt
e AT = s
4
= ´ ´ ´ ´
- -
4 5.67 10 4 4 10
8 2 2
p ( )
´ ( ) 3000
4
= ´ ´ ´ ´ 0 4 4 4 3
2 4
. p 5.67 J/s
= ´ 3.7 10
4
watt
8.
dQ
dt R
=
Dq
th
Þ R
d
dt
K
W
th
KW = = =
-
Dq
q
1
65  |  Calorimetry and Heat Tansfer
dy
A
y
19°C –10°C q
0.01 0.08
3.5 cm 2 cm
r
b
a
r+dr
Page 3

19. Calorimetry and Heat
Tansfer
Introductory Exercise 19.1.
1.
As Heat gain = Heat loss
Q Q Q
1 2 3
+ =
Þ 140 15 80 ´ ´ + ´ 0.53 m
= ´ ´ 200 1 40
Þ m =
-
=
8000 1113
80
86 g is the mass of
ice melt
\  Mass of water
= + = 200 86 286 g g g
and mass of ice
= - = 140 86 54 g g g
while final temperature of mixture is
0°C.
2.
ms ms
A B
( ) ( ) 16 12 19 16 - = -
Þ          4 3 s s
A B
=
ms ms
B C
( ) ( ) 23 19 28 23 - = -
Þ           4 5 s s
B C
=
\     ms ms
A
( ) ( ) q q - = - 12
4
5
28
or
3
4
12
4
5
28 s s
B B
( ) ( ) q q - = -
or 15 12 16 28 ( ) ( ) q q - = -
or 31 448 180 q = +
Þ      q = ° 20 26 . C
3. mL ms = Dq
Þ 80 1 cal cal/ = °C (q - ° 0 C)
Þ q = ° 80 C
4. As Heat gain = Heat loss
Þ ( ) 100 529 80 - ´ = ´ m m
\  100 529 609 ´ = m
Þ m =
´
=
100 529
609
g 86.86 g of ice will
be formed.
5. P
d
dt
d
dt
ms
dm
dt
s = = =
q
q q ( ) D D
Þ
dm
dt
P
s
=
Dq

\
dm
dt
=
´
° ´ °
500 10
4200 10
6
J/s
J/ kg C C
= ´
5
4 2
10
4
.
kg/s = ´ 12 10
4
. kg/s
ice
–15°C
ice
0°C
water
0°C
Q
1
Q
2
140 g 140 g m g
water
0°C
200 g
Q
3 water
40°C
200 g
A
12°C
B
19°C
C
28°C
16°C 23°C
0°C
Introductory Excersise 19.2
1. Rest of the liquid will be heated due to
conduction and not convection.
2.
dQ
dt
k r d
dr
=
× - 4
2
p q ( )
\
dQ
dt
dr
r
k d × = -
2
4p q
or
dQ
dt
dr
r
k d
a
b
T
T
2
4
1
2
ò ò
= - p q
or
dQ
dt a b
k T T
1 1
4
2 1
-
æ
è
ç
ö
ø
÷
= - - p ( )
Þ
dQ
dt
k T T
a b
kab
T T
b a
=
-
-
=
-
-
4
1 1
4
1 2 1 2
p
p
( )
3.
d Q
d t
kA
t
=
Dq

