Chapter 28 : Heat Transfer - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 11

JEE : Chapter 28 : Heat Transfer - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


28.1
CHAPTER 28
HEAT TRANSFER
1. t
1
= 90°C, t
2
= 10°C
l = 1 cm = 1 × 10
–3
m
A = 10 cm × 10 cm = 0.1 × 0.1 m
2
= 1 × 10
–2
m
2
K = 0.80 w/m-°C
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
2
2 1
10 1
80 10 1 10 8
?
? ?
?
? ? ? ?
= 64 J/s = 64 × 60 3840 J.
2. t = 1 cm = 0.01 m, A = 0.8 m
2
?
1
= 300, ?
2
= 80
K = 0.025, 
t
Q
= 
l
) ( KA
2 1
? ? ?
=  
01 . 0
) 30030 ( 8 . 0 025 . 0 ? ?
= 440 watt.
3. K = 0.04 J/m-5°C, A =  1.6 m
2
t
1
= 97°F = 36.1°C t
2
= 47°F = 8.33°C
l = 0.5 cm = 0.005 m
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
3
2
10 5
78 . 27 6 . 1 10 4
?
?
?
? ? ?
= 356 J/s
4. A = 25 cm
2
= 25 × 10
–4
m
2
l = 1 mm = 10
–3
m
K = 50 w/m-°C
t
Q
= Rate of conversion of water into steam
= 
min 1
10 26 . 2 10 100
6 3
? ? ?
?
= 
60
10 26 . 2 10
6 1
? ?
?
= 0.376 × 10
4
t
Q
= 
l
) ( KA
2 1
? ? ?
? 0.376 ×10
4
= 
3
4
10
) 100 ( 10 25 50
?
?
? ? ? ? ?
? ? = 
4
4 3
10 25 50
10 376 . 0 10
?
?
? ?
? ?
= 
25 50
376 . 0 10
5
?
?
= 30.1 ˜ 30 ?
5. K = 46 w/m-s°C
l = 1 m
A = 0.04 cm
2
= 4 × 10
–6
m
2
L
fussion ice
= 3.36 × 10
5
j/Kg
t
Q
= 
1
100 10 4 46
6
? ? ?
?
= 5.4 × 10
–8
kg ˜ 5.4 × 10
–5
g.
6. A = 2400 cm
2
= 2400 × 10
–4
m
2
l = 2 mm = 2 × 10
–3
m
K = 0.06 w/m-°C
?
1
= 20°C
?
2
= 0°C
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
3
4
10 2
20 10 2400 06 . 0
?
?
?
? ? ?
= 24 × 6 × 10
–1
× 10 = 24 × 6 = 144 J/sec
Rate in which ice melts = 
t
m
= 
L t
Q
?
= 
5
10 4 . 3
144
?
Kg/h = 
5
10 4 . 3
3600 144
?
?
Kg/s = 1.52 kg/s.  
7. l = 1 mm = 10
–3
m m = 10 kg
A = 200 cm
2
= 2 × 10
–2
m
2
L
vap
= 2.27 × 10
6
J/kg
K = 0.80 J/m-s-°C
0°C
100°C
10
10 cm
1 cm
Page 2


28.1
CHAPTER 28
HEAT TRANSFER
1. t
1
= 90°C, t
2
= 10°C
l = 1 cm = 1 × 10
–3
m
A = 10 cm × 10 cm = 0.1 × 0.1 m
2
= 1 × 10
–2
m
2
K = 0.80 w/m-°C
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
2
2 1
10 1
80 10 1 10 8
?
? ?
?
? ? ? ?
= 64 J/s = 64 × 60 3840 J.
2. t = 1 cm = 0.01 m, A = 0.8 m
2
?
1
= 300, ?
2
= 80
K = 0.025, 
t
Q
= 
l
) ( KA
2 1
? ? ?
=  
01 . 0
) 30030 ( 8 . 0 025 . 0 ? ?
= 440 watt.
3. K = 0.04 J/m-5°C, A =  1.6 m
2
t
1
= 97°F = 36.1°C t
2
= 47°F = 8.33°C
l = 0.5 cm = 0.005 m
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
3
2
10 5
78 . 27 6 . 1 10 4
?
?
?
? ? ?
= 356 J/s
4. A = 25 cm
2
= 25 × 10
–4
m
2
l = 1 mm = 10
–3
m
K = 50 w/m-°C
t
Q
= Rate of conversion of water into steam
= 
min 1
10 26 . 2 10 100
6 3
? ? ?
?
= 
60
10 26 . 2 10
6 1
? ?
?
= 0.376 × 10
4
t
Q
= 
l
) ( KA
2 1
? ? ?
? 0.376 ×10
4
= 
3
4
10
) 100 ( 10 25 50
?
?
? ? ? ? ?
? ? = 
4
4 3
10 25 50
10 376 . 0 10
?
?
? ?
? ?
= 
25 50
376 . 0 10
5
?
?
= 30.1 ˜ 30 ?
5. K = 46 w/m-s°C
l = 1 m
A = 0.04 cm
2
= 4 × 10
–6
m
2
L
fussion ice
= 3.36 × 10
5
j/Kg
t
Q
= 
1
100 10 4 46
6
? ? ?
?
= 5.4 × 10
–8
kg ˜ 5.4 × 10
–5
g.
6. A = 2400 cm
2
= 2400 × 10
–4
m
2
l = 2 mm = 2 × 10
–3
m
K = 0.06 w/m-°C
?
1
= 20°C
?
2
= 0°C
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
3
4
10 2
20 10 2400 06 . 0
?
?
?
? ? ?
= 24 × 6 × 10
–1
× 10 = 24 × 6 = 144 J/sec
Rate in which ice melts = 
t
m
= 
L t
Q
?
= 
5
10 4 . 3
144
?
