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where the sum in the denominator is over the layers. If
L
g
is the thickness of a glass layer,
L
a
is the thickness of the air layer,
k
g
is the thermal conductivity of glass, and
k
a
is the
thermal conductivity of air, then the denominator is
22
.
g
ga
ag
a
ga
a
g
LL
k
L
k
L
L
kk
k
k
k
+
=+
=
¦
Therefore, the heat conducted per unit area occurs at the following rate:
()
()
()
()
()
()
(
)
()
cond
3
2
51.1 C 0.026W m K 1.0W m K
2
2 3.0 10 m 0.026W m K
0.075m 1.0W m K
18W m .
HC
a
g
ga
ag
TT
k
k
P
AL
k
L
k
−
−
°⋅
⋅
==
+
×⋅
+⋅
=
59. (a) We use
cond
H
C
TT
Pk
A
L
−
=
with the conductivity of glass given in Table 186 as 1.0 W/m·K. We choose to use the
Celsius scale for the temperature: a temperature difference of
()
72 F
20 F
92 F
HC
TT
−=°
−
−
°
=
°
is equivalent to
5
9
(92)
51.1C
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 Spring '08
 Any
 Physics, Conductivity

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