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1 month ago

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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?

1 answer:

Yuliya22 [3.3K]1 month ago

3 0

Answer:

-13.18°C

Explanation:

To solve this issue, we must examine the principles associated with the rate of thermal conduction.

This rate is defined by the equation

$\frac{Q}{t} = \frac{kA\Delta T}{d}$

Where,

Q = Amount of heat transferred

t = time

k = Thermal conductivity constant

A = Area of cross-section

$\Delta T =$ Temperature difference across the material

d = Material thickness

The scenario indicates a heat loss that is double the initial value, which means

$Q_2 = 2*Q_1$

Substituting values yields,

$kA\frac{\Delta T_m}{x} = 2*kA\frac{\Delta T}{x}$

$\frac{\Delta T}{x}=2*\frac{\Delta T}{x}$

$\frac{T_i-T_o}{x} = 2\frac{T_1-T_2}{x}$

$\frac{28.9-T_o}{x} = 2\frac{28.9-7.86}{x}$

Solving for $T_o$ ,

$T_o = -13.18$

Thus, when the heat lost per second is doubled, the temperature on the external surface of the window is -13.18°C.

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