Question

A glass windowpane in a home is 0.62 cm thick and has dimensions of 1.5 m × 1.7 m. On a certain day, the indoor temperature is 28°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

The rate of energy transferred through glass window, dQ/dt = (KA/t) ΔT

where k = coefficient of thermal conductivity = 0.96 w/mK,

Area, A = l x b = 2.55 m²

Thickness, t = 0.62 cm = 0.62 * 10^-2 m

ΔT is temperature change =28 -0 = 28 deg

dQ/dt = (KA/t) ΔT

= ([(0.96 * 2.55) / 0.62*10^-2]) * 28

= 4249.72 W

The rate at which energy is transferred by heat through the glass is 4249.72 W

b) 1 day = 24 hours = 24 * 60 min =1440 min

= 1440 * 60 sec

= 86400 sec

The energy transferred in one day, dQ =  4249.72 W *  86400 sec

= 367175808 J

Hence, the energy is lost through the window in one day = 367175808 J