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

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* 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(dT)/(dt)`

`Q = K* \pi R^(2)((T_(o) - T_(i)))/(t' - t)`

`1500 = 0.63* \pi * (0.25)^(2)((20 - (- 12)))/(t' - (3* 10^(- 4)))`

`t' - (3* 10^(- 4)) = 0.63* \pi * (0.25)^(2)(32)/(1500)`

t' = 2.94 mm