Heat Transfer & Newton’s Law of Cooling Practice Questions - DPP for NEET

 Page 1


1. (a)
dQ KA
dt
Dq
=
l
For both rods K, A and Dq are same
Þ
1 dQ
dt
µ
l
So,
( /)
( /)
semi circular
straight
dQ dt
dQ dt
straight
semi circular
=
l
l
22 r
r
==
pp
.
2. (b) Suppose thickness of each wall is x then
combination A
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
12 1
()2 ()
2
s
KA KA
xx
q -q q -q
=
Q
224
(2 )3
s
KK
KK
KK
´´
==
+
and 
12
( ) 36 q-q =°
A B
K 2K
x x
q
1
q
2
q
Þ
1
4
36
2 ()
3
2
KA
KA
xx
´
q -q
=
Hence temperature difference across wall A is
1
() q -q = 12° C
3. (a) Heat developed by the heater
22
(200)
.
20 4.2
Vtt
H
RJ
´
==
´
Heat conducted by the glass
0.2 1 (20 )
0.002
t
H
´ ´ -q
=
Hence 
2
(200) 0.2 (20 )
20 4.2 0.002
tt ´ ´ -q
=
´
Þ q = 15.24° C
4. ( b) If suppose K
Ni
 = K Þ K
Al
 = 3K and K
Cu
 = 6K
Since all metal bars are connected in series.
So, 
Combination Cu Al Ni
Q Q QQ
t t tt
æö æ ö æö æ ö
= ==
ç÷ ç ÷ ç÷ ç ÷
èø è ø èø è ø
and 
3 1 11
eq Cu Al Ni
K K KK
= ++
1 1 19
636 K K KK
= + +=
ÞK
eq
 = 2K
Cu Ni Al
100° C 0° C
Q Q
2
q 1
q
25 cm 10 cm 15 cm
Hence, if 
Combination Cu
QQ
tt
æ ö æö
=
ç ÷ ç÷
è ø èø
Þ
1
(100 0)
(100 ) eq
Cu
Combination Cu
KA
KA
-
-q
=
ll
Þ
1
6 (100 ) 2 (100 0)
(25 10 15) 25
KA KA -q -
=
++
Þ
1
q = 83.33° C
Similar if 
Combination Al
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
2
3 ( 0) 2 (100 0)
50 15
KA KA q- -
=
Þ
2
q = 20° C
5. ( b) Let the temperature of junction be q then according to
following figure.
100°C
50°C
20°C
K
2K
3K
H
q
H
1
H
2
Page 2


1. (a)
dQ KA
dt
Dq
=
l
For both rods K, A and Dq are same
Þ
1 dQ
dt
µ
l
So,
( /)
( /)
semi circular
straight
dQ dt
dQ dt
straight
semi circular
=
l
l
22 r
r
==
pp
.
2. (b) Suppose thickness of each wall is x then
combination A
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
12 1
()2 ()
2
s
KA KA
xx
q -q q -q
=
Q
224
(2 )3
s
KK
KK
KK
´´
==
+
and 
12
( ) 36 q-q =°
A B
K 2K
x x
q
1
q
2
q
Þ
1
4
36
2 ()
3
2
KA
KA
xx
´
q -q
=
Hence temperature difference across wall A is
1
() q -q = 12° C
3. (a) Heat developed by the heater
22
(200)
.
20 4.2
Vtt
H
RJ
´
==
´
Heat conducted by the glass
0.2 1 (20 )
0.002
t
H
´ ´ -q
=
Hence 
2
(200) 0.2 (20 )
20 4.2 0.002
tt ´ ´ -q
=
´
Þ q = 15.24° C
4. ( b) If suppose K
Ni
 = K Þ K
Al
 = 3K and K
Cu
 = 6K
Since all metal bars are connected in series.
So, 
Combination Cu Al Ni
Q Q QQ
t t tt
æö æ ö æö æ ö
= ==
ç÷ ç ÷ ç÷ ç ÷
èø è ø èø è ø
and 
3 1 11
eq Cu Al Ni
K K KK
= ++
1 1 19
636 K K KK
= + +=
ÞK
eq
 = 2K
Cu Ni Al
100° C 0° C
Q Q
2
q 1
q
25 cm 10 cm 15 cm
Hence, if 
Combination Cu
QQ
tt
æ ö æö
=
ç ÷ ç÷
è ø èø
Þ
1
(100 0)
(100 ) eq
Cu
Combination Cu
KA
KA
-
-q
=
ll
Þ
1
6 (100 ) 2 (100 0)
(25 10 15) 25
KA KA -q -
=
++
Þ
1
q = 83.33° C
Similar if 
Combination Al
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
2
3 ( 0) 2 (100 0)
50 15
KA KA q- -
=
Þ
2
q = 20° C
5. ( b) Let the temperature of junction be q then according to
following figure.
