Question

In a house the temperature at the surface of a window is 28.9 °C. The temperature outside at the window surface is 7.89 °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

-13.18°C

Step-by-step explanation:

To develop the problem it is necessary to consider the concepts related to the thermal conduction rate.

Its definition is given by the function

`(Q)/(t) = (kA\Delta T)/(d)`

Where,

Q = The amount of heat transferred

t = time

k = Thermal conductivity constant

A = Cross-sectional area

`\Delta T =` The difference in temperature between one side of the material and the other

d= thickness of the material

The problem says that there is a loss of heat twice that of the initial state, that is

`Q_2 = 2*Q_1`

Replacing,

`kA(\Delta T_m)/(x) = 2*kA(\Delta T)/(x)`

`(\Delta T)/(x)=2*(\Delta T)/(x)`

`(T_i-T_o)/(x) = 2(T_1-T_2)/(x)`

`(28.9-T_o)/(x) = 2(28.9-7.86)/(x)`

Solvinf for
`T_o`,

`T_o = -13.18`

Therefore the temprature at the outside windows furface when the heat lost per second doubles is -13.18°C