Kg/h = 
5
10 4 . 3
3600 144
?
?
Kg/s = 1.52 kg/s.  
7. l = 1 mm = 10
–3
m m = 10 kg
A = 200 cm
2
= 2 × 10
–2
m
2
L
vap
= 2.27 × 10
6
J/kg
K = 0.80 J/m-s-°C
0°C
100°C
10
10 cm
1 cm
Heat Transfer
28.2
dQ = 2.27 × 10
6
× 10,
dt
dQ
= 
5
7
10
10 27 . 2 ?
= 2.27 × 10
2
J/s
Again we know 
dt
dQ
= 
3
2
10 1
) T 42 ( 10 2 80 . 0
?
?
?
? ? ? ?
So, 
3
3
10
) T 42 ( 10 2 8
?
?
? ? ?
= 2.27 × 10
2
? 16 × 42 – 16T = 227 ? T = 27.8 ˜ 28°C
8. K = 45 w/m-°C
l = 60 cm = 60 × 10
–2
m
A = 0.2 cm
2
= 0.2 × 10
–4
m
2
Rate of heat flow,
=
?
) ( KA
2 1
? ? ?
= 
2
4
10 60
20 10 2 . 0 45
?
?
?
? ? ?
= 30 × 10
–3
0.03 w
9. A = 10 cm
2
, h = 10 cm
t
Q
?
?
= 
?
) ( KA
2 1
? ? ?
= 
3
3
10 1
30 10 200
?
?
?
? ?
= 6000
Since heat goes out from both surfaces. Hence net heat coming out.
= 
t
Q
?
?
= 6000 × 2 = 12000, 
t
Q
?
?
= 
t
MS
?
? ?
? 6000 × 2 = 10
–3
× 10
–1
× 1000 × 4200 × 
t ?
? ?
?
t ?
? ?
= 
420
72000
= 28.57
So, in 1 Sec. 28.57°C is dropped
Hence for drop of 1°C 
57 . 28
1
sec. = 0.035 sec. is required
10. l = 20 cm = 20× 10
–2
m
A = 0.2 cm
2
  = 0.2 × 10
–4
m
2
?
1
= 80°C, ?
2
= 20°C, K = 385
(a) 
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
2
4
10 20
) 20 80 ( 10 2 . 0 385
?
?
?
? ? ?
= 385 × 6 × 10
–4
×10 = 2310 × 10
–3
= 2.31
(b) Let the temp of the 11 cm point be ?
l ?
? ?
= 
tKA
Q
?
l ?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
?
2
10 11
20
?
?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
? ? – 20 = 
2
4
10 11
2 . 0 385
10 31 . 2
?
? ?
?
?
= 33
? ? = 33 + 20 = 53 ?
11. Let the point to be touched be ‘B’
No heat will flow when, the temp at that point is also 25°C
i.e. Q
AB
= Q
BC
So, 
x 100
) 25 100 ( KA
?
?
= 
x
) 0 25 ( KA ?
? 75 x = 2500 – 25 x ? 100 x = 2500   ? x = 25 cm from the end with 0°C
Q 1 = 40°
Q 2 = 20°
11 cm
80°C 20°C
B
C A
100 cm
100–x x
Page 3


28.1
CHAPTER 28
HEAT TRANSFER
1. t
1
= 90°C, t
2
= 10°C
l = 1 cm = 1 × 10
–3
m
A = 10 cm × 10 cm = 0.1 × 0.1 m
2
= 1 × 10
–2
m
2
K = 0.80 w/m-°C
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
2
2 1
10 1
80 10 1 10 8
?
? ?
?
? ? ? ?
= 64 J/s = 64 × 60 3840 J.
2. t = 1 cm = 0.01 m, A = 0.8 m
2
?
1
= 300, ?
2
= 80
K = 0.025, 
t
Q
= 
l
) ( KA
2 1
? ? ?
=  
01 . 0
) 30030 ( 8 . 0 025 . 0 ? ?
= 440 watt.
3. K = 0.04 J/m-5°C, A =  1.6 m
2
t
1
= 97°F = 36.1°C t
2
= 47°F = 8.33°C
l = 0.5 cm = 0.005 m
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
3
2
10 5
78 . 27 6 . 1 10 4
?
?
?
? ? ?
= 356 J/s
4. A = 25 cm
2
= 25 × 10
–4
m
2
l = 1 mm = 10
–3
m
K = 50 w/m-°C
t
Q
= Rate of conversion of water into steam
= 
min 1
10 26 . 2 10 100
6 3
? ? ?
?
= 
60
10 26 . 2 10
6 1
? ?
?
= 0.376 × 10
4
t
Q
= 
l
) ( KA
2 1
? ? ?
? 0.376 ×10
4
= 
3
4
10
) 100 ( 10 25 50
?
?
? ? ? ? ?
? ? = 
4
4 3
10 25 50
10 376 . 0 10
?
?
? ?
? ?
= 
25 50
376 . 0 10
5
?
?
= 30.1 ˜ 30 ?
5. K = 46 w/m-s°C
l = 1 m
A = 0.04 cm
2
= 4 × 10
–6
m
2
L
fussion ice
= 3.36 × 10
5
j/Kg
t
Q
= 
1
100 10 4 46
6
? ? ?
?
= 5.4 × 10
–8
kg ˜ 5.4 × 10
–5
g.
6. A = 2400 cm
2
= 2400 × 10
–4
m
2
l = 2 mm = 2 × 10
–3
m
K = 0.06 w/m-°C
?
1
= 20°C
?
2
= 0°C
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
3
4
10 2
20 10 2400 06 . 0
?
?
?
? ? ?
= 24 × 6 × 10
–1
× 10 = 24 × 6 = 144 J/sec
Rate in which ice melts = 
t
m
= 
L t
Q
?
