A window pane has a thickness 0.5 cm and area 0.8 m^2. The temperature difference between outside and inside surface is 20°C. The rate of heat flow through the pane is ( Thermal conductivity of glass being 0.8J/s m k)
a)2.56 KJ/s
b)3.80KJ/s
c)4.50 KJ/s
d)521J/s
Answers
Given A window pane has a thickness 0.5 cm and area 0.8 m^2. The temperature difference between outside and inside surface is 20°C. The rate of heat flow through the pane is ( Thermal conductivity of glass being 0.8J/s m k)
We have, the heat transfer is given by the equation
The rate of heat transfer is q/t , k is thermal conductivity of glass, surface area of object is A, difference in temperature is (T2 - T1), d is the thickness.
q / t = k A(T2 - T1) / d
= 0.8 x 0.8 x 20 / 0.5 x 10^-2
= 2,560 J /s
= 2.560 K J/s
The rate of flow is 2.56 K J/s
Given conditions ⇒
Thickness of the window pane(l) = 0.5 cm. = 0.005 m.
Area = 0.8 m²
Temperature difference(ΔT) = 20 °C or 20 K.
Thermal Conductivity (K) = 0.8 W/mk
Using the formula,
Power = KAΔT/l
∴ Power = 0.8 × 0.8 × 20/0.005
∴ Power = 2560 W.
∴ Power = 2.560 kW.
Hence, Option (a). is correct.
Hope it helps.