85.

(a) Recalling that a Watt is a Joule-per-second, and that a change in Celsius temperature is equivalent

(numerically) to a change in Kelvin temperature, we convert the value of k to SI units, using
Eq. 19-12.

2.9

× 10

−3

cal

cm

·C

◦

·s



4.186 J

1 cal



100 cm

1 m



= 1.2

W

m

·K

.

(b) Now, a change in Celsius is equivalent to five-ninths of a Fahrenheit change, so

2.9

× 10

−3

cal

cm

·C

◦

·s



0.003969 Btu

1 cal



5 C

◦

9 F

◦



3600 s

1 h



30.48 cm

1 ft



= 0.70

Btu

ft

·F

◦

·h

.

(c) Using Eq. 19-33, we obtain

R =

L

k

=

0.0064 m

1.2 W/m

·K

= 0.0053 m

2

·K/W .