A transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of \(70^{\circ} \mathrm{C}\) is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of \(1 \mathrm{~mm}\) and thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the solid plate is \(13 \mathrm{~mm}\) thick and has a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Inside the heated chamber, the convection heat transfer coefficient is \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the bottom surface of the solid plate is maintained at \(52^{\circ} \mathrm{C}\), determine the temperature inside the heated chamber and the surface temperature of the transparent film. Assume thermal contact resistance is negligible.

Thermal conductivity is a fundamental property of material that quantifies its ability to conduct heat. It is represented by the symbol \(k\) and typically has units of \(W/(m\cdot K)\). In the context of the given exercise, thermal conductivity plays a vital role in determining how fast or slow heat is conducted through the film and the solid plate.

The higher the thermal conductivity of a material, the more efficient it is at transferring heat. In our exercise, the thermal conductivity of the film is quite low (\(k_{film} = 0.05\, W/(m\cdot K)\)), indicating it is a poor conductor of heat, while the solid plate, with a \(k_{plate} = 1.2\, W/(m\cdot K)\), conducts heat relatively well. This difference has a direct impact on the thermal resistances and the temperatures within the system. To improve understanding, students could compare materials with different thermal conductivities to observe how they would affect the system's temperature distribution.

The convection heat transfer coefficient, denoted \(h\), measures the efficiency with which heat is transferred between a surface and a fluid moving past it. This coefficient is expressed in \(W/(m^2\cdot K)\) and is influenced by various factors such as fluid velocity, viscosity, temperature, and the characteristics of the surface.

In our exercise, the convection heat transfer coefficient is \(70\, W/(m^2\cdot K)\), which is a measure of the heat exchanged between the surface of the transparent film and the air inside the heated chamber. For students looking to deepen their understanding, exploring how changing \(h\) affects the overall heat transfer in real-world scenarios can be insightful. Also, understanding the effects of different cooling or heating mechanisms, such as fans or radiators, on the convection heat transfer coefficient can offer practical insights into thermal management.

Thermal resistance is analogous to electrical resistance in that it quantifies the opposition to heat flow through a specific material or interface and is calculated using the formula \(R = L/(kA)\) for conductive resistance, and \(R_{conv} = 1/(hA)\) for convective resistance. Here, \(L\) is the thickness of the material, \(k\) is its thermal conductivity, \(A\) is the cross-sectional area, and \(h\) represents the convection heat transfer coefficient.

Understanding thermal resistance is crucial for analyzing and designing systems in which temperature control is needed, as seen in our exercise. Factors like thickness (\(L\)), material properties (\(k\)), and the nature of the heat transfer process (conduction or convection) all affect the thermal resistance. A practical exercise for students would be to calculate the thermal resistance of various materials and arrangements to see how they could improve or impede heat flow. By mastering thermal resistance concepts, students can better tackle problems involving heat transfer calculations and design more efficient thermal systems.