10.3 Solutions
435
10.20 (a) The rate of flow of heat through the composite slab (Fig. 10.8) is
given by
−
dQ
dt
=
(T 1 − T 2 )
d
n
l
k i A i
(1)
(b) Rewriting (1)
−
dQ
dt
=
(T 1 − T 2 )
d
n
1
A i
k i A i
A i
(2)
The equivalent conductivity of the system is
k eq =
k i A i
A i
(3)
Fig. 10.8 Heat flow in a
composite slab made of n
slabs in parallel
10.21 T =
(k 1 T 1 /d 1 ) + (k 2 T 2 /d 2 )
(k 1 /d 1 ) + (k 2 /d 2 )
as d 1 = d 2
T =
k 1 T 1 + k 2 T 2
k 1 + k 2
=
92 × 100 + 16 × 0
92 + 16
= 85.18
◦ C
10.22 Heat transferred/second
−
dQ
dt
= k A
(θ 2 − θ 1 )
d
= kπr
2 (θ 2 − θ 1 )
d
= 90π(0.01)
2 (100 − 0)
0.2
= 14.137 J/s
Heat required to melt 0.05 kg of ice
= 0.05 × 8 × 10
4
= 4000 cal = 16720 J
Time required = 16720/14.137 = 1183 s.
435
10.20 (a) The rate of flow of heat through the composite slab (Fig. 10.8) is
given by
−
dQ
dt
=
(T 1 − T 2 )
d
n
l
k i A i
(1)
(b) Rewriting (1)
−
dQ
dt
=
(T 1 − T 2 )
d
n
1
A i
k i A i
A i
(2)
The equivalent conductivity of the system is
k eq =
k i A i
A i
(3)
Fig. 10.8 Heat flow in a
composite slab made of n
slabs in parallel
10.21 T =
(k 1 T 1 /d 1 ) + (k 2 T 2 /d 2 )
(k 1 /d 1 ) + (k 2 /d 2 )
as d 1 = d 2
T =
k 1 T 1 + k 2 T 2
k 1 + k 2
=
92 × 100 + 16 × 0
92 + 16
= 85.18
◦ C
10.22 Heat transferred/second
−
dQ
dt
= k A
(θ 2 − θ 1 )
d
= kπr
2 (θ 2 − θ 1 )
d
= 90π(0.01)
2 (100 − 0)
0.2
= 14.137 J/s
Heat required to melt 0.05 kg of ice
= 0.05 × 8 × 10
4
= 4000 cal = 16720 J
Time required = 16720/14.137 = 1183 s.