Question

A room has a window fitted with a single 1.0 m × 2.0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and the outside is 40°C. (b) The glass is now replaced by two glasspanes, each having a thickness of 1 mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = 1.0 J s⁻¹ m⁻¹°C⁻¹ and that of air = 0.025 J s⁻¹ m⁻¹°C⁻¹.

Answer 1

(a)

Length, l = 2 mm = 0.0002 m

$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{heat}=\frac{∆T}{l/KA} \\ =\frac{KA.∆T}{l} \\ =\frac{1\times 2\times \left(40-32\right)}{2\times {10}^{-3}} \\ =8000\mathrm{J}/\sec \end{gathered}$$

(b)

Resistance of glass, $R_{\mathrm{g}}=\frac{l}{K_{\mathrm{g}}.A}$
Resistance of air, $R_{\mathrm{A}}=\frac{l}{K_{\mathrm{A}}.A}$

From the circuit diagram, we can find that all the resistors are connected in series.

$$\begin{gathered} R_s=R_g+R_A+R_g \\ =\frac{{10}^{-3}}{2}\left(\frac{2}{K_{\mathrm{g}}}+\frac{1}{K_{\mathrm{A}}}\right) \\ =\frac{{10}^{-3}}{2}\left(\frac{2}{1}+\frac{1}{0.025}\right) \\ =\frac{{10}^{-3}}{2}\times \frac{\left(2\times 0.025+1\right)}{0.025} \end{gathered}$$
$$\begin{gathered} \mathrm{Rate}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{heat},q=\frac{∆T}{R_s} \\ =\frac{T_1-T_2}{R_s} \\ =\frac{\left(40-32\right)\times 2\times 0.025}{{40}^{-3}\times \left(2\times 0.025+1\right)} \\ =381W \end{gathered}$$