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# Chapter 19 - Calorimetry and Heat Transfer - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

## NEET : Chapter 19 - Calorimetry and Heat Transfer - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

``` 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
â€“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
â€“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 â€“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
â€“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 â€“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
â€“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 â€“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 Wienâ€™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
â€“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 â€“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 Wienâ€™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 â€˜bâ€™ 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
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