= 
5
10 4 . 3
144
?
Kg/h = 
5
10 4 . 3
3600 144
?
?
Kg/s = 1.52 kg/s.  
7. l = 1 mm = 10
–3
m m = 10 kg
A = 200 cm
2
= 2 × 10
–2
m
2
L
vap
= 2.27 × 10
6
J/kg
K = 0.80 J/m-s-°C
0°C
100°C
10
10 cm
1 cm
Heat Transfer
28.2
dQ = 2.27 × 10
6
× 10,
dt
dQ
= 
5
7
10
10 27 . 2 ?
= 2.27 × 10
2
J/s
Again we know 
dt
dQ
= 
3
2
10 1
) T 42 ( 10 2 80 . 0
?
?
?
? ? ? ?
So, 
3
3
10
) T 42 ( 10 2 8
?
?
? ? ?
= 2.27 × 10
2
? 16 × 42 – 16T = 227 ? T = 27.8 ˜ 28°C
8. K = 45 w/m-°C
l = 60 cm = 60 × 10
–2
m
A = 0.2 cm
2
= 0.2 × 10
–4
m
2
Rate of heat flow,
=
?
) ( KA
2 1
? ? ?
= 
2
4
10 60
20 10 2 . 0 45
?
?
?
? ? ?
= 30 × 10
–3
0.03 w
9. A = 10 cm
2
, h = 10 cm
t
Q
?
?
= 
?
) ( KA
2 1
? ? ?
= 
3
3
10 1
30 10 200
?
?
?
? ?
= 6000
Since heat goes out from both surfaces. Hence net heat coming out.
= 
t
Q
?
?
= 6000 × 2 = 12000, 
t
Q
?
?
= 
t
MS
?
? ?
? 6000 × 2 = 10
–3
× 10
–1
× 1000 × 4200 × 
t ?
? ?
?
t ?
? ?
= 
420
72000
= 28.57
So, in 1 Sec. 28.57°C is dropped
Hence for drop of 1°C 
57 . 28
1
sec. = 0.035 sec. is required
10. l = 20 cm = 20× 10
–2
m
A = 0.2 cm
2
  = 0.2 × 10
–4
m
2
?
1
= 80°C, ?
2
= 20°C, K = 385
(a) 
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
2
4
10 20
) 20 80 ( 10 2 . 0 385
?
?
?
? ? ?
= 385 × 6 × 10
–4
×10 = 2310 × 10
–3
= 2.31
(b) Let the temp of the 11 cm point be ?
l ?
? ?
= 
tKA
Q
?
l ?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
?
2
10 11
20
?
?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
? ? – 20 = 
2
4
10 11
2 . 0 385
10 31 . 2
?
? ?
?
?
= 33
? ? = 33 + 20 = 53 ?
11. Let the point to be touched be ‘B’
No heat will flow when, the temp at that point is also 25°C
i.e. Q
AB
= Q
BC
So, 
x 100
) 25 100 ( KA
?
?
= 
x
) 0 25 ( KA ?
? 75 x = 2500 – 25 x ? 100 x = 2500   ? x = 25 cm from the end with 0°C
Q 1 = 40°
Q 2 = 20°
11 cm
80°C 20°C
B
C A
100 cm
100–x x
Heat Transfer
28.3
12. V = 216 cm
3
a = 6 cm, Surface area = 6 a
2
= 6 × 36 m
2
t = 0.1 cm
t
Q
= 100 W, 
t
Q
= 
?
) ( KA
2 1
? ? ?
? 100 = 
2
4
10 1 . 0
5 10 36 6 K
?
?
?
? ? ? ?
? K = 
1
10 5 36 6
100
?
? ? ?
= 0.9259 W/m°C ˜ 0.92 W/m°C
13. Given ?
1
= 1°C, ?
2
  = 0°C
K = 0.50 w/m-°C, d = 2 mm = 2 × 10
–3
m
A = 5 × 10
–2
m
2
, v = 10 cm/s = 0.1 m/s 
Power = Force × Velocity = Mg × v
Again Power = 
dt
dQ
= 
d
) ( KA
2 1
? ? ?
So, Mgv = 
d
) ( KA
2 1
? ? ?
? M = 
dvg
) ( KA
2 1
? ? ?
= 
10 10 10 2
1 5 10 5
1 3
2 1
? ? ?
? ? ? ?
? ?
? ?
= 12.5 kg.   
14. K = 1.7 W/m-°C ƒ
w
= 1000 Kg/m
3
L
ice
= 3.36 × 10
5
J/kg T = 10 cm = 10 × 10
–2
m
(a) 
t
Q
= 
?
) ( KA
2 1
? ? ?
  ?
t
?
= 
Q
) ( KA
2 1
? ? ?
= 
mL
) ( KA
2 1
? ? ?
= 
L ƒ At
) ( KA
w
2 1
? ? ?
= 
5 2
10 36 . 3 1000 10 10
)] 10 ( 0 [ 7 . 1
? ? ? ?
? ? ?
?
= 
7
10
36 . 3
17
?
? = 5.059 × 10
–7 
˜ 5 × 10
–7 
m/sec 
(b) let us assume that x length of ice has become formed to form a small strip of ice of length dx, dt time 
is required.
dt
dQ
= 
x
) ( KA ? ?
?
dt
dmL
= 
x
) ( KA ? ?
?
dt
L ƒ Adx ?
= 
x
) ( KA ? ?
?
dt
L ƒ dx ?
= 
x
) ( K ? ?
? dt = 
) ( K
L ƒ xdx
? ?
?
?
?
t
0
dt = 
?
? ?
?
t
0
xdx
) ( K
L ƒ
? t = 
l
o
2
2
x
) ( K
L ƒ
?
?
?
?
?
?
?
?
? ?
?
= 
2
l
K
L ƒ
2
? ?
