In a house the temperature at the surface of a window is 20.1 °C. The temperature outside at the window surface is 2.72 °C. 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 in degrees Celsius at the outside window surface when the heat lost per second doubles?

Question

In a house the temperature at the surface of a window is 20.1 °C. The temperature outside at the window surface is 2.72 °C. 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 in degrees Celsius at the outside window surface when the heat lost per second doubles?

Answer 1

To find the temperature at the outside window surface when the heat lost per second doubles, we need to understand the relationship between heat loss and temperature difference.

Heat loss through a window via conduction is directly proportional to the temperature difference between the inside and outside surfaces of the window. This relationship is described by Newton's Law of Cooling:

Q = h * A * (T_inside - T_outside),

where Q is the rate of heat loss (heat lost per second), h is the heat transfer coefficient, A is the area of the window, T_inside is the temperature inside the house, and T_outside is the temperature outside.

In this case, we know that the initial heat loss rate, Q_initial, is related to the initial temperature difference:

Q_initial = h * A * (T_inside - 2.72).

Let's call the temperature at the outside window surface when the heat lost per second doubles as T_new_outside. In that case, the heat loss rate, Q_new, will be twice the initial heat loss rate:

Q_new = 2 * Q_initial.

Substituting the expression for Q_initial, we have:

2 * h * A * (T_inside - 2.72) = h * A * (T_inside - T_new_outside).

Now, we can simplify the equation by canceling out the area and heat transfer coefficient:

2 * (T_inside - 2.72) = T_inside - T_new_outside.

Expanding the left side:

2 * T_inside - 5.44 = T_inside - T_new_outside.

Rearranging the equation:

2 * T_inside - T_inside = -T_new_outside + 5.44,

T_inside = -T_new_outside + 5.44.

Finally, rearranging the equation to solve for T_new_outside:

T_new_outside = 5.44 - T_inside.

Since the temperature inside the house is given as 20.1 °C, we can substitute this value into the equation:

T_new_outside = 5.44 - 20.1.

Calculating the value:

T_new_outside = -14.66 °C.

Therefore, the temperature at the outside window surface when the heat lost per second doubles is approximately -14.66 °C.

Answer 2

In a house the temperature at the surface of a window is 22.4 °C. The temperature outside at the window surface is 2.66 °C. 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 in degrees Celsius at the outside window surface when the heat lost per second doubles?