Question

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.17 m2 and whose thickness is 2.3 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 15.00 m2 and 0.080 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window? (The thermal conductivity for glass window is kG = 0.80 J/(s·m·C°), and the thermal conductivity for Styrofoam is kS = 0.010 J/(s·m·C°).)

Answer 1

Answer: 96.93%

Explanation: The basic equation for the rate of steady conduction heat transfer is called Fourier's Law of Conduction.

Mathematically:

$q=-kA\frac{dT}{dx}$

where:

• q= rate of  heat transfer

• k= thermal conductivity of the material (negative sign just denotes that the heat transfer occurs in a direction of decreasing temperature)

• dT= temperature difference across the surface (which drives the heat transfer)

• A= surface area through which the heat transfer occurs

• dx= thickness of the material through which heat gets transferred

In this question we have two materials through which heat transfer occurs through a wall, one is the Styrofoam of which the wall is made and the other is the glass pane of window.

According to the language of question we consider that we are given the area of Styrofoam after removing the area of glass window pane on it, otherwise we would subtract the area of glass pane from the area of wall.

Given data:

• area of glass, $A_{g}=0.17 m^{2}$

• thickness of glass, $x_{g}=2.3\times10^{-3} m$

• thermal conductivity of glass, $k_{g}=0.8Wm^{-1} \degree C^{-1}$

• area of Styrofoam, $A_{s}=15 m^{2}$

• thickness of Styrofoam, $x_{s}=0.08 m$

• thermal conductivity of Styrofoam, $k_{s}=0.01Wm^{-1} \degree C^{-1}$

Heat transfer through the glass pane of window:

Putting the given values in Fourier's eqn.

$q_{g}= 0.8 \times 0.17 \times \frac{dT}{2.3\times10^{-3}} \\\\q_{g}= 59.13 (dT) W$

Heat transfer through the Styrofoam wall:

$q_{s}= 0.01 \times 15 \times \frac{dT}{0.08} \\\\q_{s}= 1.875 (dT) W$

Total heat transfer 'q':

$q=q_{s}+q_{g}$

$q= 1.875(dT)+59.13(dT)$

q= 61.005(dT) W

Now,

% of Heat lost by window= $\frac{q_{g} }{q} \times 100$

$=\frac{59.13 (dT)}{61.005(dT)}$

=96.93%