Windows are often a source of heat loss during the winter. The use of double panes, where there is a gap filled with gas between window panes, can lower heat losses. For a window that is 1 m wide and 2 m tall, calculate the heat loss rate when: a) there is only a glass pane (5 mm thick) with an inside temperature of 15°, an outside temperature of - 20°C, and a thermal conductivity of 1.4 W/m/K. b) there is an air gap (10 mm thick) with an inside edge temperature of 10°C, an outside edge temperature of - 15°C, and a thermal conductivity of 0.024 W/m/K. Be sure to use each step of the problem solving process.

Question

Chemistry

Emilie Booth

20 May, 06:26

Windows are often a source of heat loss during the winter. The use of double panes, where there is a gap filled with gas between window panes, can lower heat losses. For a window that is 1 m wide and 2 m tall, calculate the heat loss rate when: a) there is only a glass pane (5 mm thick) with an inside temperature of 15°, an outside temperature of - 20°C, and a thermal conductivity of 1.4 W/m/K. b) there is an air gap (10 mm thick) with an inside edge temperature of 10°C, an outside edge temperature of - 15°C, and a thermal conductivity of 0.024 W/m/K. Be sure to use each step of the problem solving process.

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Answers (1)

Know the Answer?

Jake Hull · Answer 1

a) Q=19600W b) Q=60W

Explanation:

q=k. (T₀-T₁) / x

a)

T₀=inside temperature (K) = 273+15=288K

T₁=outside temperature (K) = 273-20=253K

x=5x10⁻³m

k=1.4W/m. K

∴q=9800W/m² ⇒ Q=q. Surface

Q=9800W/m². 1m. 2m = 19600W.

b)

T₀=inside temperature (K) = 273+10=273K

T₁=outside temperature (K) = 273-15=258K

x=10x10⁻³m

k=0.024W/m. K

∴q=60W/m² ⇒ Q=q. Surface

Q=60W/m². 1m. 2m = 120W.