100°C
50°C
20°C
K
2K
3K
H
q
H
1
H
2
DPP/ P 23
67
12
H HH =+
Þ
3 (100) KA ´ ´ -q
l
2 ( 50) ( 20) KA KA q - q-
=+
ll
Þ 300 – 3 q = 3q – 120
Þ q = 70° C
6. ( d) Wein's displacement law is
m
Tb l=
Þ m
b
T
l=
6
2.88 10
1000
2880
´
== nm
Energy distribution with wavelength will be as follows
2
U
1
U
3
U
( ) nm l 1499
900
499
1500
1000
500
E
l
From the graph it is clear that
U
2 
> U
1
7. (a) According to Newton law of cooling
12 12
0
2
K
t
q - q q +q éù
= -q
êú
ëû
80° C
1
2
3
5 min
10 min
15 min
64°C
52°C
? q=
For first process :
0
(80 64) 80 64
52
K
-+éù
= -q
êú
ëû
....... (i)
For second process :
0
(80 52) 80 52
102
K
-+éù
= -q
êú
ëû
....... (ii)
For third process :
0
(80 ) 80
152
K
- q +q éù
= -q
êú
ëû
....... (iii)
On solving equation (i) and (ii) we get 
1
15
K = and 
0
q
= 24° C. Putting these values in equation (iii) we get 
q
= 42.7° C.
8. (c)
12
()
Q
t
KA
=
q -q
l
12 12
() ()
mL VL
KA KA
r
==
q -q q -q
ll
5 10
5 0.92 80
2
0.004 10 3600
A
A
+
´ ´ ´´
=
´ ´´
= 19.1 hours.
9. (c)
Q KA
t
Dq
=
l
Þ
2
() mL Kr
t
p Dq
=
l
Þ Rate of melting of ice 
2
m Kr
t
æö
µ
ç÷
èø l
Since for second rod K  becomes 
1
,
4
thr becomes
double and length becomes half, so rate of melting will
be twice i.e. 
21
2 2 0.1
mm
tt
æ ö æö
= =´
ç ÷ ç÷
è ø èø
 = 0.2 gm/sec.
10. (d)
dQ KA
d
dt
=q
l
0.01 1
30
0.05
´
=´ = 6 J/sec
Heat transferred in one day (86400 sec)
Q = 6 × 86400 = 518400 J
Now Q = mL
Þ
3
518400
334 10
Q
m
L
==
´
 = 1.552 kg = 1552 g.
11. ( b)
44
0
()
dTA
TT
dt mcJ
s
=-
[In the given problem fall in temperature of body dT =
(200 – 100) = 100 K, temp. of surrounding T
0
 = 0 K,
Initial temperature of body T = 200 K].
2
44
3
100 4
(200 0 )
4
3
r
dt
r cJ
sp
=-
pr
Þ
6
10
48
r cJ
dts
-
r
=´
s
Page 3


1. (a)
dQ KA
dt
Dq
=
l
For both rods K, A and Dq are same
Þ
1 dQ
dt
µ
l
So,
( /)
( /)
semi circular
straight
dQ dt
dQ dt
straight
semi circular
=
l
l
22 r
r
==
pp
.
2. (b) Suppose thickness of each wall is x then
combination A
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
12 1
()2 ()
2
s
KA KA
xx
q -q q -q
=
Q
224
(2 )3
s
KK
KK
KK
´´
==
+
and 
12
( ) 36 q-q =°
A B
K 2K
x x
q
1
q
2
q
Þ
1
4
36
2 ()
3
2
KA
KA
xx
´
q -q
=
Hence temperature difference across wall A is
1
() q -q = 12° C
3. (a) Heat developed by the heater
22
(200)
.
