Consider a 2-m-long and 0.7-m-wide stainless-steel plate whose thickness is \(0.1 \mathrm{~m}\). The left surface of the plate is exposed to a uniform heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\) while the right surface of the plate is exposed to a convective environment at \(0^{\circ} \mathrm{C}\) with \(h=400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The thermal conductivity of the stainless steel plate can be assumed to vary linearly with temperature range as \(k(T)=k_{o}(1+\beta T)\) where \(k_{o}=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\beta=9.21 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\). The stainless steel plate experiences a uniform volumetric heat generation at a rate of \(8 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). Assuming steady state one-dimensional heat transfer, determine the temperature distribution along the plate thickness.

Since we're assuming steady-state, one-dimensional heat conduction, we can use the following simplified form of the heat conduction equation: $$\frac{d}{dx}(k(T)\frac{dT}{dx})+q_{gen}=0$$ where \(x\) is the distance along the thickness of the plate, \(k(T)\) is the thermal conductivity as a function of temperature, \(\frac{dT}{dx}\) is the temperature gradient along the \(x\) direction, and \(q_{gen}\) is the volumetric heat generation rate.

Now, let's substitute the given parameters into the equation: $$\frac{d}{dx}(k_{o}(1+\beta T)\frac{dT}{dx})+q_{gen}=0$$ with \(k_{o}=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=9.21 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\), and \(q_{gen}=8 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\).

Let's simplify the differential equation and solve for the temperature distribution \(T(x)\): $$\frac{d}{dx}(48(1+9.21 \times 10^{-4} T)\frac{dT}{dx})=-8 \times 10^{5}$$ To solve this equation, we can first multiply both sides by \(dx\) and integrate: $$\int{48(1+9.21 \times 10^{-4} T)d\left(\frac{dT}{dx}\right)}=-8 \times 10^{5}\int{dx}$$ This integration yields a first-degree polynomial in \(T(x)\): $$48(1+9.21 \times 10^{-4} T(x))\frac{dT(x)}{dx} = C -8 \times 10^{5}x$$ where \(C\) is the integration constant.

Now we need to apply the boundary conditions to determine the constant \(C\) and the temperature distribution \(T(x)\). At the left surface of the plate (\(x=0\)), the heat flux is given as \(2000 \mathrm{~W} / \mathrm{m}^{2}\). We can use Fourier's law of heat conduction to relate the heat flux to the temperature gradient at the left surface: $$q_{left}=-k(T(0)) \frac{dT}{dx}|_{x=0}$$ Rearranging and substituting the given heat flux, $$\frac{dT}{dx}|_{x=0}=-\frac{2000}{k(T(0))}$$ Now, we substitute this into the polynomial equation derived earlier when \(x=0\), $$48(1+9.21 \times 10^{-4} T(0))\left(-\frac{2000}{k(T(0))}\right) = C$$ At the right surface of the plate (\(x=0.1 \mathrm{~m}\)), we have a convective boundary condition at \(0^{\circ} \mathrm{C}\) with \(h=400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). We apply Newton's law of cooling to find the temperature T(0.1): $$q_{right}=h(T(0.1)-T_{\infty})$$ Substituting the given values, we obtain the temperature at the right surface, $$T(0.1) = 0.1^{\circ}\mathrm{C}$$ Now we can solve the polynomial equation with these boundary conditions to find the temperature distribution \(T(x)\) along the plate thickness. The detailed solution of this equation may require numerical methods such as finite difference or finite element analysis. In summary, it is essential to understand the given conditions, set up the governing differential equation, and apply the boundary conditions to determine the temperature distribution along the plate thickness. The solution may require numerical methods for complex problems.