A glass windowpane in a home is 0.62 cm thick and has dimensions 1.0 m x 2.0 m. On a certain day, the indoor temperature is 25 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?
Heatrate= Coefficinet of transmission*Area/length
look up the coefficient.
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To calculate the rate of energy transfer through the glass, we can use the formula for heat transfer:
Q = k * A * ΔT / d
Where:
Q is the rate of energy transfer
k is the thermal conductivity of the material (glass)
A is the area of the glass windowpane
ΔT is the temperature difference between the indoor and outdoor temperatures
d is the thickness of the glass windowpane
a) To find the rate of energy transfer, we need to find the temperature difference (ΔT) first:
ΔT = T_in - T_out = 25°C - 0°C = 25°C
Next, we need to find the thermal conductivity of glass. The thermal conductivity of glass can vary, but for the purposes of this example, we will use a value of 0.84 W/(m·K).
Now we can calculate the rate of energy transfer:
Q = k * A * ΔT / d
Q = 0.84 W/(m·K) * (1.0 m * 2.0 m) * 25°C / 0.0062 m
Q ≈ 8,548.39 W
Q ≈ 8.55 kW
So the rate at which energy is transferred by heat through the glass is approximately 8.55 kW.
b) To calculate the total energy lost through the window in one day, we need to know the amount of time (t) the window is exposed to the temperature difference.
Assuming the window is exposed to this temperature difference for 24 hours (1 day), we can use the formula:
Energy = Q * t
Energy = 8.55 kW * 24 hours
Energy ≈ 205.2 kWh
Therefore, the amount of energy lost through the window in one day, assuming constant indoor and outdoor temperatures, is approximately 205.2 kWh.
To answer these questions, we need to use the formula for heat transfer through a material. The formula is given by:
Q = (k * A * ΔT) / d
where:
Q is the rate of heat transfer (in watts),
k is the thermal conductivity of the material (in watts per meter per degree Celsius),
A is the cross-sectional area of the material (in square meters),
ΔT is the temperature difference across the material (in degrees Celsius), and
d is the thickness of the material (in meters).
a) To find the rate at which energy is transferred by heat through the glass, we need to calculate Q using the given information.
Let's plug in the values:
k = thermal conductivity of glass (varies based on the type of glass; for example, around 0.8 W/m°C for single-pane window glass)
A = 1.0 m * 2.0 m = 2.0 m^2 (cross-sectional area of the glass)
ΔT = difference in temperature = (25°C - 0°C) = 25°C
d = 0.62 cm = 0.62 cm / 100 = 0.0062 m (thickness of the glass)
Now we can calculate Q:
Q = (k * A * ΔT) / d
Q = (0.8 W/m°C * 2.0 m^2 * 25°C) / 0.0062 m
Q ≈ 2580.65 watts
Therefore, the rate at which energy is transferred by heat through the glass is approximately 2580.65 watts.
b) To find out how much energy is lost through the window in one day, we need to calculate the total energy transferred over time. Since the temperatures inside and outside remain constant, the rate of heat transfer remains the same throughout the day.
Let's convert one day into seconds:
1 day = 24 hours = 24 hours * 60 minutes/hour * 60 seconds/minute = 86,400 seconds
Now, we can calculate the total energy transferred:
Energy = Rate of heat transfer * Time
Energy = 2580.65 watts * 86,400 seconds
Energy ≈ 223,540,160 joules
Therefore, approximately 223,540,160 joules of energy is lost through the window in one day, assuming the indoor and outdoor temperatures remain constant.