# A double-pane window used for insulating a room thermally: Heat Conduction

Problem: A double-pane window used for insulating a room thermally from outside 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. Calculate the rate of heat flow through the window pane. Also find the temperatures of the other interfaces. Given thermal conductivities of glass and air as 0.8 and 0.08 M/(m-K), respectively.  (IIT JEE 1997)

Answer: The rate of heat flow through the window pane is 41.5 W. The temperatures of the interfaces are 26.48°C and 0.52°C.

Solution: Given, room temperature Tr = 27°C, atmosphere temperature Ta = 0°C, glass thickness xg = m, air-space thickness xa = 0.05 m, area A = 1 m2, thermal conductivity of the glass Kg = 0.8 W/(m-K), and thermal conductivity of the air Ka = 0.08 W/(m-K). Let the room-side glass-air interface temperature be T1 and the out-side glass-air interface temperature be T2.

The rate of heat flow through the room-side glass is \begin{align} \label{gsa:eqn:1} \frac{\mathrm{d}Q_\text{g1}}{\mathrm{d}t}=\frac{K_g A (T_r-T_1)}{x_g}, \end{align} and through the air-space is \begin{align} \label{gsa:eqn:2} \frac{\mathrm{d}Q_\text{a}}{\mathrm{d}t}=\frac{K_a A (T_1-T_2)}{x_a}, \end{align} and through the out-side glass is \begin{align} \label{gsa:eqn:3} \frac{\mathrm{d}Q_\text{g2}}{\mathrm{d}t}=\frac{K_g A (T_2-T_a)}{x_g}. \end{align} In the steady state, ${\mathrm{d}Q_\text{g1}}/{\mathrm{d}t}={\mathrm{d}Q_\text{a}}/{\mathrm{d}t}$. Use first and second equations to get \begin{align} \label{gsa:eqn:4} K_g x_a (T_r-T_1)=K_a x_g(T_1-T_2). \end{align} Also, ${\mathrm{d}Q_\text{g2}}/{\mathrm{d}t}={\mathrm{d}Q_\text{a}}/{\mathrm{d}t}$ gives \begin{align} \label{gsa:eqn:5} T_r-T_1=T_2-T_a. \end{align} Solve last two equations to get \begin{align} T_1&=\frac{(K_g x_a +K_a x_g)T_r+K_a x_g T_a}{2K_a x_g+K_g x_a}\\ &=26.48^\circ\,\mathrm{C}\nonumber\\ T_2&=0.52^\circ\,\mathrm{C}.\nonumber \end{align} Substitute these values in second equation to get the rate of heat flow \begin{align} \frac{\mathrm{d}Q_\text{a}}{\mathrm{d}t}&=\frac{K_a A (T_1-T_2)}{x_a} \nonumber\\ &=\frac{0.08(1) (26.48-0.52)}{0.05}\\ &=41.5\,\mathrm{W}.\nonumber \end{align}