Two plates A and B have thermal conductivities 84 \(Wm^{-1}K^{-1}\) and 126 \(Wm^{-1}K^{-1}\) respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of A and B are kept at 100 °C and 0 ºC respectively, then the temperature of the surface of contact in steady state is ________°C.
To solve for the temperature of the surface of contact in steady state, we need to apply the concept of thermal resistance in a series circuit. Given the thermal conductivities \(k_A = 84 \, Wm^{-1}K^{-1}\) and \(k_B = 126 \, Wm^{-1}K^{-1}\), and the fact that both plates have the same area \(A\) and thickness \(d\), we can define the thermal resistances as follows:
Resistance of plate A: \(R_A = \frac{d}{k_A \cdot A}\)
Resistance of plate B: \(R_B = \frac{d}{k_B \cdot A}\)
The total thermal resistance \(R_{total} = R_A + R_B = \frac{d}{A}\left(\frac{1}{k_A} + \frac{1}{k_B}\right)\)
In steady state, the heat flow \(Q\) through both plates is constant: \(Q = \frac{\Delta T}{R_{total}}\)
Let \(T\) be the temperature at the contact surface. The temperature drop across plate A is \((100 - T)\), and across plate B is \((T - 0)\). Therefore, we have:
