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to the rate across layer 2 (
P
2
).
Using Eq. 1837 and canceling out the common factor of
area
A
, we obtain
T
H

T
c
(
L
1
/
k
1
+
L
2
/
k
2
+
L
3
/
k
3
)
=
Δ
T
2
(
L
2
/
k
2
)
¡
45 C°
(1 + 7/9 + 35/80)
=
Δ
T
2
(7/9)
which leads to
Δ
T
2
= 15.8 °C.
(b) We expect (and this is supported by the result in the next part) that greater
conductivity should mean a larger rate of conductive heat transfer.
(c) Repeating the calculation above with the new value for
k
2
, we have
45 C°
(1 + 7/11 + 35/80)
=
Δ
T
2
(7/11)
which leads to
Δ
T
2
= 13.8 °C.
This is less than our part (a) result which implies that the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Heat

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