?
Putting values
? t = 
? ?
2 10 7 . 1
10 10 10 36 . 3 1000
2
2 5
? ?
? ? ? ?
?
= 
6
10
17 2
36 . 3
?
?
sec. = 
3600 17 2
10 36 . 3
6
? ?
?
hrs = 27.45 hrs ˜ 27.5 hrs.
15. let ‘B’ be the maximum level upto which ice is formed. Hence the heat conducted at that point from both 
the levels is the same.
Let AB = x
i.e. ice
t
Q
= water
t
Q
?
x
10 A K
ice
? ?
= 
) x 1 (
4 A K
water
?
? ?
?
x
10 7 . 1 ?
= 
x 1
4 10 5
1
?
? ?
?
?
x
17
= 
x 1
2
?
? 17 – 17 x = 2x ? 19 x = 17 ? x = 
19
17
= 0.894 ˜ 89 cm
M
–0°C
10 cm
0°C
x
dx
1 cm
x
1–x
A
C
–10°C
4°C
Page 4


28.1
CHAPTER 28
HEAT TRANSFER
1. t
1
= 90°C, t
2
= 10°C
l = 1 cm = 1 × 10
–3
m
A = 10 cm × 10 cm = 0.1 × 0.1 m
2
= 1 × 10
–2
m
2
K = 0.80 w/m-°C
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
2
2 1
10 1
80 10 1 10 8
?
? ?
?
? ? ? ?
= 64 J/s = 64 × 60 3840 J.
2. t = 1 cm = 0.01 m, A = 0.8 m
2
?
1
= 300, ?
2
= 80
K = 0.025, 
t
Q
= 
l
) ( KA
2 1
? ? ?
=  
01 . 0
) 30030 ( 8 . 0 025 . 0 ? ?
= 440 watt.
3. K = 0.04 J/m-5°C, A =  1.6 m
2
t
1
= 97°F = 36.1°C t
2
= 47°F = 8.33°C
l = 0.5 cm = 0.005 m
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
3
2
10 5
78 . 27 6 . 1 10 4
?
?
?
? ? ?
= 356 J/s
4. A = 25 cm
2
= 25 × 10
–4
m
2
l = 1 mm = 10
–3
m
K = 50 w/m-°C
t
Q
= Rate of conversion of water into steam
= 
min 1
10 26 . 2 10 100
6 3
? ? ?
?
= 
60
10 26 . 2 10
6 1
? ?
?
= 0.376 × 10
4
t
Q
= 
l
) ( KA
2 1
? ? ?
? 0.376 ×10
4
= 
3
4
10
) 100 ( 10 25 50
?
?
? ? ? ? ?
? ? = 
4
4 3
10 25 50
10 376 . 0 10
?
?
? ?
? ?
= 
25 50
376 . 0 10
5
?
?
= 30.1 ˜ 30 ?
5. K = 46 w/m-s°C
l = 1 m
A = 0.04 cm
2
= 4 × 10
–6
m
2
L
fussion ice
= 3.36 × 10
5
j/Kg
t
Q
= 
1
100 10 4 46
6
? ? ?
?
= 5.4 × 10
–8
kg ˜ 5.4 × 10
–5
g.
6. A = 2400 cm
2
= 2400 × 10
–4
m
2
l = 2 mm = 2 × 10
–3
m
K = 0.06 w/m-°C
?
1
= 20°C
?
2
= 0°C
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
3
4
10 2
20 10 2400 06 . 0
?
?
?
? ? ?
= 24 × 6 × 10
–1
× 10 = 24 × 6 = 144 J/sec
Rate in which ice melts = 
t
m
= 
L t
Q
?
= 
5
10 4 . 3
144
?
Kg/h = 
5
10 4 . 3
3600 144
?
?
Kg/s = 1.52 kg/s.  
7. l = 1 mm = 10
–3
m m = 10 kg
A = 200 cm
2
= 2 × 10
–2
m
2
L
vap
= 2.27 × 10
6
J/kg
K = 0.80 J/m-s-°C
0°C
100°C
10
10 cm
1 cm
Heat Transfer
28.2
dQ = 2.27 × 10
6
× 10,
dt
dQ
= 
5
7
10
10 27 . 2 ?
= 2.27 × 10
2
J/s
Again we know 
dt
dQ
= 
3
2
10 1
) T 42 ( 10 2 80 . 0
?
?
?
? ? ? ?
So, 
3
3
10
) T 42 ( 10 2 8
?
?
? ? ?
= 2.27 × 10
2
? 16 × 42 – 16T = 227 ? T = 27.8 ˜ 28°C
8. K = 45 w/m-°C
l = 60 cm = 60 × 10
–2
m
A = 0.2 cm
2
= 0.2 × 10
–4
m
2
Rate of heat flow,
=
?
) ( KA
2 1
? ? ?
= 
2
4
10 60
20 10 2 . 0 45
?
?
?
? ? ?
= 30 × 10
–3
0.03 w
9. A = 10 cm
2
, h = 10 cm
t
Q
?
?
= 
?
) ( KA
2 1
? ? ?
= 
3
3
10 1
30 10 200
?
?
?
? ?
= 6000
Since heat goes out from both surfaces. Hence net heat coming out.
= 
t
Q
?
?
= 6000 × 2 = 12000, 
t
Q
?
?
= 
t
MS
?
? ?
? 6000 × 2 = 10
–3
× 10
–1
× 1000 × 4200 × 
t ?
? ?
?
t ?
? ?
= 
420
72000
= 28.57
So, in 1 Sec. 28.57°C is dropped
Hence for drop of 1°C 
57 . 28
1
sec. = 0.035 sec. is required
10. l = 20 cm = 20× 10
–2
m
A = 0.2 cm
2
  = 0.2 × 10
–4
m
2
?
