The outer faces of a rectangular slab made of equal thickness of iron and brass are maintained at 100°C and 0°C respectively. The temperature at the interface is
(Thermal conductivity of iron and brass are 0.2 and 0.3 respectively.)
To find the temperature at the interface of the rectangular slab, we use the concept of thermal conduction through two materials in series.
Given:
- Thermal conductivity of iron, \(K_{\text{iron}} = 0.2 \, \text{W/mK}\)
- Thermal conductivity of brass, \(K_{\text{brass}} = 0.3 \, \text{W/mK}\)
- Temperature at one side of the iron slab, \(T_{\text{iron}} = 100^\circ C\)
- Temperature at one side of the brass slab, \(T_{\text{brass}} = 0^\circ C\)
- Thicknesses of both slabs are equal.
Since the slabs are in series and have equal thicknesses, the **heat flux** through both materials is the same. Using the formula for heat conduction, the temperature at the interface \(T_{\text{interface}}\) can be calculated using the ratio of thermal conductivities:
\[
\frac{T_{\text{iron}} - T_{\text{interface}}}{T_{\text{interface}} - T_{\text{brass}}} = \frac{K_{\text{brass}}}{K_{\text{iron}}}
\]
Substitute the known values:
\[
\frac{100 - T_{\text{interface}}}{T_{\text{interface}} - 0} = \frac{0.3}{0.2} = 1.5
\]
Now solve for \(T_{\text{interface}}\):
\[
100 - T_{\text{interface}} = 1.5 T_{\text{interface}}
\]
\[
100 = 2.5 T_{\text{interface}}
\]
\[
T_{\text{interface}} = \frac{100}{2.5} = 40^\circ C
\]
Thus, the temperature at the interface is \(40^\circ C\).