Question

A 1500 W portable heater is needed to replace the heat lost through a circular window (radius 25 cm) in the wall of a home that is made of a single pane of glass 0.30 cm thick. Thermal conductivity for the glass is 0.63 W/mK. The temperature inside the house is 20°C and outside is -12°C. What is the thickness of the window?

Answer 1

Answer:

The thickness of the window pane is 2.94 mm

Solution:

As per the question:

Power of the heater, P = 1500 W

Thickness of the glass pane, t = 0.30 cm = $3\times 10^{- 4}\ m$

Radius, R = 25 cm = 0.25 m

Thermal Conductivity of glass, K = 0.63 W/ mK

Inside Temperature, $T_{in} = 20^{\circ}$

Outside Temperature, $T_{o} = - 12^{\circ}$

Now,

Power of the heater = Heat Loss, Q

Thus

Q = 1500 W

We know that:

$Q = KA\frac{dT}{dt}$

$Q = K\times \pi R^{2}\frac{(T_{o} - T_{i})}{t' - t}$

$1500 = 0.63\times \pi \times (0.25)^{2}\frac{(20 - (- 12))}{t' - (3\times 10^{- 4})}$

$t' - (3\times 10^{- 4}) = 0.63\times \pi \times (0.25)^{2}\frac{32}{1500}$

t' = 2.94 mm