1
= 80°C, ?
2
= 20°C, K = 385
(a) 
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
2
4
10 20
) 20 80 ( 10 2 . 0 385
?
?
?
? ? ?
= 385 × 6 × 10
–4
×10 = 2310 × 10
–3
= 2.31
(b) Let the temp of the 11 cm point be ?
l ?
? ?
= 
tKA
Q
?
l ?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
?
2
10 11
20
?
?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
? ? – 20 = 
2
4
10 11
2 . 0 385
10 31 . 2
?
? ?
?
?
= 33
? ? = 33 + 20 = 53 ?
11. Let the point to be touched be ‘B’
No heat will flow when, the temp at that point is also 25°C
i.e. Q
AB
= Q
BC
So, 
x 100
) 25 100 ( KA
?
?
= 
x
) 0 25 ( KA ?
? 75 x = 2500 – 25 x ? 100 x = 2500   ? x = 25 cm from the end with 0°C
Q 1 = 40°
Q 2 = 20°
11 cm
80°C 20°C
B
C A
100 cm
100–x x
Heat Transfer
28.3
12. V = 216 cm
3
a = 6 cm, Surface area = 6 a
2
= 6 × 36 m
2
t = 0.1 cm
t
Q
= 100 W, 
t
Q
= 
?
) ( KA
2 1
? ? ?
? 100 = 
2
4
10 1 . 0
5 10 36 6 K
?
?
?
? ? ? ?
? K = 
1
10 5 36 6
100
?
? ? ?
= 0.9259 W/m°C ˜ 0.92 W/m°C
13. Given ?
1
= 1°C, ?
2
  = 0°C
K = 0.50 w/m-°C, d = 2 mm = 2 × 10
–3
m
A = 5 × 10
–2
m
2
, v = 10 cm/s = 0.1 m/s 
Power = Force × Velocity = Mg × v
Again Power = 
dt
dQ
= 
d
) ( KA
2 1
? ? ?
So, Mgv = 
d
) ( KA
2 1
? ? ?
? M = 
dvg
) ( KA
2 1
? ? ?
= 
10 10 10 2
1 5 10 5
1 3
2 1
? ? ?
? ? ? ?
? ?
? ?
= 12.5 kg.   
14. K = 1.7 W/m-°C ƒ
w
= 1000 Kg/m
3
L
ice
= 3.36 × 10
5
J/kg T = 10 cm = 10 × 10
–2
m
(a) 
t
Q
= 
?
) ( KA
2 1
? ? ?
  ?
t
?
= 
Q
) ( KA
2 1
? ? ?
= 
mL
) ( KA
2 1
? ? ?
= 
L ƒ At
) ( KA
w
2 1
? ? ?
= 
5 2
10 36 . 3 1000 10 10
)] 10 ( 0 [ 7 . 1
? ? ? ?
? ? ?
?
= 
7
10
36 . 3
17
?
? = 5.059 × 10
–7 
˜ 5 × 10
–7 
m/sec 
(b) let us assume that x length of ice has become formed to form a small strip of ice of length dx, dt time 
is required.
dt
dQ
= 
x
) ( KA ? ?
?
dt
dmL
= 
x
) ( KA ? ?
?
dt
L ƒ Adx ?
= 
x
) ( KA ? ?
?
dt
L ƒ dx ?
= 
x
) ( K ? ?
? dt = 
) ( K
L ƒ xdx
? ?
?
?
?
t
0
dt = 
?
? ?
?
t
0
xdx
) ( K
L ƒ
? t = 
l
o
2
2
x
) ( K
L ƒ
?
?
?
?
?
?
?
?
? ?
?
= 
2
l
K
L ƒ
2
? ?
?
Putting values
? t = 
? ?
2 10 7 . 1
10 10 10 36 . 3 1000
2
2 5
? ?
? ? ? ?
?
= 
6
10
17 2
36 . 3
?
?
sec. = 
3600 17 2
10 36 . 3
6
? ?
?
hrs = 27.45 hrs ˜ 27.5 hrs.
15. let ‘B’ be the maximum level upto which ice is formed. Hence the heat conducted at that point from both 
the levels is the same.
Let AB = x
i.e. ice
t
Q
= water
t
Q
?
x
10 A K
ice
? ?
= 
) x 1 (
4 A K
water
?
? ?
?
x
10 7 . 1 ?
= 
x 1
4 10 5
1
?
? ?
?
?
x
17
= 
x 1
2
?
? 17 – 17 x = 2x ? 19 x = 17 ? x = 
19
17
= 0.894 ˜ 89 cm
M
–0°C
10 cm
0°C
x
dx
1 cm
x
1–x
A
C
–10°C
4°C
Heat Transfer
28.4
16. K
AB
= 50 j/m-s-°c ?
A
= 40°C
K
BC
= 200 j/m-s-°c ?
B
= 80°C
K
AC
= 400 j/m-s-°c ?
C
= 80°C
Length = 20 cm = 20 × 10
–2
m
A = 1 cm
2
= 1 × 10
–4
m
2
(a) 
t
Q
AB
= 
l
) ( A K
A B AB
? ? ? ?
= 
2
4
10 20
40 10 1 50
?
?
?
? ? ?
  = 1 W.
(b) 
t
Q
AC
= 
l
) ( A K
A C AC
? ? ? ?
= 
2
4
10 20
40 10 1 400
?
?
?
? ? ?
  = 800 × 10
–2
= 8 
(c) 
t
Q
BC
= 
l
) ( A K
C B BC
? ? ? ?
= 
2
4
10 20
0 10 1 200
?
?
?
? ? ?
  = 0
17. We know Q = 
d
) ( KA
2 1
? ? ?
Q
1
= 
1
2 1
d
) ( KA ? ? ?
, Q
2
= 
2
2 1
d
) ( KA ? ? ?
