Heat Transfer - Conduction and Convection: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Heat Transfer - Conduction and Convection. Understanding the underlying formula is key to arriving at the correct answer below:

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 \(40^\circ C\).