Þ  k
dQ
dt
t
A
= ×
Dq
\   Unit of k
k
= = watt
m
m -
W/m - K
2
4.
K A
l
K A
l
1 1
1
2 2
2
D D q q
=
001 19
3 5
10
2
. ( )
.
( ) -
=
+ q q 0.08
or 2 19 28 10 ( ) ( ) - = + q q
or 38 280 30 - = q
or q = - = - °
242
30
8.07 C
dQ
dt
=
´ ´ +
´
-
0 . 0 1 8 .1
3.5
1 19
10
2
( )
= 7.74 W / m
2
5.
dQ
dt
dm
dt
L = × = ´ ´
0.44 kg
s
2.256
300
10
6
J/kg
= 3308 J/s . 8
= =
´ ´ -
´
-
kA
t
q q 50.2 0.15
1.2
( ) 100
10
2
= - 627.5 ( ) q 100
Þ q - = = 100
3308.8
627.5
5.27
Þ q = ° 105.27 C
6.
dQ
dt
kA
y
dm
dt
L =
- -
= ×
[ ( )] 0 q
= × = × r r
dV
dt
L A
dy
dt
L
Þ
kA
y
AL
dy
dt
q
r =
Þ
dy
dt
k
L y
=
q
r
(Proved)
7.
dQ
dt
e AT = s
4
= ´ ´ ´ ´
- -
4 5.67 10 4 4 10
8 2 2
p ( )
´ ( ) 3000
4
= ´ ´ ´ ´ 0 4 4 4 3
2 4
. p 5.67 J/s
= ´ 3.7 10
4
watt
8.
dQ
dt R
=
Dq
th
Þ R
d
dt
K
W
th
KW = = =
-
Dq
q
1
65  |  Calorimetry and Heat Tansfer
dy
A
y
19°C –10°C q
0.01 0.08
3.5 cm 2 cm
r
b
a
r+dr
AIEEE Corner
¢ Subjective Questions (Level-1)
1. ice Water Water steam
0 C 0 C 100 C 100 C ° ° ° °
¾® ¾® ¾®
Q Q Q
1 2 3
Q Q Q Q = + +
1 2 3
= + + mL ms mL
f v
Dq
= + ´ + 10 80 1 100 540 [ ]
= ´ 10 720 cal = 7200 cal
2. 10 g of wa ter at 40°C do not have
suf fi cient heat en ergy to melt 15 g of ice
at 0°C , so there will be a mix ture  of
ice-wa ter at 0°C. Let the mass of ice left
is mg.
\     ( ) 15 80 10 1 40 - ´ = ´ ´ m
15 5 - = m Þ m = 10 g
\ Mass of ice = 10 g
and mass of water = + = ( ) 10 5 15 g g
3. 4 60 55 1 55 50 ´ - = ´ ´ - s s
P R
( ) ( )
Þ 4s s
P R
=
1 60 55 1 55 50 ´ - = ´ - s s
P Q
( ) ( )
Þ s s
P Q
=
1 60 1 50 ´ - = ´ - s s
Q R
( ) ( ) q q
or s s
P P
( ) ( ) 60 4 50 - = - q q
260 5 = q Þ q = = °
260
5
52 C
4.
dQ
dt
m
=
´ ´
´
336 10
4 60
3
J/ kg
s
= 1400 J/ kg
= 1400 m W / kg
=
×
=
´ -
´
m s
t
m D q q 4 2 0 0 0
2 6 0
( ) c
s
\
1 4 00 2 60
4200
´ ´
= q
Þ q = ° 40 C
5. Q mv ms mL = ´ = +
1
2
1
2
2
Dq
Þ v s L = + 4 ( ) Dq
\ v = ´ + ´ 4 125 300 2 5 10
4
( . )
= ´ + ´ 4 3 75 10
4
( . ) 2.5
= ´ ´ 4 10
4
6.25 = 500 m/s
6. h q mg h ms D D =
\ D
D
q
h
= =
´ ´
= °
g h
s
0.4 0.5
400
C
10
800
1
= ´ °
-
2.5 C 10
3
7.
K A
l
K A
l
1 2
0 100 ( ) ( ) q q -
=
-
Þ ( ) K K K
1 2 2
100 + = q
\ q =
+
=
´
´
= °
100 100 46
390 46
1055
2
1 2
K
K K
. C
8. i i i
CD AC CB
= -

KA
l
KA
l
KA
l
( ) ( )
/
( )
/
q q q -
=
-
-
- 25 100
2
0
2
or q q q - = - - 25 2 100 2 ( )
or 5 225 q =  Þ q = ° 45 C
\ i
R
CD
= =
-
=
Dq
th
45 25
5
4 W
9. i i i
A C D
= +
KA T
l
KA T
l
KA T
l
( ) ( )
/
( )
/
1 3 2
3 2 3 2
-
=
-
+
- q q q
Þ T T T
1 3 2
2
3
2
3
- = - + - q q q ( ) ( )
or T T T
1 2 3
2
3
1
4
3
+ + = +
æ
è
ç
ö
ø
÷
( ) q
Þ q =
+ + T T T
1 2 3
2
3
7 3
( )
/
=
+ + 3 2
7
1 2 3
T T T ( )
10.
K A
l
K A
l
( ) ( ) 2 0 0 2
1 1 2
-
=
- q q q
Calorimetry and Heat Tansfer | 66
T
t
0°C
q°C
5 1 7
Page 4

