A \(1200-W\) iron is left on the iron board with its base exposed to ambient air at \(26^{\circ} \mathrm{C}\). The base plate of the iron has a thickness of \(L=0.5 \mathrm{~cm}\), base area of \(A=150 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. The outer surface of the base plate whose emissivity is \(\varepsilon=0.7\), loses heat by convection to ambient air with an average heat transfer coefficient of \(h=\) \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as well as by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=295 \mathrm{~K}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the temperature of the outer surface of the plate by solving the differential equation, and (c) evaluate the outer surface temperature.

Thermal conductivity is a key concept when discussing heat conduction. It is a material-specific trait that tells us how well the material conducts heat. Mathematically, it is denoted by the symbol \(k\) and is measured in watts per meter-kelvin (\(W/m\cdot K\)). The higher the thermal conductivity, the more efficient the material is at transferring heat. In the context of the iron base plate in the exercise, the thermal conductivity is given as \(k=18 W/m\cdot K\), indicating it transfers heat effectively.

One interesting point to be noted here is that when we talk about thermal conductivity in a steady one-dimensional heat transfer situation, it assumes a constant value within the material, implying uniform heat transfer characteristics across the material. Therefore, in the solved problem, the rate of heat flow through the iron's base plate is consistent, leading to the establishment of a steady temperature profile that can be determined by solving appropriate mathematical equations under given boundary conditions.

In the realm of heat transfer, the term 'steady' implies that temperature does not change with time, while 'one-dimensional' specifies that heat flows in only one direction. Steady one-dimensional heat transfer is a simplified scenario that is often used to model heat transfer in systems where temperature gradients exist in only one dimension and remain constant over time.

In our discussed exercise, the heat transfer through the thickness of the iron's base plate represents a classic example of steady one-dimensional heat transfer. This assumption simplifies the problem, allowing us to use the basic heat conduction equation \(\frac{d^{2} T}{d x^{2}}=0\) to describe the system. The solution to this differential equation points to a linear temperature gradient within the base plate, making the analysis far more manageable and the computation of temperature profiles straightforward once the boundary conditions are addressed.

Boundary conditions are essential for solving differential equations in heat transfer problems as they define how the system interacts with its surroundings. They're set based on the physical constraints and forces acting at the boundaries of the system. There are generally two types of boundary conditions in heat transfer: temperature-based (Dirichlet) and flux-based (Neumann).

In the example of the iron's base plate, we have a flux-based boundary condition at the inner surface where heat is being generated by the resistance heater. This is expressed by a uniform heat flux that determines the temperature gradient at that surface. Conversely, at the base plate's outer surface, the heat loss due to convection and radiation to the surroundings sets the boundary condition, which matches the heat flux leaving the plate. Adequately defining and applying these boundary conditions is critical for accurately assessing the temperature distribution within the base plate. Understanding and implementing these conditions correctly allows us to predict how the plate's temperature will react to the internal generation of heat and external cooling influences.