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 Wm^(-1)K^(-1)` and of air `0.08 Wm^(-1)K^(-1)`. Answer the following questions.
(a) Calculate the temperature of the inner glass-air interface.
(b) Calculate the temperature of the outer glass-air interface.
(c) Calculate the rate of flow of heat through the window pane.

Options

A. `41.5 W`
B. `31.5 W`
C. `21.5 W`
D. `11.5 W`

Explanation

To solve the given problem step by step, we will calculate the temperature of the inner glass-air interface, the outer glass-air interface, and the rate of heat flow through the window pane.

### Given Data:
- Area of each glass sheet, \( A = 1 \, m^2 \)
- Thickness of each glass sheet, \( L_g = 0.01 \, m \)
- Thickness of stagnant air space, \( L_a = 0.05 \, m \)
- Temperature at room-glass interface, \( T_1 = 27^\circ C \)
- Temperature at glass-outdoor interface, \( T_4 = 0^\circ C \)
- Thermal conductivity of glass, \( K_g = 0.8 \, Wm^{-1}K^{-1} \)
- Thermal conductivity of air, \( K_a = 0.08 \, Wm^{-1}K^{-1} \)

### (a) Calculate the temperature of the inner glass-air interface \( T_2 \)

1. **Calculate the thermal resistances:**
- For the first glass sheet:
\[ R_1 = \frac{L_g}{K_g \cdot A} = \frac{0.01}{0.8 \cdot 1} = 0.0125 \, K/W \]
- For the air layer:
\[ R_2 = \frac{L_a}{K_a \cdot A} = \frac{0.05}{0.08 \cdot 1} = 0.625 \, K/W \]
- For the second glass sheet:
\[ R_3 = \frac{L_g}{K_g \cdot A} = \frac{0.01}{0.8 \cdot 1} = 0.0125 \, K/W \]

2. **Total thermal resistance \( R_{total} \):**
\[ R_{total} = R_1 + R_2 + R_3 = 0.0125 + 0.625 + 0.0125 = 0.650 \, K/W \]

3. **Calculate the temperature difference \( \Delta T \):**
\[ \Delta T = T_1 - T_4 = 27 - 0 = 27 \, K \]

4. **Calculate the rate of heat flow \( H \):**
\[ H = \frac{\Delta T}{R_{total}} = \frac{27}{0.650} \approx 41.54 \, W \]

5. **Calculate the temperature of the inner glass-air interface \( T_2 \):**
- Heat flow through the first glass:
\[ \Delta T_1 = H \cdot R_1 = 41.54 \cdot 0.0125 \approx 0.51925 \, K \]
- Therefore, \( T_2 = T_1 - \Delta T_1 \):
\[ T_2 = 27 - 0.51925 \approx 26.48 \, ^\circ C \]

### (b) Calculate the temperature of the outer glass-air interface \( T_3 \)

1. **Calculate the temperature drop across the air layer:**
\[ \Delta T_2 = H \cdot R_2 = 41.54 \cdot 0.625 \approx 25.96 \, K \]
- Therefore, \( T_3 = T_2 - \Delta T_2 \):
\[ T_3 = 26.48 - 25.96 \approx 0.52 \, ^\circ C \]

### (c) Calculate the rate of flow of heat through the window pane

From our earlier calculations, we already found the rate of heat flow \( H \):
\[ H \approx 41.54 \, W \]

### Final Answers:
(a) The temperature of the inner glass-air interface \( T_2 \approx 26.48 \, ^\circ C \)
(b) The temperature of the outer glass-air interface \( T_3 \approx 0.52 \, ^\circ C \)
(c) The rate of flow of heat through the window pane \( H \approx 41.54 \, W \)