19. Calorimetry and Heat
Tansfer
Introductory Exercise 19.1.
1.
As Heat gain = Heat loss
Q Q Q
1 2 3
+ =
Þ 140 15 80 ´ ´ + ´ 0.53 m
= ´ ´ 200 1 40
Þ m =
-
=
8000 1113
80
86 g is the mass of
ice melt
\  Mass of water
= + = 200 86 286 g g g
and mass of ice
= - = 140 86 54 g g g
while final temperature of mixture is
0°C.
2.
ms ms
A B
( ) ( ) 16 12 19 16 - = -
Þ          4 3 s s
A B
=
ms ms
B C
( ) ( ) 23 19 28 23 - = -
Þ           4 5 s s
B C
=
\     ms ms
A
( ) ( ) q q - = - 12
4
5
28
or
3
4
12
4
5
28 s s
B B
( ) ( ) q q - = -
or 15 12 16 28 ( ) ( ) q q - = -
or 31 448 180 q = +
Þ      q = ° 20 26 . C
3. mL ms = Dq
Þ 80 1 cal cal/ = °C (q - ° 0 C)
Þ q = ° 80 C
4. As Heat gain = Heat loss
Þ ( ) 100 529 80 - ´ = ´ m m
\  100 529 609 ´ = m
Þ m =
´
=
100 529
609
g 86.86 g of ice will
be formed.
5. P
d
dt
d
dt
ms
dm
dt
s = = =
q
q q ( ) D D
Þ
dm
dt
P
s
=
Dq

\
dm
dt
=
´
° ´ °
500 10
4200 10
6
J/s
J/ kg C C
= ´
5
4 2
10
4
.
kg/s = ´ 12 10
4
. kg/s
ice
–15°C
ice
0°C
water
0°C
Q
1
Q
2
140 g 140 g m g
water
0°C
200 g
Q
3 water
40°C
200 g
A
12°C
B
19°C
C
28°C
16°C 23°C
0°C
Introductory Excersise 19.2
1. Rest of the liquid will be heated due to
conduction and not convection.
2.
dQ
dt
k r d
dr
=
× - 4
2
p q ( )
\
dQ
dt
dr
r
k d × = -
2
4p q
or
dQ
dt
dr
r
k d
a
b
T
T
2
4
1
2
ò ò
= - p q
or
dQ
dt a b
k T T
1 1
4
2 1
-
æ
è
ç
ö
ø
÷
= - - p ( )
Þ
dQ
dt
k T T
a b
kab
T T
b a
=
-
-
=
-
-
4
1 1
4
1 2 1 2
p
p
( )
3.
d Q
d t
kA
t
=
Dq

Þ  k
dQ
dt
t
A
= ×
Dq
\   Unit of k
k
= = watt
m
m -
W/m - K
2
4.
K A
l
K A
l
1 1
1
2 2
2
D D q q
=
001 19
3 5
10
2
. ( )
.
( ) -
=
+ q q 0.08
or 2 19 28 10 ( ) ( ) - = + q q
or 38 280 30 - = q
or q = - = - °
242
30
8.07 C
dQ
dt
=
´ ´ +
´
-
0 . 0 1 8 .1
3.5
1 19
10
2
( )
= 7.74 W / m
2
5.
dQ
dt
dm
dt
L = × = ´ ´
0.44 kg
s
2.256
300
10
6
J/kg
= 3308 J/s . 8
= =
´ ´ -
´
-
kA
t
q q 50.2 0.15
1.2
( ) 100
10
2
= - 627.5 ( ) q 100
Þ q - = = 100
3308.8
627.5
5.27
Þ q = ° 105.27 C
6.
dQ
dt
kA
y
dm
dt
L =
- -
= ×
[ ( )] 0 q
= × = × r r
dV
dt
L A
dy
dt
L
Þ
kA
y
AL
dy
dt
q
r =
Þ
dy
dt
k
L y
=
q
r
(Proved)
7.
dQ
dt
e AT = s
4
= ´ ´ ´ ´
- -
4 5.67 10 4 4 10
8 2 2
p ( )
´ ( ) 3000
4
= ´ ´ ´ ´ 0 4 4 4 3
2 4
. p 5.67 J/s
= ´ 3.7 10
4
watt
8.
dQ
dt R
=
Dq
th
Þ R
d
dt
K
W
th
KW = = =
-
Dq
q
1
65  |  Calorimetry and Heat Tansfer
dy
A
y
19°C –10°C q
0.01 0.08
3.5 cm 2 cm
r
b
a
r+dr
AIEEE Corner
¢ Subjective Questions (Level-1)
1. ice Water Water steam
0 C 0 C 100 C 100 C ° ° ° °
¾® ¾® ¾®
Q Q Q
1 2 3
Q Q Q Q = + +
1 2 3
= + + mL ms mL
f v
Dq
= + ´ + 10 80 1 100 540 [ ]
= ´ 10 720 cal = 7200 cal
2. 10 g of wa ter at 40°C do not have
suf fi cient heat en ergy to melt 15 g of ice
at 0°C , so there will be a mix ture  of
ice-wa ter at 0°C. Let the mass of ice left
is mg.
\     ( ) 15 80 10 1 40 - ´ = ´ ´ m
15 5 - = m Þ m = 10 g
\ Mass of ice = 10 g
and mass of water = + = ( ) 10 5 15 g g
3. 4 60 55 1 55 50 ´ - = ´ ´ - s s
P R
( ) ( )
Þ 4s s
P R
=
1 60 55 1 55 50 ´ - = ´ - s s
P Q
( ) ( )
Þ s s
P Q
=
1 60 1 50 ´ - = ´ - s s
Q R
( ) ( ) q q
or s s
P P
( ) ( ) 60 4 50 - = - q q
260 5 = q Þ q = = °
260
5
52 C
4.
dQ
dt
m
=
´ ´
´
336 10
4 60
3
J/ kg
s
= 1400 J/ kg
= 1400 m W / kg
=
×
=
´ -
´
m s
t
m D q q 4 2 0 0 0
2 6 0
( ) c
s
\
1 4 00 2 60
4200
´ ´
= q
Þ q = ° 40 C
5. Q mv ms mL = ´ = +
1
2
1
2
2
Dq
Þ v s L = + 4 ( ) Dq
\ v = ´ + ´ 4 125 300 2 5 10
4
( . )
= ´ + ´ 4 3 75 10
4
( . ) 2.5
= ´ ´ 4 10
4
6.25 = 500 m/s
6. h q mg h ms D D =
\ D
D
q
h
= =
´ ´
= °
g h
s
0.4 0.5
400
C
10
800
1
= ´ °
-
2.5 C 10
3
7.
K A
l
K A
l
1 2
0 100 ( ) ( ) q q -
=
-
Þ ( ) K K K
1 2 2
100 + = q
\ q =
+
=
´
´
= °
100 100 46
390 46
1055
2
1 2
K
K K
. C
8. i i i
CD AC CB
= -

