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Question

A double-pane window used for insulating a room thermally from outside consists of two glass sheets each of area 1m2 and thickness 0.01m separated by 0.05m thick stagnant air space.In the steady state, the room-glass interface and the glass-outdoor interface are at constanttemperatures of 27oC and 0oC respectively.The thermal conductivity of glass is 0.8 and of air 0.08Wm1K1

The temperatures of the inner glass-air interface is

A
2.5oC
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B
2.0oC
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C
1.5oC
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D
0.5oC
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Solution

The correct option is D 0.5oC
T2 = temperature of glass-1 and air interface
T3 = temperature of air and glass-2 interface
The equation of heat flow are:
Q1t=K1A(T1T2)d1=0.8×1(27T2)0.01
Q2t=K2A(T2T3)d2=0.08×1(T2T3)0.05
Q3t=K3A(T3T4)d3=0.8×1(T3)0.01

In steady state: q1t=q2t=q3t

On equating (1)(2) and 13, we find the following :
1350=51T2T3----------(a)
27T2=T3----------(b)

The temperature of outer glass interface, T2=26.480C
So, from equation (b), we find temperature of inner glass interface T3=27T2=2726.48=0.520C

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