Consider a large plane wall of thickness \(I_{\text {, }}\) thermal conductivity \(k\), and surface area \(A\). The left surface of the wall is exposed to the ambient air at \(T_{\infty}\) with a heat transfer coefficient of \(h\) while the right surface is insulated. The variation of temperature in the wall for steady one-dimensional heat conduction with no heat generation is (a) \(T(x)=\frac{h(L-x)}{k} T_{\infty}\) (b) \(T(x)=\frac{k}{h(x+0.5 L)} T_{\infty}\) (c) \(T(x)=\left(1-\frac{x h}{k}\right) T_{\infty}\) (d) \(T(x)=(L-x) T_{\infty}\) (e) \(T(x)=T_{\infty}\)

Since the system is in the steady state, there is no change in temperature with time and no internal heat generation. So, the heat flux \(q''\) through the wall must be constant and determined by the surface heat transfer at the left surface. This can be calculated using convective heat transfer equation at the left surface (x=0): $$ q'' = h(T(0) - T_{\infty}) $$

Since the heat flux is constant due to steady-state condition and no heat generation, the temperature gradient in the wall should also be constant. Therefore, the change in temperature with respect to position in the wall should be constant: $$ \frac{dT}{dx}=C $$ Where \(C\) is a constant value.

We should check the given options against the boundary conditions: 1. At x = 0, the left surface should have a temperature \(T_{0}\) which can be calculated from the heat flux equation mentioned in step 1: $$ T(0)=T_{0} = T_{\infty} +\frac{q''}{h} $$ 2. At x = L, the right surface is insulated, so there should be no heat flux through the surface. This means that the derivative of the temperature with respect to x should be zero at the right surface: $$ \frac{dT}{dx}(L) = 0 $$

- Option (a): At x=0, \(T(0)=0\), which does not meet the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) = -\frac{h}{k}\neq0\), which also does not meet the boundary condition 2. So, this option is not correct. - Option (b): At x=0, \(T(0)=\frac{k}{0.5hL}T_{\infty}\), which does not meet the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) \) is not zero (derivative calculation needed), which does not meet the boundary condition 2. So, this option is also not correct. - Option (c): At x=0, \(T(0)= T_{\infty}\), which meets the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) = -\frac{h}{k}\neq0\), which does not meet the boundary condition 2. So, this option is not correct. - Option (d): At x=0, \(T(0)=(L) T_{\infty}\), which does not meet the boundary condition 1. Derivative calculation is not needed, as it already doesn't follow boundary condition 1. - Option (e): Since we've eliminated the other options, this must be the correct choice. However, let's verify: At x=0, \(T(0)=T_{\infty}\), which meets the boundary condition 1. At x=L, \(\frac{dT}{dx}(L) = 0\), which meets the boundary condition 2. Therefore, the correct equation for the temperature distribution in the wall, under given conditions, is: $$T(x) = T_{\infty}$$