KA
l
KA
l
KA
l
( ) ( )
/
( )
/
q q q -
=
-
-
- 25 100
2
0
2
or q q q - = - - 25 2 100 2 ( )
or 5 225 q =  Þ q = ° 45 C
\ i
R
CD
= =
-
=
Dq
th
45 25
5
4 W
9. i i i
A C D
= +
KA T
l
KA T
l
KA T
l
( ) ( )
/
( )
/
1 3 2
3 2 3 2
-
=
-
+
- q q q
Þ T T T
1 3 2
2
3
2
3
- = - + - q q q ( ) ( )
or T T T
1 2 3
2
3
1
4
3
+ + = +
æ
è
ç
ö
ø
÷
( ) q
Þ q =
+ + T T T
1 2 3
2
3
7 3
( )
/
=
+ + 3 2
7
1 2 3
T T T ( )
10.
K A
l
K A
l
( ) ( ) 2 0 0 2
1 1 2
-
=
- q q q
Calorimetry and Heat Tansfer | 66
T
t
0°C
q°C
5 1 7
=
- 3 100
2
KA
l
( ) q
\ 200 2 3 100
1 1 2 2
- = - = - q q q q ( ) ( )
Þ  3 2 200
1 2
q q - =

q q
q
1 2
2
3 500
11 1300
+ =
- = -
Þ q
2
1300
11
118 2 = = ° . C
q q
1 2
1
3
200 2 145 45 = + = ° [ ] . C
11. 25
400 10 100
12
4
=
´ -
-
( )
/
q
+
´ -
-
400 10 0
1 2
4
( )
/
q
25 8 10 100
2
= ´ - +
-
[ ] q q
or 312.5 = - 2 100 q
Þ  q = =
412.5
2
206 25 .
\ Dq
1
= 106.25  and  Dq
2
206 25 = .
\
D
D
q
1
1 2 l
=
°
= °
106.25 C
m
212.5 C/m
/
and
D
D
q
2
1 2 l
=
°
=
206.25 C
m
412.5
/
°C/m
12.
dQ
dt
e AT = = ´ ´
-
s
4 8
10 0.6 5.67
´ ´ ´ 2 1073
2 4
( ) ( ) 0.1
= ´ ´ ´ ´
-
0.6 5.67 10.73 ( )
4 2
10 2
= 902 W
13.
dQ
dt
e AT
æ
è
ç
ö
ø
÷
=
1
4
s and
dQ
dt
AT
æ
è
ç
ö
ø
÷
=
2
4
s
Þ e
dQ dt
dQ dt
= = =
( / )
( / )
1
2
210
700
0.3
14.
( ) 80 50
5
80 50
2
20
-
=
+
-
æ
è
ç
ö
ø
÷
c
c
Þ K =
6
45
;
( ) 60 30 6
45
-
=
t

