A window has a glass surface of 2686 cm2 and a thickness of 7.4 mm.
Find the rate of energy transfer by conduc- tion through this pane when the temperature of the inside surface of the glass is 66◦F and the outside temperature is 77 ◦F. Assume the thermal conductivity of window glass is 0.8 J/s · m ·◦ C .
Answer in units of kW
To find the rate of energy transfer by conduction through the window pane, we can use the formula:
Q = (k * A * ΔT) / d
Where:
Q is the rate of energy transfer (in watts),
k is the thermal conductivity of window glass (in J/s · m ·◦ C),
A is the surface area of the window (in square meters),
ΔT is the temperature difference between the inside and outside surfaces of the glass (in ◦C), and
d is the thickness of the window (in meters).
First, let's convert the surface area from square centimeters to square meters:
A = 2686 cm^2 = 2686 / 10000 m^2 = 0.2686 m^2
Next, let's convert the thickness from millimeters to meters:
d = 7.4 mm = 7.4 / 1000 m = 0.0074 m
Now, we can calculate the temperature difference:
ΔT = 77 ◦F - 66 ◦F = 11 ◦F = 11/1.8 ◦C = 6.11 ◦C
Finally, we can substitute the values into the formula to find the rate of energy transfer:
Q = (0.8 J/s · m ·◦ C * 0.2686 m^2 * 6.11 ◦C) / 0.0074 m
Simplifying and converting the units:
Q = (1.07288 J/s) / 0.0074 m = 144.969 J/s
To convert from watts to kilowatts, we divide by 1000:
Q = 144.969 J/s / 1000 = 0.145 kW
Therefore, the rate of energy transfer by conduction through this window pane is 0.145 kW.
To find the rate of energy transfer by conduction through this pane, we can use the following formula:
Q = k * A * (T₂ - T₁) / d
Where:
Q is the rate of energy transfer by conduction (in Watts)
k is the thermal conductivity of window glass (0.8 J/s · m ·◦C)
A is the area of the glass surface (2686 cm²)
T₂ is the outside temperature (77 ◦F)
T₁ is the inside temperature (66 ◦F)
d is the thickness of the glass pane (7.4 mm)
First, we need to convert the units of area and thickness to the appropriate units:
1 cm² = 0.0001 m²
1 mm = 0.001 m
Converting the area:
2686 cm² * 0.0001 m²/cm² = 0.2686 m²
Converting the thickness:
7.4 mm * 0.001 m/mm = 0.0074 m
Now, we can substitute the values into the formula:
Q = 0.8 J/s · m ·◦C * 0.2686 m² * (77 ◦F - 66 ◦F) / 0.0074 m
Calculating the temperature difference:
77 ◦F - 66 ◦F = 11 ◦F
Converting the temperature difference to Celsius:
11 ◦F * (5/9) = 6.11 ◦C
Substituting the values and simplifying:
Q = 0.8 J/s · m ·◦C * 0.2686 m² * 6.11 ◦C / 0.0074 m
Q = 1.892 W
Finally, converting the rate of energy transfer to kilowatts:
1 kW = 1000 W
1.892 W * (1 kW / 1000 W) = 0.001892 kW
Therefore, the rate of energy transfer by conduction through this pane is approximately 0.001892 kW.