The steady state one dimensional heat conduction in any material can be expressed as:

$(\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{x}}(k_x\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{x}})+\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{y}}(k_y\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{y}})+\frac{\text{d}\text{e}\text{l}}{\text{d}\text{e}\text{l}\text{z}}(k_z\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{z}}))+q_g=C_p\frac{\text{d}\text{e}\text{l}\text{T}}{\text{d}\text{e}\text{l}\text{t}}$

Consider a slab with a thickness of $1m$ made of material with a thermal conductivity of $10\frac{W}{m}\frac{?}{K}.$ The slab separates $2$ bodies of water, each at a constant temperature of $20C,$ sketched below. The temperature at the surface of the slab $T_S$ is related to the temperature of the water $T_w$ by: $T_S=0\mathrm{.}5(200C-T_w).$

Considering steady state one dimensional heat conduction through thickness and a heat generation of $50\frac{W}{m^3}$ :

a$)$ Simplify the generalized heat conduction equation as much as possible. Justify your reasoning.

b$)$ Find the generalsolution to the problem.

c$)$ Write the boundary conditions.

d$)$ Find the particular solution.

e$)$ Plot the through thickness temperature profile.

f$)$ Find the location of the maximum temperature and its value.