60 30
2
20
+
-
æ
è
ç
ö
ø
÷
Þ t = 9 min
¢ Objective Questions (Level-1)
1.
3 35
10
0
20
KA KA ( ) ( ) -
=
- q q

Þ 6 35 ( ) - = q q
Þ q =
´
= °
6 35
7
30 C
\ Dq
A
= - = ° 35 30 5 C
2.
T
T
S
N
N
S
= = =
l
l
350
510
0.69
According to Wien’s law
3.
dQ
dt
dQ
dt
K A
l
K
l
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
=
×
=
2 1
1
4
4
2
2
D
D
q
q
/
Þ
dm
dt
dm
dt
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
2 1
2 = 0.2 g/s
4.
dQ
dt
K
a a
a a
K
a a
a a
=
-
-
×
=
-
-
×
4 0
2
2
4 100
3 2
3 2
p q p q ( ) ( )
Þ 2 6 100 q q = - ( )
Þ q = ´ = °
6
8
100 75 C
5.
K A T T
d
K A T T
d
1 2 1 2 3 2
3
( ) ( ) -
=
-
Þ K T T K T T
1 2 1 2 3 2
1
3
( ) ( ) - = -
Þ K K
1 2
1
3
= Þ K K
1 2
1 3 : : =
6.
dQ
dt
dQ
dt
K A
l
KA
l
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
=
× ×
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
2 1
2 2
2
D D q q
= 2
Þ
d Q
d t
d Q
d t
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
=
2 1
2 8 cal/s
7.
67  |  Calorimetry and Heat Tansfer
H
100°C 0°C S
1/2 m 1/2 m
25 W
0°C dx
q, q, + dq
x
Page 5

19. Calorimetry and Heat
Tansfer
Introductory Exercise 19.1.
1.
As Heat gain = Heat loss
Q Q Q
1 2 3
+ =
Þ 140 15 80 ´ ´ + ´ 0.53 m
= ´ ´ 200 1 40
Þ m =
-
=
8000 1113
80
86 g is the mass of
ice melt
\  Mass of water
= + = 200 86 286 g g g
and mass of ice
= - = 140 86 54 g g g
while final temperature of mixture is
0°C.
2.
ms ms
A B
( ) ( ) 16 12 19 16 - = -
Þ          4 3 s s
A B
=
ms ms
B C
( ) ( ) 23 19 28 23 - = -
Þ           4 5 s s
B C
=
\     ms ms
A
( ) ( ) q q - = - 12
4
5
28
or
3
4
12
4
5
28 s s
B B
( ) ( ) q q - = -
or 15 12 16 28 ( ) ( ) q q - = -
or 31 448 180 q = +
Þ      q = ° 20 26 . C
3. mL ms = Dq
Þ 80 1 cal cal/ = °C (q - ° 0 C)
Þ q = ° 80 C
4. As Heat gain = Heat loss
Þ ( ) 100 529 80 - ´ = ´ m m
\  100 529 609 ´ = m
Þ m =
´
=
100 529
609
g 86.86 g of ice will
be formed.
5. P
d
dt
d
dt
ms
dm
dt
s = = =
q
q q ( ) D D
Þ
dm
dt
P
s
=
Dq

\
dm
dt
=
´
° ´ °
500 10
4200 10
6
J/s
J/ kg C C
= ´
5
4 2
10
4
.
kg/s = ´ 12 10
4
. kg/s
ice
–15°C
ice
0°C
water
0°C
Q
1
Q
2
140 g 140 g m g
water
0°C
200 g
Q
3 water
40°C
200 g
A
12°C
B
19°C
C
28°C
16°C 23°C
0°C
Introductory Excersise 19.2
1. Rest of the liquid will be heated due to
conduction and not convection.
2.
dQ
dt
k r d
dr
=
× - 4
2
p q ( )
\
dQ
dt
dr
r
k d × = -
2
4p q
or
dQ
dt
dr
r
k d
a
b
T
T
2
4
1
2
ò ò
= - p q
or
dQ
dt a b
k T T
1 1
4
2 1
-
æ
è
ç
ö
ø
÷
= - - p ( )
Þ
dQ
dt
k T T
a b
kab
T T
b a
=
-
-
=
-
-
4
1 1
4
1 2 1 2
p
p
( )
3.
d Q
d t
kA
t
=
Dq

