In a house the temperature at the surface of a window is 25 ¡ãC. The temperature outside at the window surface is 5.0 ¡ãC. Heat is lost through the window via conduction, and the heat lost per second has a certain value. The temperature outside begins to fall, while the conditions inside the house remain the same. As a result, the heat lost per second increases. What is the temperature at the outside window surface when the heat lost per second doubles?
To find the temperature at the outside window surface when the heat lost per second doubles, we need to understand how the heat lost through conduction is calculated.
The rate of heat loss through conduction can be represented by the formula:
Q = kA(delta T) / d
Where:
Q is the rate of heat loss (in watts),
k is the thermal conductivity of the material (in watts per meter per degree Celsius),
A is the surface area of the window (in square meters),
(delta T) is the temperature difference between the inside and outside of the window (in degrees Celsius), and
d is the thickness of the window (in meters).
In this case, we have two scenarios:
Scenario 1: Initial situation (heat lost per second)
T1 = 25 °C (inside window temperature)
T2 = 5.0 °C (outside window temperature)
Q1 = 2Q0 (doubled rate of heat loss per second)
Scenario 2: Required situation (heat doubles, unknown outside window temperature)
T1 = 25 °C (inside window temperature)
T2 = T (unknown outside window temperature)
Q2 = 4Q0 (double the rate of heat loss per second)
Now, let's compare the two scenarios by setting up the equations:
Q1 = kA(T1 - T2) / d (equation for scenario 1)
Q2 = kA(T1 - T) / d (equation for scenario 2)
Since Q1 = 2Q0 and Q2 = 4Q0, we can rewrite the equations as:
2Q0 = kA(T1 - T2) / d
4Q0 = kA(T1 - T) / d
We can now solve these equations to find the unknown temperature T:
2Q0 = kA(T1 - T2) / d (equation 1)
4Q0 = kA(T1 - T) / d (equation 2)
Divide equation 2 by equation 1 to eliminate the constant terms:
(4Q0) / (2Q0) = (kA(T1 - T) / d) / (kA(T1 - T2) / d)
Simplifying the equation gives:
2 = (T1 - T) / (T1 - T2)
Multiply both sides by (T1 - T2):
2(T1 - T2) = T1 - T
Expand the equation:
2T1 - 2T2 = T1 - T
Combine like terms:
T = T1 + T2
Therefore, the outside window temperature when the heat lost per second doubles is equal to the sum of the inside window temperature T1 and the outside window temperature T2.
Keep in mind that this analysis assumes that all other factors, such as the material properties and thickness of the window, remain constant.
To find the temperature at the outside window surface when the heat lost per second doubles, we can use the concept of heat transfer via conduction.
The rate of heat transfer through conduction is given by Fourier's law, which states that the rate of heat transfer (Q) is proportional to the temperature difference (ΔT) and the surface area (A), while inversely proportional to the thickness of the material (d) and the thermal conductivity of the material (k). Mathematically, this can be expressed as:
Q = (k * A * ΔT) / d
In this case, the temperature difference ΔT is the difference between the indoor and outdoor temperatures at the window surface.
Initially, the temperature at the outside window surface is 5.0 °C, and the temperature at the surface of the window inside the house is 25 °C. So, the initial temperature difference (ΔT1) is:
ΔT1 = 25 °C - 5.0 °C = 20 °C
Let's assume that the initial heat lost per second (Q1) is a certain value.
Now, we need to find the temperature at the outside window surface (T2) when the heat lost per second doubles, which means the new heat lost per second (Q2) will be twice the initial value (2 * Q1).
To find T2, we can rearrange Fourier's law equation:
Q2 = (k * A * ΔT2) / d
Where ΔT2 is the new temperature difference between the indoor and outdoor temperatures at the window surface.
Since Q2 is twice Q1, we can write:
2 * Q1 = (k * A * ΔT2) / d
Simplifying the equation:
ΔT2 = (2 * Q1 * d) / (k * A)
Now, we can substitute the initial temperature difference ΔT1 into this equation:
(2 * Q1 * d) / (k * A) = 20 °C
We have two unknowns in this equation, T2 and Q1, so we can't solve for both. If the question provided the value or relationship between Q1 and Q2, we could find T2. Without that information, it is not possible to determine the temperature at the outside window surface when the heat lost per second doubles.