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 = [tex]3\times 10^{- 4}\ m[/tex]
Radius, R = 25 cm = 0.25 m
Thermal Conductivity of glass, K = 0.63 W/ mK
Inside Temperature, [tex]T_{in} = 20^{\circ}[/tex]
Outside Temperature, [tex]T_{o} = - 12^{\circ}[/tex]
Now,
Power of the heater = Heat Loss, Q
Thus
Q = 1500 W
We know that:
[tex]Q = KA\frac{dT}{dt}[/tex]
[tex]Q = K\times \pi R^{2}\frac{(T_{o} - T_{i})}{t' - t}[/tex]
[tex]1500 = 0.63\times \pi \times (0.25)^{2}\frac{(20 - (- 12))}{t' - (3\times 10^{- 4})}[/tex]
[tex]t' - (3\times 10^{- 4}) = 0.63\times \pi \times (0.25)^{2}\frac{32}{1500}[/tex]
t' = 2.94 mm