Þ  k
dQ
dt
t
A
= ×
Dq
\   Unit of k
k
= = watt
m
m -
W/m - K
2
4.
K A
l
K A
l
1 1
1
2 2
2
D D q q
=
001 19
3 5
10
2
. ( )
.
( ) -
=
+ q q 0.08
or 2 19 28 10 ( ) ( ) - = + q q
or 38 280 30 - = q
or q = - = - °
242
30
8.07 C
dQ
dt
=
´ ´ +
´
-
0 . 0 1 8 .1
3.5
1 19
10
2
( )
= 7.74 W / m
2
5.
dQ
dt
dm
dt
L = × = ´ ´
0.44 kg
s
2.256
300
10
6
J/kg
= 3308 J/s . 8
= =
´ ´ -
´
-
kA
t
q q 50.2 0.15
1.2
( ) 100
10
2
= - 627.5 ( ) q 100
Þ q - = = 100
3308.8
627.5
5.27
Þ q = ° 105.27 C
6.
dQ
dt
kA
y
dm
dt
L =
- -
= ×
[ ( )] 0 q
= × = × r r
dV
dt
L A
dy
dt
L
Þ
kA
y
AL
dy
dt
q
r =
Þ
dy
dt
k
L y
=
q
r
(Proved)
7.
dQ
dt
e AT = s
4
= ´ ´ ´ ´
- -
4 5.67 10 4 4 10
8 2 2
p ( )
´ ( ) 3000
4
= ´ ´ ´ ´ 0 4 4 4 3
2 4
. p 5.67 J/s
= ´ 3.7 10
4
watt
8.
dQ
dt R
=
Dq
th
Þ R
d
dt
K
W
th
KW = = =
-
Dq
q
1
65  |  Calorimetry and Heat Tansfer
dy
A
y
19°C –10°C q
0.01 0.08
3.5 cm 2 cm
r
b
a
r+dr
AIEEE Corner
¢ Subjective Questions (Level-1)
1. ice Water Water steam
0 C 0 C 100 C 100 C ° ° ° °
¾® ¾® ¾®
Q Q Q
1 2 3
Q Q Q Q = + +
1 2 3
= + + mL ms mL
f v
Dq
= + ´ + 10 80 1 100 540 [ ]
= ´ 10 720 cal = 7200 cal
2. 10 g of wa ter at 40°C do not have
suf fi cient heat en ergy to melt 15 g of ice
at 0°C , so there will be a mix ture  of
ice-wa ter at 0°C. Let the mass of ice left
is mg.
\     ( ) 15 80 10 1 40 - ´ = ´ ´ m
15 5 - = m Þ m = 10 g
\ Mass of ice = 10 g
and mass of water = + = ( ) 10 5 15 g g
3. 4 60 55 1 55 50 ´ - = ´ ´ - s s
P R
( ) ( )
Þ 4s s
P R
=
1 60 55 1 55 50 ´ - = ´ - s s
P Q
( ) ( )
Þ s s
P Q
=
1 60 1 50 ´ - = ´ - s s
Q R
( ) ( ) q q
or s s
P P
( ) ( ) 60 4 50 - = - q q
260 5 = q Þ q = = °
260
5
52 C
4.
dQ
dt
m
=
´ ´
´
336 10
4 60
3
J/ kg
s
= 1400 J/ kg
= 1400 m W / kg
=
×
=
´ -
´
m s
t
m D q q 4 2 0 0 0
2 6 0
( ) c
s
\
1 4 00 2 60
4200
´ ´
= q
Þ q = ° 40 C
5. Q mv ms mL = ´ = +
1
2
1
2
2
Dq
Þ v s L = + 4 ( ) Dq
\ v = ´ + ´ 4 125 300 2 5 10
4
( . )
= ´ + ´ 4 3 75 10
4
( . ) 2.5
= ´ ´ 4 10
4
6.25 = 500 m/s
6. h q mg h ms D D =
\ D
D
q
h
= =
´ ´
= °
g h
s
0.4 0.5
400
C
10
800
1
= ´ °
-
2.5 C 10
3
7.
K A
l
K A
l
1 2
0 100 ( ) ( ) q q -
=
-
Þ ( ) K K K
1 2 2
100 + = q
\ q =
+
=
´
´
= °
100 100 46
390 46
1055
2
1 2
K
K K
. C
8. i i i
CD AC CB
= -

