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An engineer is investigating energy loss through windows. The windowpane of interest is 0.650 cm thick, has dimensions of 1.19 m × 2.25 m, and has a thermal conductivity of 0.8 w/(m·°C). On a given cold day, the outside temperature is 0°C and the temperature of the interior surface of the glass is 24.00. (a) Determine the rate (in W) at which heat energy is transferred through the window (b) Determine the amount of energy (in J) transferred through the window in one day, assuming the temperature on the surfaces remains constant.

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Answer:

7908.92307 W

683330953.248 J

Explanation:

k = Heat conduction coefficient = 0.8 W/(m·°C)

A = Area = [tex]1.19\times 2.25\ m^2[/tex]

l = Thickness = 0.65 cm

[tex]T_2[/tex] = 24°C

[tex]T_1[/tex] = 0°C

Rate of heat transfer is given by

[tex]Q=\frac{kA(T_2-T_1)}{l}\\\Rightarrow Q=\frac{0.8\times 1.19\times 2.25(24-0)}{0.65\times 10^{-2}}\\\Rightarrow Q=7908.92307\ W[/tex]

The rate of heat transfer is 7908.92307 W

Amount of energy is given by

[tex]E=Qt\\\Rightarrow E=7908.92307\times 24\times 3600\\\Rightarrow E=683330953.248\ J[/tex]

The energy transferred through the window in one day is 683330953.248 J