20 4.2
Vtt
H
RJ
´
==
´
Heat conducted by the glass
0.2 1 (20 )
0.002
t
H
´ ´ -q
=
Hence 
2
(200) 0.2 (20 )
20 4.2 0.002
tt ´ ´ -q
=
´
Þ q = 15.24° C
4. ( b) If suppose K
Ni
 = K Þ K
Al
 = 3K and K
Cu
 = 6K
Since all metal bars are connected in series.
So, 
Combination Cu Al Ni
Q Q QQ
t t tt
æö æ ö æö æ ö
= ==
ç÷ ç ÷ ç÷ ç ÷
èø è ø èø è ø
and 
3 1 11
eq Cu Al Ni
K K KK
= ++
1 1 19
636 K K KK
= + +=
ÞK
eq
 = 2K
Cu Ni Al
100° C 0° C
Q Q
2
q 1
q
25 cm 10 cm 15 cm
Hence, if 
Combination Cu
QQ
tt
æ ö æö
=
ç ÷ ç÷
è ø èø
Þ
1
(100 0)
(100 ) eq
Cu
Combination Cu
KA
KA
-
-q
=
ll
Þ
1
6 (100 ) 2 (100 0)
(25 10 15) 25
KA KA -q -
=
++
Þ
1
q = 83.33° C
Similar if 
Combination Al
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
2
3 ( 0) 2 (100 0)
50 15
KA KA q- -
=
Þ
2
q = 20° C
5. ( b) Let the temperature of junction be q then according to
following figure.
100°C
50°C
20°C
K
2K
3K
H
q
H
1
H
2
DPP/ P 23
67
12
H HH =+
Þ
3 (100) KA ´ ´ -q
l
2 ( 50) ( 20) KA KA q - q-
=+
ll
Þ 300 – 3 q = 3q – 120
Þ q = 70° C
6. ( d) Wein's displacement law is
m
Tb l=
Þ m
b
T
l=
6
2.88 10
1000
2880
´
== nm
Energy distribution with wavelength will be as follows
2
U
1
U
3
U
( ) nm l 1499
900
499
1500
1000
500
E
l
From the graph it is clear that
U
2 
> U
1
7. (a) According to Newton law of cooling
12 12
0
2
K
t
q - q q +q éù
= -q
êú
ëû
80° C
1
2
3
5 min
10 min
15 min
64°C
52°C
? q=
For first process :
0
(80 64) 80 64
52
K
-+éù
= -q
êú
ëû
....... (i)
For second process :
0
(80 52) 80 52
102
K
-+éù
= -q
êú
ëû
....... (ii)
For third process :
0
(80 ) 80
152
K
- q +q éù
= -q
êú
ëû
....... (iii)
On solving equation (i) and (ii) we get 
1
15
K = and 
0
q
= 24° C. Putting these values in equation (iii) we get 
q
= 42.7° C.
8. (c)
12
()
Q
t
KA
=
q -q
l
12 12
() ()
mL VL
KA KA
r
==
q -q q -q
ll
5 10
5 0.92 80
2
0.004 10 3600
A
A
+
´ ´ ´´
=
´ ´´
= 19.1 hours.
9. (c)
Q KA
t
Dq
=
l
Þ
2
() mL Kr
t
p Dq
=
l
Þ Rate of melting of ice 
2
m Kr
t
æö
µ
ç÷
èø l
Since for second rod K  becomes 
1
,
4
thr becomes
double and length becomes half, so rate of melting will
be twice i.e. 
21
2 2 0.1
mm
tt
æ ö æö
= =´
ç ÷ ç÷
è ø èø
 = 0.2 gm/sec.
10. (d)
dQ KA
d
dt
=q
l
0.01 1
30
0.05
´
=´ = 6 J/sec
Heat transferred in one day (86400 sec)
Q = 6 × 86400 = 518400 J
Now Q = mL
Þ
3
518400
334 10
Q
m
L
==
´
 = 1.552 kg = 1552 g.
11. ( b)
44
0
()
dTA
TT
dt mcJ
s
=-
[In the given problem fall in temperature of body dT =
(200 – 100) = 100 K, temp. of surrounding T
0
 = 0 K,
Initial temperature of body T = 200 K].