KA
l
KA
l
KA
l
( ) ( )
/
( )
/
q q q -
=
-
-
- 25 100
2
0
2
or q q q - = - - 25 2 100 2 ( )
or 5 225 q =  Þ q = ° 45 C
\ i
R
CD
= =
-
=
Dq
th
45 25
5
4 W
9. i i i
A C D
= +
KA T
l
KA T
l
KA T
l
( ) ( )
/
( )
/
1 3 2
3 2 3 2
-
=
-
+
- q q q
Þ T T T
1 3 2
2
3
2
3
- = - + - q q q ( ) ( )
or T T T
1 2 3
2
3
1
4
3
+ + = +
æ
è
ç
ö
ø
÷
( ) q
Þ q =
+ + T T T
1 2 3
2
3
7 3
( )
/
=
+ + 3 2
7
1 2 3
T T T ( )
10.
K A
l
K A
l
( ) ( ) 2 0 0 2
1 1 2
-
=
- q q q
Calorimetry and Heat Tansfer | 66
T
t
0°C
q°C
5 1 7
=
- 3 100
2
KA
l
( ) q
\ 200 2 3 100
1 1 2 2
- = - = - q q q q ( ) ( )
Þ  3 2 200
1 2
q q - =

q q
q
1 2
2
3 500
11 1300
+ =
- = -
Þ q
2
1300
11
118 2 = = ° . C
q q
1 2
1
3
200 2 145 45 = + = ° [ ] . C
11. 25
400 10 100
12
4
=
´ -
-
( )
/
q
+
´ -
-
400 10 0
1 2
4
( )
/
q
25 8 10 100
2
= ´ - +
-
[ ] q q
or 312.5 = - 2 100 q
Þ  q = =
412.5
2
206 25 .
\ Dq
1
= 106.25  and  Dq
2
206 25 = .
\
D
D
q
1
1 2 l
=
°
= °
106.25 C
m
212.5 C/m
/
and
D
D
q
2
1 2 l
=
°
=
206.25 C
m
412.5
/
°C/m
12.
dQ
dt
e AT = = ´ ´
-
s
4 8
10 0.6 5.67
´ ´ ´ 2 1073
2 4
( ) ( ) 0.1
= ´ ´ ´ ´
-
0.6 5.67 10.73 ( )
4 2
10 2
= 902 W
13.
dQ
dt
e AT
æ
è
ç
ö
ø
÷
=
1
4
s and
dQ
dt
AT
æ
è
ç
ö
ø
÷
=
2
4
s
Þ e
dQ dt
dQ dt
= = =
( / )
( / )
1
2
210
700
0.3
14.
( ) 80 50
5
80 50
2
20
-
=
+
-
æ
è
ç
ö
ø
÷
c
c
Þ K =
6
45
;
( ) 60 30 6
45
-
=
t

60 30
2
20
+
-
æ
è
ç
ö
ø
÷
Þ t = 9 min
¢ Objective Questions (Level-1)
1.
3 35
10
0
20
KA KA ( ) ( ) -
=
- q q

Þ 6 35 ( ) - = q q
Þ q =
´
= °
6 35
7
30 C
\ Dq
A
= - = ° 35 30 5 C
2.
T
T
S
N
N
S
= = =
l
l
350
510
0.69
According to Wien’s law
3.
dQ
dt
dQ
dt
K A
l
K
l
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
=
×
=
2 1
1
4
4
2
2
D
D
q
q
/
Þ
dm
dt
dm
dt
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
2 1
2 = 0.2 g/s
4.
dQ
dt
K
a a
a a
K
a a
a a
=
-
-
×
=
-
-
×
4 0
2
2
4 100
3 2
3 2
p q p q ( ) ( )
Þ 2 6 100 q q = - ( )
Þ q = ´ = °
6
8
100 75 C
5.
K A T T
d
K A T T
d
1 2 1 2 3 2
3
( ) ( ) -
=
-
Þ K T T K T T
1 2 1 2 3 2
1
3
( ) ( ) - = -
Þ K K
1 2
1
3
= Þ K K
1 2
1 3 : : =
6.
dQ
dt
dQ
dt
K A
l
KA
l
æ
è
ç
ö
ø
÷
æ
è
ç
ö
ø
÷
=
× ×
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
2 1
2 2
2
D D q q
= 2
Þ
d Q
d t
d Q
d t
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
=
2 1
2 8 cal/s
7.
67  |  Calorimetry and Heat Tansfer
H
100°C 0°C S
1/2 m 1/2 m
25 W
0°C dx
q, q, + dq
x
P
dQ
dt
dx
K ax A d
dx
= =
×
=
+ q q
0
1 ( )
\
dx
ax
K A
P
d
l
1
0
0
0
100
+
=
ò ò
q
1
1
10 10
1
1
0
2 4
0
100
a
ax
l
ln ( )| | + =
´
× =
-
q
Þ ln ( ) ln 1 1 1 + - = al
Þ ln ( ) ln 1 1 1 + - = al
Þ ln ( ) 1 1 + = al
or 1
1
+ = al e
or l
a
e e = - = - =
1
1 1 1 7 ( ) . m
8.
l
l
2
1
1
2
2
3
= =
T
T
Þ l l
2
2
3
=
m