2
1
Q
Q
= 
r 2
) ( KA
r
) ( KA
1 1
1 1
? ? ?
?
? ? ?
=
r
r 2
?
= 
?
2
  [d
1
= ?r, d
2
= 2r] ?
18. The rate of heat flow per sec.
= 
dt
dQ
A
= 
dt
d
KA
?
The rate of heat flow per sec.
= 
dt
dQ
B
= 
dt
d
KA
B
?
This part of heat is absorbed by the red.
t
Q
= 
dt
ms ? ?
where 
dt
d ?
= Rate of net temp. variation
?
dt
msd ?
= 
dt
d
KA
dt
d
KA
B A
?
?
?
?
dt
d
ms
?
= ?
?
?
?
?
? ?
?
?
dt
d
dt
d
KA
B A
?
dt
d
4 . 0
?
? = 200 × 1 × 10
–4
(5 – 2.5) °C/cm
?
dt
d
4 . 0
?
? = 200 × 10
-4
× 2.5
?
dt
d ?
= 
2
4
10 4 . 0
10 5 . 2 200
?
?
?
? ?
°C/m = 1250 × 10
–2
= 12.5 °C/m
19. Given
K
rubber
= 0.15 J/m-s-°C T
2
- T
1
= 90°C
We know for radial conduction in a Cylinder
t
Q
= 
) R / R ln(
) T T ( Kl 2
1 2
1 2
? ?
= 
) 1 / 2 . 1 ln(
90 10 50 10 15 14 . 3 2
1 2
? ? ? ? ? ?
? ?
= 232.5 ˜ 233 j/s.
20.
dt
dQ
= Rate of flow of heat
Let us consider a strip at a distance r from the center of thickness dr.
dt
dQ
= 
dr
d rd 2 K ? ? ? ?
[d ? = Temperature diff across the thickness dr]
r
r
50 cm
120°C
Page 5


28.1
CHAPTER 28
HEAT TRANSFER
1. t
1
= 90°C, t
2
= 10°C
l = 1 cm = 1 × 10
–3
m
A = 10 cm × 10 cm = 0.1 × 0.1 m
2
= 1 × 10
–2
m
2
K = 0.80 w/m-°C
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
2
2 1
10 1
80 10 1 10 8
?
? ?
?
? ? ? ?
= 64 J/s = 64 × 60 3840 J.
2. t = 1 cm = 0.01 m, A = 0.8 m
2
?
1
= 300, ?
2
= 80
K = 0.025, 
t
Q
= 
l
) ( KA
2 1
? ? ?
=  
01 . 0
) 30030 ( 8 . 0 025 . 0 ? ?
= 440 watt.
3. K = 0.04 J/m-5°C, A =  1.6 m
2
t
1
= 97°F = 36.1°C t
2
= 47°F = 8.33°C
l = 0.5 cm = 0.005 m
t
Q
= 
l
) ( KA
2 1
? ? ?
= 
3
2
10 5
78 . 27 6 . 1 10 4
?
?
?
? ? ?
= 356 J/s
4. A = 25 cm
2
= 25 × 10
–4
m
2
l = 1 mm = 10
–3
m
K = 50 w/m-°C
t
Q
= Rate of conversion of water into steam
= 
min 1
10 26 . 2 10 100
6 3
? ? ?
?
= 
60
10 26 . 2 10
6 1
? ?
?
= 0.376 × 10
4
t
Q
= 
l
) ( KA
2 1
? ? ?
? 0.376 ×10
4
= 
3
4
10
) 100 ( 10 25 50
?
?
? ? ? ? ?
? ? = 
4
4 3
10 25 50
10 376 . 0 10
?
?
? ?
? ?
= 
25 50
376 . 0 10
5
?
?
= 30.1 ˜ 30 ?
5. K = 46 w/m-s°C
l = 1 m
A = 0.04 cm
2
= 4 × 10
–6
m
2
L
fussion ice
= 3.36 × 10
5
j/Kg
t
Q
= 
1
100 10 4 46
6
? ? ?
?
= 5.4 × 10
–8
kg ˜ 5.4 × 10
–5
g.
6. A = 2400 cm
2
= 2400 × 10
–4
m
2
l = 2 mm = 2 × 10
–3
m
K = 0.06 w/m-°C
?
1
= 20°C
?
2
= 0°C
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
3
4
10 2
20 10 2400 06 . 0
?
?
?
? ? ?
= 24 × 6 × 10
–1
× 10 = 24 × 6 = 144 J/sec
Rate in which ice melts = 
t
m
= 
L t
Q
?
= 
5
10 4 . 3
144
?
Kg/h = 
5
10 4 . 3
3600 144
?
?
Kg/s = 1.52 kg/s.  
7. l = 1 mm = 10
–3
m m = 10 kg
A = 200 cm
2
= 2 × 10
–2
m
2
L
vap
= 2.27 × 10
6
J/kg
K = 0.80 J/m-s-°C
0°C
100°C
10
10 cm
1 cm
Heat Transfer
28.2
dQ = 2.27 × 10
6
× 10,
dt
dQ
= 
5
7
10
10 27 . 2 ?
= 2.27 × 10
2
J/s
Again we know 
dt
dQ
= 
3
2
10 1
) T 42 ( 10 2 80 . 0
?
?
?
? ? ? ?
So, 
3
3
10
) T 42 ( 10 2 8
?
?
? ? ?
= 2.27 × 10
2
? 16 × 42 – 16T = 227 ? T = 27.8 ˜ 28°C
8. K = 45 w/m-°C
l = 60 cm = 60 × 10
–2
m
A = 0.2 cm
2
= 0.2 × 10
–4
m
2
Rate of heat flow,
=
?
) ( KA
2 1
? ? ?
= 
2
4
10 60
20 10 2 . 0 45
?
?
?
