Question

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.

Answer 1

The rate of heat loss through a single glass pane window is 19600 W, while the rate of heat loss through a double-paned window with an air gap is significantly lower at 120 W.

Step-by-step explanation:

To calculate the rate of heat loss through a window, we apply the formula for heat conduction: Q/t = kA(Tinside - Toutside) / d, where Q/t is the heat transfer rate in watts (W), k is the thermal conductivity in watts per meter-kelvin (W/m·K), A is the area in square meters (m²), Tinside and Toutside are the inside and outside temperatures in degrees Celsius (°C), and d is the thickness of the material in meters (m).

Part (a): Single Glass Pane

Given:
k = 1.4 W/m·K
A = 1 m x 2 m = 2 m²
Tinside = 15°C
Toutside = -20°C
d = 5 mm = 0.005 m
Q/t = (1.4)(2)((15) - (-20)) / 0.005 = 19600 W

Part (b): Glass with Air Gap

Given:
k = 0.024 W/m·K for air gap
A = 1 m x 2 m = 2 m²
Tinside = 10°C
Toutside = -15°C
d = 10 mm = 0.01 m
Q/t = (0.024)(2)((10) - (-15)) / 0.01 = 120 W

The use of an air gap significantly lowers the heat loss rate, showing the value of double-paned windows for insulation.

Answer 2

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

Step-by-step 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.