Question

The brick wall of a house has the square S = 20 m2 and the thickness d = 37 cm. The room inside the house is heated with the electrical heater with power P = 1 kWt. Define the distance from the inner surface of the wall to the plane inside the brick wall where the temperature equals 0t = °0 С, if the temperature outside is t =− ° 26 С. The heat conductivity of the brick is λ = 0,4 Wt/(m⋅К). Other heat losses shall be neglected. The room temperature is the same all over the room.

Answer 1

The thickness of the wall is much less then the its measurements, and the room temperature is the same all over the room. We can to consider the heat transfer as a flat task. That is

(1) $\lambda\frac{\Delta T}{ \Delta x}=\frac{P}{S}$ , where $\Delta T=T_0-T_{outside}$ , $\Delta x =d$. $T_0$ is the room temperature.

Furthermore $T(x)$ is the line function of $x$

(2) $T(x)=T_0\cdot \frac{d-x}{d}+T_{outside}\cdot \frac{x}{d}$

satisfying the boundary conditions

$T(x=0)=T_0; T(x=d)=T_{outside}$.

First we should find the room tempreture $T_0$. From (1) we determine

$T_0=T_{outside}+\frac{P\cdot d}{S\cdot \lambda}=-26^oC+\frac{10^3Wt\cdot 0.37m}{20 m^2\cdot 0.4 Wt\cdot m^{-1}K^{-1}}=-26^oC+46.25^oC=20.25^oC$

In the calculations, we convert all quantities to the same dimension of the SI base units and took into account that the Kelvin degree and the Celsius degree are equal in magnitude. Now from (2) we can find the answer to the main question of the problem. Solve the equation

$T(x)=T_0\cdot \frac{d-x}{d}+T_{outside}\cdot \frac{x}{d}=0^oC$ .

We have

$x=\frac{T_0}{T_0-T_{outside}}\cdot d=\frac{20.25}{46.25}\cdot 0.37m=0.162m=16.2cm$

Answer: The distance from the inner surface of the wall to the plane inside the brick wall where the temperature equals °0 С, is 16.2cm.