? ? ?
= 30 × 10
–3
0.03 w
9. A = 10 cm
2
, h = 10 cm
t
Q
?
?
= 
?
) ( KA
2 1
? ? ?
= 
3
3
10 1
30 10 200
?
?
?
? ?
= 6000
Since heat goes out from both surfaces. Hence net heat coming out.
= 
t
Q
?
?
= 6000 × 2 = 12000, 
t
Q
?
?
= 
t
MS
?
? ?
? 6000 × 2 = 10
–3
× 10
–1
× 1000 × 4200 × 
t ?
? ?
?
t ?
? ?
= 
420
72000
= 28.57
So, in 1 Sec. 28.57°C is dropped
Hence for drop of 1°C 
57 . 28
1
sec. = 0.035 sec. is required
10. l = 20 cm = 20× 10
–2
m
A = 0.2 cm
2
  = 0.2 × 10
–4
m
2
?
1
= 80°C, ?
2
= 20°C, K = 385
(a) 
t
Q
= 
?
) ( KA
2 1
? ? ?
= 
2
4
10 20
) 20 80 ( 10 2 . 0 385
?
?
?
? ? ?
= 385 × 6 × 10
–4
×10 = 2310 × 10
–3
= 2.31
(b) Let the temp of the 11 cm point be ?
l ?
? ?
= 
tKA
Q
?
l ?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
?
2
10 11
20
?
?
? ?
= 
4
10 2 . 0 385
31 . 2
?
? ?
? ? – 20 = 
2
4
10 11
2 . 0 385
10 31 . 2
?
? ?
?
?
= 33
? ? = 33 + 20 = 53 ?
11. Let the point to be touched be ‘B’
No heat will flow when, the temp at that point is also 25°C
i.e. Q
AB
= Q
BC
So, 
x 100
) 25 100 ( KA
?
?
= 
x
) 0 25 ( KA ?
? 75 x = 2500 – 25 x ? 100 x = 2500   ? x = 25 cm from the end with 0°C
Q 1 = 40°
Q 2 = 20°
11 cm
80°C 20°C
B
C A
100 cm
100–x x
Heat Transfer
28.3
12. V = 216 cm
3
a = 6 cm, Surface area = 6 a
2
= 6 × 36 m
2
t = 0.1 cm
t
Q
= 100 W, 
t
Q
= 
?
) ( KA
2 1
? ? ?
? 100 = 
2
4
10 1 . 0
5 10 36 6 K
?
?
?
? ? ? ?
? K = 
1
10 5 36 6
100
?
? ? ?
= 0.9259 W/m°C ˜ 0.92 W/m°C
13. Given ?
1
= 1°C, ?
2
  = 0°C
K = 0.50 w/m-°C, d = 2 mm = 2 × 10
–3
m
A = 5 × 10
–2
m
2
, v = 10 cm/s = 0.1 m/s 
Power = Force × Velocity = Mg × v
Again Power = 
dt
dQ
= 
d
) ( KA
2 1
? ? ?
So, Mgv = 
d
) ( KA
2 1
? ? ?
? M = 
dvg
) ( KA
2 1
? ? ?
= 
10 10 10 2
1 5 10 5
1 3
2 1
? ? ?
? ? ? ?
? ?
? ?
= 12.5 kg.   
14. K = 1.7 W/m-°C ƒ
w
= 1000 Kg/m
3
L
ice
= 3.36 × 10
5
J/kg T = 10 cm = 10 × 10
–2
m
(a) 
t
Q
= 
?
) ( KA
2 1
? ? ?
  ?
t
?
= 
Q
) ( KA
2 1
? ? ?
= 
mL
) ( KA
2 1
? ? ?
= 
L ƒ At
) ( KA
w
2 1
? ? ?
= 
5 2
10 36 . 3 1000 10 10
)] 10 ( 0 [ 7 . 1
? ? ? ?
? ? ?
?
= 
7
10
36 . 3
17
?
? = 5.059 × 10
–7 
˜ 5 × 10
–7 
m/sec 
(b) let us assume that x length of ice has become formed to form a small strip of ice of length dx, dt time 
is required.
dt
dQ
= 
x
) ( KA ? ?
?
dt
dmL
= 
x
) ( KA ? ?
?
dt
L ƒ Adx ?
= 
x
) ( KA ? ?
?
dt
L ƒ dx ?
= 
x
) ( K ? ?
? dt = 
) ( K
L ƒ xdx
? ?
?
?
?
t
0
dt = 
?
? ?
?
t
0
xdx
) ( K
L ƒ
? t = 
l
o
2
2
x
) ( K
L ƒ
?
?
?
?
?
?
?
?
? ?
?
= 
2
l
K
L ƒ
2
? ?
?
Putting values
? t = 
? ?
2 10 7 . 1
10 10 10 36 . 3 1000
2
2 5
? ?
? ? ? ?
?
= 
6
10
17 2
36 . 3
?
?
sec. = 
3600 17 2
10 36 . 3
6
? ?
?
hrs = 27.45 hrs ˜ 27.5 hrs.
15. let ‘B’ be the maximum level upto which ice is formed. Hence the heat conducted at that point from both 
the levels is the same.
Let AB = x
i.e. ice
t
Q
= water
t
Q
?
x
10 A K
ice
? ?
= 
) x 1 (
4 A K
water
?
? ?
?
x
10 7 . 1 ?
= 
x 1
4 10 5
1
?
? ?
?
?
x
17
= 
x 1
2
?
? 17 – 17 x = 2x ? 19 x = 17 ? x = 
19
17
= 0.894 ˜ 89 cm
M
–0°C
10 cm
0°C
x
dx
1 cm
x
1–x
A
C
–10°C
4°C
Heat Transfer
28.4
16. K
AB
= 50 j/m-s-°c ?
