A double pan window used for insulating a room thermally from outside consists of two glass sheets each of area $1 m^{2}$ and thickness 0.01m 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 $27 ° C$ and $0 ° C$ respectively. The thermal conductivity of glass is $0.8 W m^{- 1} K^{- 1}$ and of air $0.08 W m^{- 1} K^{- 1} .$ Calculate the temperature of the inner glass-air interface.

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answer is 26.5.

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Detailed Solution

Let $T_{2} a n d T_{3}$ be the temperatures of the glass 1-air interface and air-glass 2 interface respectively The equations of heat flow are $\frac{d Q_{1}}{d t} = \frac{d Q_{2}}{d t} = \frac{d Q_{3}}{d t} \Rightarrow \frac{d Q}{d t} = \frac{K_{g} A (T_{1} - T_{2})}{d_{g}} = \frac{K_{a} A (T_{2} - T_{3})}{d_{a}} = \frac{K_{g} A (T_{3} - T_{4})}{d_{g}} ... (A)$

$\Rightarrow \frac{d Q}{d t} = \frac{K_{g} A (300 - T_{2})}{d_{g}} = \frac{K_{a} A (T_{2} - T_{3})}{d_{a}} = \frac{K_{g} A (T_{3} - 273)}{d_{g}} \Rightarrow 50 (300 - T_{2}) = T_{2} - T_{3} ......... (i)$