Thermal Resistance Analysis

The problem asks for the thickness of an aluminium plate ($L_2$) whose thermal resistance is equal to the thermal resistance of an interface between two aluminium plates.

Interface Resistance Calculation

The thermal resistance of an interface is inversely proportional to the thermal contact conductance ($h_c$). We consider the resistance per unit area ($R''_{interface}$):

$ R''_{interface} = \frac{1}{h_c} $

Given $h_c = 10000\ W/m^2\ ^\circ C$, the interface resistance per unit area is:

$ R''_{interface} = \frac{1}{10000\ W/m^2\ ^\circ C} = 0.0001\ m^2\ ^\circ C/W $

Plate Resistance Calculation

The thermal resistance of a plate (per unit area, $R''_{plate}$) is given by its thickness ($L$) divided by its thermal conductivity ($k$):

$ R''_{plate} = \frac{L}{k} $

We need to find the thickness $L_2$ for a second aluminium plate such that its resistance equals the interface resistance:

$ R''_{plate2} = R''_{interface} $

$ \frac{L_2}{k} = \frac{1}{h_c} $

Determining Plate Thickness ($L_2$)

Substitute the known values for the thermal conductivity of aluminium ($k = 237\ W/m\ ^\circ C$) and the thermal contact conductance ($h_c = 10000\ W/m^2\ ^\circ C$):

$ L_2 = \frac{k}{h_c} $

$ L_2 = \frac{237\ W/m\ ^\circ C}{10000\ W/m^2\ ^\circ C} $

$ L_2 = 0.0237\ m $

Convert the thickness from meters to centimeters:

$ L_2 = 0.0237\ m \times \frac{100\ cm}{1\ m} = 2.37\ cm $

Therefore, the thickness of the aluminium plate whose thermal resistance is equal to the thermal resistance of the interface is 2.37 cm.