Compare the R-values of:
(a) Two pieces of 3-mm thick glass in thermal contact.
(b) Two pieces of 3-mm thick glass with a 1-cm air space between them.

Question

Quizplus · Accepted Answer

The Answer is

ext { The thermal conductivity of glass is given as } 59 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } } ext { and for air it is }

2.3 \frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } }

. Since

R

is defined as

R = l / k

where

l

is in

\mathrm { cm }

and

k

is in

\frac { \mathrm { W } \cdot \mathrm { cm } } { \mathrm { m } ^ { 2 } \cdot { } ^ { \circ } \mathrm { C } }

, then a

3 \mathrm {~mm} ( 0.3 \mathrm {~cm} )

piece of glass will have an

R

-value of

R = \frac { 0.3 } { 59 } = 0.005

and a

1 - \mathrm { cm }

layer of air will have an

R

-value of

R = \frac { 1.0 } { 2.3 } = 0.435

. Since

R _ { ext {total } } = \sum R

then
for (a)

R _ { ext {total } } = 0.005 + 0.005 = 0.01

and
for (b)

R _ { ext {tota } } = 0.005 + 0.435 + 0.005 = 0.445

.
This emphasizes the importance of the air space in insulating properties.

Compare the R-Values Of