2
44
3
100 4
(200 0 )
4
3
r
dt
r cJ
sp
=-
pr
Þ
6
10
48
r cJ
dts
-
r
=´
s
68
DPP/ P 23
6
4.2
10
48
rc
-
r
= ´´
s
77
80 72
r c rc
µs µs
rr
=
ss
; [As J = 4.2]
12. (c) Suppose temperature difference between A and B is
100° C and 
AB
q >q
H H
B A
C
D
H/2
H/2
H/2
H/2
Heat current will flow from A to B via path ACB and
ADB. Since all the rod are identical so
() ()
AC AD
Dq = Dq
(Because heat current ; H
R
Dq
= here R = same for all)
Þ
A C AD
q -q = q -q
Þ
CD
q =q
i.e. temperature difference between C and D is zero.
T
B A
C
D
R
R
R
R
2T
13. (a) Initially at t = 0
Ra te of coolin g (R) µ Fall in temperature of body (q – q
0
)
Þ
1 1 0
22 0
R
R
q -q
=
q -q
100 40 3
80 40 2
-
==
-
14. (c)
60 50 60 50
25
102
K
-+æö
=-
ç÷
èø
........... (i)
50 50
25
102
K
-q +q æö
=-
ç÷
èø
.......... (ii)
On dividing, we get
10 60
50
=
-qq
Þ q
 = 42.85° C
15. (d)
44
0
( ).
dA
TT
dt mc
qs
=- If the liquids put in exactly similar
calorimeters and identical surrounding then we can
consider T
0
 and A constant then
44
0
() TT d
dt mc
- q
µ
If we consider that equal masses of liquid (m) are taken
at the same temperature then
1 d
dtc
q
µ
So for same rate of cooling c should be equal which is
not possible because liquids are of different nature.
Again from equation (i),
44
0
() TT d
dt mc
- q
µ
Þ
44
0
() TT d
dt Vc
- q
µ
r
Now if we consider that equal volume of liquid (V) are
taken at the same temperature then
1 d
dtc
q
µ
r
.
So for same rate of cooling multiplication of r × c for
two liquids of different nature can be possible. So,
option (d) may be correct.
16. (d) For cooking utensils, low specific heat is preferred for
it's material as it should need less heat to raise it's
temperature and it should have high conductivity,
because, it should transfer heat quickly .
17. (b)
11 12
1
() KA Q
t
q -q æö
=
ç÷
èø l
and 
22 12
2
() KA Q
t
q -q æö
=
ç÷
èø l
Given, 
12
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
1 1 22
K A KA =
18. (a) Convection may be stopped
19. (d) Heated fluid becomes less dense than the cold fluid
above it
20. (c) According to Kirchoff's law , the ratio of emissive power
to absorptive power is same for all bodies is equal to
Page 4


1. (a)
dQ KA
dt
Dq
=
l
For both rods K, A and Dq are same
Þ
1 dQ
dt
µ
l
So,
( /)
( /)
semi circular
straight
dQ dt
dQ dt
straight
semi circular
=
l
l
22 r
r
==
pp
.
2. (b) Suppose thickness of each wall is x then
combination A
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
12 1
()2 ()
2
s
KA KA
xx
q -q q -q
=
Q
224
(2 )3
s
KK
KK
KK
´´
==
+
and 
12
( ) 36 q-q =°
A B
K 2K
x x
q
1
q
2
q
Þ
1
4
36
2 ()
3
2
KA
KA
xx
´
q -q
=
Hence temperature difference across wall A is
1
() q -q = 12° C
3. (a) Heat developed by the heater
22
(200)
.
20 4.2
Vtt
H
RJ
´
==
´
Heat conducted by the glass
0.2 1 (20 )
0.002
t
H
´ ´ -q
=
Hence 
2
(200) 0.2 (20 )
20 4.2 0.002
tt ´ ´ -q
=
´
Þ q = 15.24° C
4. ( b) If suppose K
Ni
 = K Þ K
Al
 = 3K and K
Cu
 = 6K
Since all metal bars are connected in series.
