Question

Two models of a windowpane are made. In one model, two identical glass panes of thickness 3 mm are separated with an air gap of 3 mm. This composite system is fixed in the window of a room. The other model consists of a single glass pane of thickness 6 mm, the temperature difference being the same as for the first model. The ratio of the heat flow for the double pane to that for the single pane is ($K_{glass}=2.5\times {10}^{-4}$ cal / m. ${}^{\circ }C$ and $K_{air}=6.2\times {10}^{-6}$ cal/s.m. ${}^{\circ }C$)

• A. 1/20
• B. 1/70
• C. 31/1312
• D. 31/656

Answer 1

The correct option is D 31/656

Thermal Resistance$R_A=2R_g+R_{Air}$

Thermal Resistance$R_B=\frac{6mm}{K_{g}A}$

$RA=\frac{2\times 3mm}{K_{g}A}+\frac{3mm}{K_{air}A}$

The ratio of heat flow

$\frac{i_A}{i_B}=\frac{\Delta T\times R_b}{R_A\times \Delta T}$

$=\frac{6}{K_{g}A\times (\frac{6}{K_{g}A}+\frac{3}{K_{air}A})}$

$=\frac{2}{K_g(\frac{2}{K_g}+\frac{1}{K_{air}})}$

$=\frac{2\times K_{air}}{{{2K}_{air}+K}_g}$

$=\frac{2\times 6.2\times {10}^{-6}}{2\times 6.2\times {10}^{-6}+2.5\times {10}^{-4}}$

$=\frac{2\times 6.2\times {10}^{-2}}{2\times 6.2\times {10}^{-2}+2.5}$

$=\frac{0.124}{0.124+2.5}=\frac{0.124}{2.624}$

$=\frac{124}{2624}=\frac{31}{656}$