A double-plane window consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass interface and the glass outdoor interface are at constant temperature of 27°C and 0°C respectively. The rate of heat flow through the window pane is (Given, Kglass = 0.8 W m-1 K-1, Kair= 0.08 W m-1 K-1)

Question

A double-plane window consists of two glass sheets each of area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state, the room glass interface and the glass outdoor interface are at constant temperature of 27°C and 0°C respectively. The rate of heat flow through the window pane is (Given, Kglass = 0.8 W m-1 K-1, Kair= 0.08 W m-1 K-1) (A) 41.5 W (B) 31.5 W (C) 21.5 W (D) 11.5 W

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/0b9ff62c3d1a58dfc2808aba7f159a98-.jpeg" /> Total thermal resistance is R=(t1/K1A1)+(t2/K2A2)+(.t1/K1A1) R=2 × (0.01/0.8 ×1)+(0.05/0.08×1) =0.65 W-1 K ∴ Heat current, H=(Δ T/R)=(27-0/0.65) =41.5 W

$41.5 W$

$31.5 W$

$21.5 W$

$11.5 W$

Total thermal resistance is
$R = \frac{t _{1}}{K _{1} A _{1}} + \frac{t _{2}}{K _{2} A _{2}} + \frac{. t _{1}}{K _{1} A _{1}}$
$R = 2 \times \frac{0.01}{0.8 \times 1} + \frac{0.05}{0.08 \times 1}$
$= 0.65 W^{- 1} K$
$∴$ Heat current, $H = \frac{Δ T}{R} = \frac{27 - 0}{0.65}$
$= 41.5 W$

41.5W

31.5W

21.5W

11.5W

Total thermal resistance is R=K1​A1​t1​​+K2​A2​t2​​+K1​A1​.t1​​ R=2×0.8×10.01​+0.08×10.05​ =0.65W−1K ∴ Heat current, H=RΔT​=0.6527−0​ =41.5W

$41.5 W$

$31.5 W$

$21.5 W$

$11.5 W$

Total thermal resistance is
$R = \frac{t _{1}}{K _{1} A _{1}} + \frac{t _{2}}{K _{2} A _{2}} + \frac{. t _{1}}{K _{1} A _{1}}$
$R = 2 \times \frac{0.01}{0.8 \times 1} + \frac{0.05}{0.08 \times 1}$
$= 0.65 W^{- 1} K$
$∴$ Heat current, $H = \frac{Δ T}{R} = \frac{27 - 0}{0.65}$
$= 41.5 W$