So, 
Combination Cu Al Ni
Q Q QQ
t t tt
æö æ ö æö æ ö
= ==
ç÷ ç ÷ ç÷ ç ÷
èø è ø èø è ø
and 
3 1 11
eq Cu Al Ni
K K KK
= ++
1 1 19
636 K K KK
= + +=
ÞK
eq
 = 2K
Cu Ni Al
100° C 0° C
Q Q
2
q 1
q
25 cm 10 cm 15 cm
Hence, if 
Combination Cu
QQ
tt
æ ö æö
=
ç ÷ ç÷
è ø èø
Þ
1
(100 0)
(100 ) eq
Cu
Combination Cu
KA
KA
-
-q
=
ll
Þ
1
6 (100 ) 2 (100 0)
(25 10 15) 25
KA KA -q -
=
++
Þ
1
q = 83.33° C
Similar if 
Combination Al
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
2
3 ( 0) 2 (100 0)
50 15
KA KA q- -
=
Þ
2
q = 20° C
5. ( b) Let the temperature of junction be q then according to
following figure.
100°C
50°C
20°C
K
2K
3K
H
q
H
1
H
2
DPP/ P 23
67
12
H HH =+
Þ
3 (100) KA ´ ´ -q
l
2 ( 50) ( 20) KA KA q - q-
=+
ll
Þ 300 – 3 q = 3q – 120
Þ q = 70° C
6. ( d) Wein's displacement law is
m
Tb l=
Þ m
b
T
l=
6
2.88 10
1000
2880
´
== nm
Energy distribution with wavelength will be as follows
2
U
1
U
3
U
( ) nm l 1499
900
499
1500
1000
500
E
l
From the graph it is clear that
U
2 
> U
1
7. (a) According to Newton law of cooling
12 12
0
2
K
t
q - q q +q éù
= -q
êú
ëû
80° C
1
2
3
5 min
10 min
15 min
64°C
52°C
? q=
For first process :
0
(80 64) 80 64
52
K
-+éù
= -q
êú
ëû
....... (i)
For second process :
0
(80 52) 80 52
102
K
-+éù
= -q
êú
ëû
....... (ii)
For third process :
0
(80 ) 80
152
K
- q +q éù
= -q
êú
ëû
....... (iii)
On solving equation (i) and (ii) we get 
1
15
K = and 
0
q
= 24° C. Putting these values in equation (iii) we get 
q
= 42.7° C.
8. (c)
12
()
Q
t
KA
=
q -q
l
12 12
() ()
mL VL
KA KA
r
==
q -q q -q
ll
5 10
5 0.92 80
2
0.004 10 3600
A
A
+
´ ´ ´´
=
´ ´´
= 19.1 hours.
9. (c)
Q KA
t
Dq
=
l
Þ
2
() mL Kr
t
p Dq
=
l
Þ Rate of melting of ice 
2
m Kr
t
æö
µ
ç÷
èø l
Since for second rod K  becomes 
1
,
4
thr becomes
double and length becomes half, so rate of melting will
be twice i.e. 
21
2 2 0.1
mm
tt
æ ö æö
= =´
ç ÷ ç÷
è ø èø
 = 0.2 gm/sec.
10. (d)
dQ KA
d
dt
=q
l
0.01 1
30
0.05
´
=´ = 6 J/sec
Heat transferred in one day (86400 sec)
Q = 6 × 86400 = 518400 J
Now Q = mL
Þ
3
518400
334 10
Q
m
L
==
´
 = 1.552 kg = 1552 g.
11. ( b)
44
0
()
dTA
TT
dt mcJ
s
=-
[In the given problem fall in temperature of body dT =
(200 – 100) = 100 K, temp. of surrounding T
0
 = 0 K,
Initial temperature of body T = 200 K].
2
44
3
100 4
(200 0 )
4
3
r
dt
r cJ
sp
=-
pr
Þ
6
10
48
r cJ
dts
-
r
=´
s
68
DPP/ P 23
6
4.2
10
48
rc
-
r
= ´´
s
77
80 72
r c rc
µs µs
rr
=
ss
; [As J = 4.2]
12. (c) Suppose temperature difference between A and B is
100° C and 
AB
q >q
H H
B A
C
D
H/2
H/2
H/2
H/2
Heat current will flow from A to B via path ACB and
ADB. Since all the rod are identical so
() ()
AC AD
Dq = Dq
(Because heat current ; H
R
Dq
= here R = same for all)
Þ
A C AD
q -q = q -q
Þ
CD
q =q
i.e. temperature difference between C and D is zero.
