Question

A glass windowpane in a home is 0.62 cm thick and has dimensions of 1.4 m × 1.6 m. On a certain day, the indoor temperature is 24°C and the outdoor temperature is 0°C.

(a) What is the rate at which energy is transferred by heat through the glass? ________W

(b) How much energy is lost through the window in one day, assuming the temperatures inside and outside remain constant? _________ J

Answer 1

(a) The expression for the thermal conductivity is given as –

P = K x ∆T x A / d --------------------------(i)

Where –

P is power in watts transferred through the material
K is thermal conductivity of the material in W/mK
∆T is change in temperature across the material
A is cross-sectional area in m²
d is thickness in m

Given that –

A = 1.4 m x 1.6 m = 2.24 m^2

d = 0.62 cm = 0.0062 m

For ordinary glass, K = 0.8 W/m K

∆T = 24 deg K

Put these values in expression (i) –

P = (0.8 x 24 x 2.24) / 0.0062 = 6937 Watts or 6937 J/s

(b) Now, no. of seconds in a day = 24 x 60 x 60 = 86400 s

Therefore –

Energy lost through window in one day = 6937 J/s x 86400/day = 5.99 x 10^8 J