# Question 5116e

Jul 25, 2016

c) 35:1

#### Explanation:

The thermal conductivity of the glass window is taken to be k = 0.8 Wm^-1"^@C^-1. Let $\Delta T$ be the temperature difference between the inner and outer surfaces
Heat conducted through the window Q_g=kA(DeltaT" ")/"Thickness of glass"
$\implies {Q}_{g} = \frac{0.8 \times 2}{0.75 \times {10}^{-} 2} \Delta T$

The thermal conductivity of the walls is given to be twice of glass wool k_w = 2xx0.04 Wm^-1"^@C^-1.
Therefore, heat conducted through the walls Q_w=k_wA_w(DeltaT" ")/"Thickness of wall"#
$\implies {Q}_{w} = \frac{0.08 \times 10}{13 \times {10}^{-} 2} \Delta T$
Window to Wall Ratio$= {Q}_{g} / {Q}_{w} = \frac{\frac{0.8 \times 2}{0.75 \times {10}^{-} 2} \Delta T}{\frac{0.08 \times 10}{13 \times {10}^{-} 2} \Delta T}$

Given is the same temperature difference between them. Above expression reduces to
${Q}_{g} / {Q}_{w} = \frac{\frac{0.8 \times 2}{0.75}}{\frac{0.08 \times 10}{13}}$
$\implies {Q}_{g} / {Q}_{w} = \left(\frac{0.8 \times 2}{0.75}\right) \times \left(\frac{13}{0.08 \times 10}\right)$
$\implies {Q}_{g} / {Q}_{w} = 34. \dot{6} \approx 35$