T
B A
C
D
R
R
R
R
2T
13. (a) Initially at t = 0
Ra te of coolin g (R) µ Fall in temperature of body (q – q
0
)
Þ
1 1 0
22 0
R
R
q -q
=
q -q
100 40 3
80 40 2
-
==
-
14. (c)
60 50 60 50
25
102
K
-+æö
=-
ç÷
èø
........... (i)
50 50
25
102
K
-q +q æö
=-
ç÷
èø
.......... (ii)
On dividing, we get
10 60
50
=
-qq
Þ q
 = 42.85° C
15. (d)
44
0
( ).
dA
TT
dt mc
qs
=- If the liquids put in exactly similar
calorimeters and identical surrounding then we can
consider T
0
 and A constant then
44
0
() TT d
dt mc
- q
µ
If we consider that equal masses of liquid (m) are taken
at the same temperature then
1 d
dtc
q
µ
So for same rate of cooling c should be equal which is
not possible because liquids are of different nature.
Again from equation (i),
44
0
() TT d
dt mc
- q
µ
Þ
44
0
() TT d
dt Vc
- q
µ
r
Now if we consider that equal volume of liquid (V) are
taken at the same temperature then
1 d
dtc
q
µ
r
.
So for same rate of cooling multiplication of r × c for
two liquids of different nature can be possible. So,
option (d) may be correct.
16. (d) For cooking utensils, low specific heat is preferred for
it's material as it should need less heat to raise it's
temperature and it should have high conductivity,
because, it should transfer heat quickly .
17. (b)
11 12
1
() KA Q
t
q -q æö
=
ç÷
èø l
and 
22 12
2
() KA Q
t
q -q æö
=
ç÷
èø l
Given, 
12
QQ
tt
æö æö
=
ç÷ ç÷
èø èø
Þ
1 1 22
K A KA =
18. (a) Convection may be stopped
19. (d) Heated fluid becomes less dense than the cold fluid
above it
20. (c) According to Kirchoff's law , the ratio of emissive power
to absorptive power is same for all bodies is equal to
DPP/ P 23
69
the emissive power of a perfectly black body i.e.,
Black body
body
e
E
a
æö
=
ç÷
èø
 for a particular wave length
()
Black body
body
e
E
a
l
l
l
æö
=
ç÷
èø
Þ e aE
l ll
=
21. (a) As for a black body rate of absorption of heat is more.
Hence thermometer A shows faster rise in temperature
but finally both will acquire the atmospheric
temperature.
22. (b)
According to Stefan's law
44
1 11
E eAsT E e AsT = Þ= and 
4
2 22
E eAsT =
44
1 2 1 1 22
E E eT eT = \= Q
Þ ( )
1
1
4
4 4 4 1
21
2
e 1
T T 5802
e 81
æö
æö
= =´
ç÷ç÷
èø
èø
B
T 1934 K Þ=
And, from Wein's law 
A A BB
? T ? T ´ =´
A B BA AB
BA BA
? T ? ? TT
?T ? T
--
Þ = Þ=
B
B
1 5802 1934 3968
? 1.5µm
? 5802 5802
-
Þ = = Þ=
23. (a) According to Newton's law of cooling.
24. (a) In forced convection rate of loss of heat
0
Q
A(T T)
T
µ,
25(a), 26(c), 27(c)
Let c be the specific heat of turpentine
Mass of the solid, M = 100g
Mass of turpentine m = 200g
Water equivalent of calorimeter, W = 4g
Initial temperature of calorimeter, T
1
 = 15°C
Temperature of ball, T
2
 = 100°C
Final temperature of the liquid, T = 23°C
Specific heat of solid, c
2
 = 0.092 cal/g°C
Heat gained by turpentine and calorimeter is
mc (T – T
1
) + W (T – T
1
) = 200c (23 – 15) + 4 (23 – 15)
= (200c + 4) 8
Heat lost by the ball is
Mc
2
 (T
2
 – T) = 100 (0.092) (100 – 23)
                       = 708.4 cal.
According to the principle of calorimetry
Heat gained = Heat lost
\ (200c + 4) 8 = 708.4
1600c + 32 = 708.4
or  
708.4 32
0.42 cal/g°C
1600
c
-
==
28. (d) Equivalent thermal conductivity of two equally thick
plates in series combination is given by
K
1
K
2
12
2 11
K KK
=+
If 
12
KK <
 then
12
K KK <<
29. (b) Both statement-1 and statement-2 are true but
statement-2 is not correctly explaining the statement-2.
30. (d) According to Wein's displacement law the
1
m
T
lµ
Hence statement-1 is true but statement-2 is false.