Answer to Question #325589 in Physics for Naruto

Question

A wall with a thermal conductivity of 0.50W/m·K is maintained at 40.0°C. The heat transfer through the wall is 250.0W. The wall surface area is 1.50m2 and its thickness is 1.00cm. Determine the temperature at the other surface.

Answer 1

The Fourier's law says

$$P=\kappa\frac{\Delta T}{\Delta x}A$$

Hence, the temperature difference

$$\Delta T=\frac{P\Delta x}{\kappa A}$$

$$\Delta T=\frac{250.0*0.01}{0.50*1.50}=3.33^\circ\rm C$$

The temperature at the other surface

$$T_2=T_1\pm\Delta T=40.0\pm3.33\\ =43.3^\circ\rm C \: \rm or\; 36.7^\circ\rm C$$