Consider a 20-cm-thick large concrete plane wall \((k=0.77 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) subjected to convection on both sides with \(T_{\infty 1}=22^{\circ} \mathrm{C}\) and \(h_{1}=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the inside, and \(T_{\infty 2}=8^{\circ} \mathrm{C}\) and \(h_{2}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the outside. Assuming constant thermal conductivity with no heat generation and negligible radiation, \((a)\) express the differential equations and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the temperatures at the inner and outer surfaces of the wall.

The governing equation for steady one-dimensional heat conduction with constant thermal conductivity and no heat generation is given by: $$\frac{d^2T}{dx^2} = 0$$ The boundary conditions at the inner and outer surfaces can be expressed using Newton's law of cooling: 1. At the inner surface (x=0): $$-k \frac{dT}{dx} \Big|_{x=0} = h_{1}(T_{1} - T_{\infty 1})$$ 2. At the outer surface (x=L; L: wall thickness): $$-k \frac{dT}{dx} \Big|_{x=L} = h_{2}(T_{2} - T_{\infty 2})$$

The general solution of the given differential equation is a linear equation, i.e., $$T(x) = Ax + B$$ Now, apply the boundary conditions to determine the constants A and B. From the first boundary condition at x=0: $$-k\frac{dT}{dx} \Big|_{x=0} = k(-A) = h_{1}(T_{1} - T_{\infty 1})$$ Or, $$A = \frac{h_{1}(T_{1}-T_{\infty 1})}{k}$$ From the second boundary condition at x=L: $$-k\frac{dT}{dx} \Big|_{x=L} = k(-A) = h_{2}(T_{2} - T_{\infty 2})$$ Or, $$A = \frac{h_{2}(T_{2}-T_{\infty 2})}{k}$$ Equating the expressions for A from both boundary conditions, we get: $$\frac{h_{1}(T_{1}-T_{\infty 1})}{k} = \frac{h_{2}(T_{2}-T_{\infty 2})}{k}$$ Now, substituting the expression for A into the general solution of T(x): $$T(x) = \frac{h_{1}(T_{1}-T_{\infty 1})}{k} x + B$$ To find B, substitute x=0 and T(x) = T1 in the equation $$T_{1} - T_{\infty 1} = -B$$ So, the temperature distribution in the wall can be written as: $$T(x) = \frac{h_{1}(T_{1}-T_{\infty 1})}{k}x + (T_{1} - T_{\infty 1})$$

From the equation derived in Step 2, the temperature at any point x can be calculated. To find the temperatures at the inner and outer surfaces: 1. At the inner surface (x=0): We know that T(0) = T1, so: $$T_{1} = T_{\infty 1} + \frac{k(T_{2}-T_{\infty 2})}{h_{1}L}$$ 2. At the outer surface (x=L): The temperature at this point is T2, so: $$T_{2} = T_{\infty 1} + \frac{k(T_{1}-T_{\infty 1})}{h_{2}L}$$ Now, plug the given values for the problem \(T_{\infty 1}=22^{\circ} \mathrm{C}\), \(h_{1}=8 \mathrm{~W} / \mathrm{m}^{2}\cdot \mathrm{K}\), \(T_{\infty 2}=8^{\circ} \mathrm{C}\), \(h_{2}=12 \mathrm{~W} / \mathrm{m}^{2}\cdot \mathrm{K}\), \(L = 0.2 \mathrm{m}\), and \(k = 0.77 \mathrm{~W} /\mathrm{m} \cdot \mathrm{K}\) into these equations to find the temperatures at the inner and outer surfaces: $$T_{1} = 22 + \frac{0.77(T_{2}-8)}{8 \times 0.2}$$ $$T_{2} = 22 + \frac{0.77(T_{1}-22)}{12 \times 0.2}$$ Solve these simultaneous equations to find T1 and T2: $$T_{1} = 20.36^{\circ} \mathrm{C}$$ $$T_{2} = 15.14^{\circ} \mathrm{C}$$ So, the temperatures at the inner and outer surfaces of the wall are 20.36°C and 15.14°C, respectively.