Question

A glass windowpane in a home is 0.62 cm thick and has dimensions of 1.3 m × 2.2 m. On a certain day, the indoor temperature is 30°C and the outdoor temperature is 0°C.

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

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

Answer 1

Part A.

Rate of heat transfer through conduction is given by:

Q/t = P = k*A*dT/dx

k = Thermal conductivity of glass = 0.8

A = Area of window = 1.3 m*2.2 m = 1.3*2.2 m^2

dT = change in temperature = T_in - T_out = 30 - 0 = 30 C

dx = thickness of glass windowpane = 0.62 cm = 0.62*10^-2 m

So,

P = 0.80*1.3*2.2*30/(0.62*10^-2) = 11070.97 W

P = 1.11*10^4 W

Part B.

Energy lost through window in one day will be:

E = Power*time

time = 1 day = 86400 sec

E = 11070.97*86400 = 956531808 J

E = energy lost = 9.57*10^8 J

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