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Conduction Law

A double-pane...

Question

A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m $^{2}$ and thickness $0.01$ m separated by $0.05$ m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constant temperatures of 27 $^{o}$ C and 0 $^{o}$ C respectively.The thermal conductivity of glass is $0.8$ and of air 0.08Wm $^{- 1}$ K $^{- 1}$
The temperature of the outer glass-air interface is

{26.5}^{o} C

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{25.5}^{o} C

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{24.5}^{o} C

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{23.5}^{o} C

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Solution

The correct option is A ${26.5}^{o} C$
$T_{2}$ = temperature of glass-1 and air interface
$T_{3}$ = temperature of air and glass-2 interface
The equation of heat flow are:
$\frac{Q_{1}}{t} = \frac{K_{1} A (T_{1} - T_{2})}{d_{1}} = \frac{0.8 \times 1 (27 - T_{2})}{0.01}$
$\frac{Q_{2}}{t} = \frac{K_{2} A (T_{2} - T_{3})}{d_{2}} = \frac{0.08 \times 1 (T_{2} - T_{3})}{0.05}$

$\frac{Q_{3}}{t} = \frac{K_{3} A (T_{3} - T_{4})}{d_{3}} = \frac{0.8 \times 1 (T_{3})}{0.01}$

In steady state: $\frac{q_{1}}{t} = \frac{q_{2}}{t} = \frac{q_{3}}{t}$
On equating $(1) (2)$ and $13$ ,we find the following:
$1350 = 51 T_{2} - T_{3}$ ----------(a)
$27 - T_{2} = T_{3}$ ----------(b)
The temperature of outer glass interface $T_{2} = {26.48}^{0} C$

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