9. Heat re quired to boil 1 g of ice is 180 cal
while 1 g of steam can re lease 540 cal
dur ing condenstion. So, temperture of
the mix ture will be 100°C with 2/3 g
steam and 4.3 g wa ter.
10. T T T
1 2 3
< < as temperature of a body
de creases in rate of cool ing also
de creases such that time in creases for
equal temperature dif fer ence.
11. Con duc t ion is max i m um for which
thermal re sis tance is min i m um, as
R
l
r
th
µ
2
then for
(a) 50 (b) 25  (c) 100 (d) 33.33,
So option ‘b’ has minimum resistance.
12. Slope of temperature versus heat graph
gives in crease of spe cific heat or heat
ca pac ity and the por tion DE is the
gas e ous state .
13. d Q m s d t =  = maT dT
3
Þ
Q
m
a
T
a a
= = - =
4 4
16 1
15
4
4
1
2
| ( )
14. Re sis tance be comes 1/4th in par al lel of
that in se ries, so times taken will also
be com e 1/4t h i e , 12/4 = 3 min.
15. ms ms
1 2
12 8 ´ = ´  Þ s s
1 2
2 3 : : =
16.
KA T T
l
KA T T
l
c c
( ) ( ) -
=
-
2
2
Þ
T
T T
T
c
c
2
2
2
+ = +
Þ
3
2
1 2
2
T T
c
=
+
Þ                T T
c
=
+
3
1 2
17. P = - = ( ) 1000 160 840 W W
=
´ ´ 2 4200 50
t
\ t =
´
= =
42 10
840
500 8 20
4
s s min
18.
dQ
dt
KA T T
x
KA T T
x
=
-
=
- ( ) ( )
2 1
2
4
Þ T T T T
2 1
1
2
1
2
- = -
Þ T T T
2 1
1
2
3
2
+ =
Þ T T T T T = +
æ
è
ç
ö
ø
÷
= +
2
3
1
2
1
3
2
2 1 2 1
( )
\
dQ
dt
KA
x
T T T = - +
é
ë
ê
ù
û
ú
2 2 1
1
3
2 ( )
= - - ´
KA
x
T T T [ ] 3 2
1
3
2 2 1
= - ´
KA
x
T T ( )
2 1
1
3
Þ      f =
1
3
19. Dq µ
1
K
Þ
D
D
q
q
A
B
B
A
K
K
= =
1
2
Þ D D q q
A B
= = °
1
2
18 C
¢ More than One Correct Options
20. Amount of heat ra di ated or absorbed
de pends upon. Sur face type, sur face
area, sur face tem per a ture and
tem p er a tu re of sur roun d ing, so (a) and
(b) are correct.
Calorimetry and Heat Tansfer | 68
```

## DC Pandey Solutions for NEET Physics

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## FAQs on DC Pandey Solutions: Calorimetry and Heat Transfer - DC Pandey Solutions for NEET Physics

 1. How do I calculate the heat transferred in a calorimetry experiment?
Ans. To calculate the heat transferred in a calorimetry experiment, you can use the formula Q = mcΔT, where Q represents the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
 2. What is the purpose of calorimetry in studying heat transfer?
Ans. Calorimetry is used to measure the amount of heat transferred during a chemical reaction or a physical change. It helps in studying heat transfer by providing accurate measurements of the heat exchanged between substances, allowing for the analysis of energy flow and the determination of specific heat capacities.
 3. How does a calorimeter work in measuring heat transfer?
Ans. A calorimeter works by isolating a substance of interest within a closed container and measuring the temperature change that occurs during a chemical reaction or a physical change. The heat transfer is determined by the change in temperature and the specific heat capacity of the substance being studied.
 4. What are the different methods of heat transfer studied in calorimetry?
Ans. The different methods of heat transfer studied in calorimetry include conduction, convection, and radiation. Conduction refers to the transfer of heat through direct contact, convection involves the transfer of heat through the movement of fluids, and radiation is the transfer of heat through electromagnetic waves.
 5. How can calorimetry be applied in real-world scenarios?
Ans. Calorimetry has various real-world applications, such as determining the energy content of food through bomb calorimetry, studying the efficiency of engines, analyzing the cooling rates of materials, and investigating the thermal properties of substances. It is also used in industries like pharmaceuticals, chemistry, and materials science for research and development purposes.

## DC Pandey Solutions for NEET Physics

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