In a house the temperature at the surface of a window is 22.3 °C. The temperature outside at the window surface is 2.82 °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 22.3 °C. The temperature outside at the window surface is 2.82 °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 answer this question, we first need to understand the relationship between temperature and heat transfer via conduction through the window. According to Fourier's Law of Heat Conduction, the rate of heat transfer through a material is directly proportional to the temperature difference across the material.

In this case, the heat lost per second is directly related to the temperature difference between the inside and outside surfaces of the window. As the temperature outside begins to fall, the temperature difference across the window increases, resulting in a higher rate of heat loss.

Now, we are given that the temperature on the inside surface of the window remains the same at 22.3 °C. Let's call the initial temperature on the outside surface of the window T1, which is 2.82 °C.

To find the temperature on the outside window surface when the heat lost per second doubles, we need to find T2. Since the rate of heat loss is directly proportional to the temperature difference, we can use this relationship to set up an equation:

Heat lost per second (initial) = Heat lost per second (final) * 2

(Temperature inside - Temperature outside initial) = (Temperature inside - Temperature outside final) * 2

(22.3 °C - T1) = (22.3 °C - T2) * 2

Now, we can solve this equation to find T2.

22.3 °C - T1 = 44.6 °C - 2T2

-T1 + 2T2 = 44.6 °C - 22.3 °C

2T2 = 22.3 °C + T1

T2 = (22.3 °C + T1) / 2

Substituting the given value of T1, we get:

T2 = (22.3 °C + 2.82 °C) / 2

T2 = 12.06 °C

Therefore, when the heat lost per second doubles, the temperature on the outside window surface will be approximately 12.06 °C.