A
= 40°C
K
BC
= 200 j/m-s-°c ?
B
= 80°C
K
AC
= 400 j/m-s-°c ?
C
= 80°C
Length = 20 cm = 20 × 10
–2
m
A = 1 cm
2
= 1 × 10
–4
m
2
(a) 
t
Q
AB
= 
l
) ( A K
A B AB
? ? ? ?
= 
2
4
10 20
40 10 1 50
?
?
?
? ? ?
  = 1 W.
(b) 
t
Q
AC
= 
l
) ( A K
A C AC
? ? ? ?
= 
2
4
10 20
40 10 1 400
?
?
?
? ? ?
  = 800 × 10
–2
= 8 
(c) 
t
Q
BC
= 
l
) ( A K
C B BC
? ? ? ?
= 
2
4
10 20
0 10 1 200
?
?
?
? ? ?
  = 0
17. We know Q = 
d
) ( KA
2 1
? ? ?
Q
1
= 
1
2 1
d
) ( KA ? ? ?
, Q
2
= 
2
2 1
d
) ( KA ? ? ?
2
1
Q
Q
= 
r 2
) ( KA
r
) ( KA
1 1
1 1
? ? ?
?
? ? ?
=
r
r 2
?
= 
?
2
  [d
1
= ?r, d
2
= 2r] ?
18. The rate of heat flow per sec.
= 
dt
dQ
A
= 
dt
d
KA
?
The rate of heat flow per sec.
= 
dt
dQ
B
= 
dt
d
KA
B
?
This part of heat is absorbed by the red.
t
Q
= 
dt
ms ? ?
where 
dt
d ?
= Rate of net temp. variation
?
dt
msd ?
= 
dt
d
KA
dt
d
KA
B A
?
?
?
?
dt
d
ms
?
= ?
?
?
?
?
? ?
?
?
dt
d
dt
d
KA
B A
?
dt
d
4 . 0
?
? = 200 × 1 × 10
–4
(5 – 2.5) °C/cm
?
dt
d
4 . 0
?
? = 200 × 10
-4
× 2.5
?
dt
d ?
= 
2
4
10 4 . 0
10 5 . 2 200
?
?
?
? ?
°C/m = 1250 × 10
–2
= 12.5 °C/m
19. Given
K
rubber
= 0.15 J/m-s-°C T
2
- T
1
= 90°C
We know for radial conduction in a Cylinder
t
Q
= 
) R / R ln(
) T T ( Kl 2
1 2
1 2
? ?
= 
) 1 / 2 . 1 ln(
90 10 50 10 15 14 . 3 2
1 2
? ? ? ? ? ?
? ?
= 232.5 ˜ 233 j/s.
20.
dt
dQ
= Rate of flow of heat
Let us consider a strip at a distance r from the center of thickness dr.
dt
dQ
= 
dr
d rd 2 K ? ? ? ?
[d ? = Temperature diff across the thickness dr]
r
r
50 cm
120°C
Heat Transfer
28.5
? C = 
dr
d rd 2 K ? ? ? ?
?
?
?
?
?
? ?
?
dr
d
c
?
r
dr
C = K2 ?d d ?
Integrating 
?
?
2
1
r
r
r
dr
C = K2 ?d 
?
?
?
?
2
1
d ? ? ?
2
1
r
r
r log C = K2 ?d ( ?
2
– ?
1
)
? C(log r
2
– log r
1
) = K2 ?d ( ?
2
– ?
1
) ? C log
?
?
?
?
?
?
?
?
1
2
r
r
= K2 ?d ( ?
2
– ?
1
)
? C = 
) r / r log(
) ( d 2 K
1 2
1 2
? ? ? ?
21. T
1
> T
2
A = ?(R
2
2
– R
1
2
)
So, Q = 
l
) T T ( KA
1 2
?
= 
l
) T T )( R R ( KA
1 2
2
1
2
2
? ?
Considering a concentric cylindrical shell of radius ‘r’ and thickness 
‘dr’. The radial heat flow through the shell
H = 
dt
dQ
= – KA
dt
d ?
[(-)ve because as r – increases ?
decreases]
A = 2 ?rl H = –2 ?rl K 
dt
d ?
or 
?
2
1
R
R
r
dr
= 
?
?
?
?
2
1
T
T
d
H
LK 2
Integrating and simplifying we get
H = 
dt
dQ
= 
) R / R ( Loge
) T T ( KL 2
1 2
1 2
? ?
= 
) R / R ln(
) T T ( KL 2
1 2
1 2
? ?
?
22. Here the thermal conductivities are in series,
?
2
2 1 2
1
2 1 1
2
2 1 2
1
2 1 1
l
) ( A K
l
) ( A K
l
) ( A K
l
) ( A K
? ? ?
?
? ? ?
? ? ?
?
? ? ?
= 
2 1
2 1
l l
) ( KA
?
? ? ?
?
2
2
1
1
2
2
1
1
l
K
l
K
l
K
l
K
?
?
  = 
2 1
l l
K
?
?
1 2 2 1
2 1
l K l K
K K
?
= 
2 1
l l
K
?
? K = 
1 2 2 1
2 1 2 1
l K l K
) l l )( K K (
?
?
23. K
Cu
= 390 w/m-°C K
St
= 46 w/m-°C
Now, Since they are in series connection,
So, the heat passed through the crossections in the same.
So, Q
1
= Q
2
Or 
l
) 0 ( A K
Cu
? ? ? ?
= 
l
) 100 ( A K
St
? ? ? ?
? 390( ? – 0) = 46 × 100 – 46 ??? 436 ? = 4600
? ? = 
436
4600
= 10.55 ˜ 10.6°C ?
l
T 2
T 1
R 2
R 1
L 1
L 2
Steel Cu 100°C
?°C
0°C
r 1
dr
r
r 2
Read More
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