a) A 1-D, composite wall without any heat generation is shown below with the given wall properties. For the steady state conditions listed, calculate by hand and then plot the temperature profile, T(x) in the wall. T Wall A AB Wall B kB Wall C kc R' Air TR R BC T TR X KA = 50 W/m/K CpA = 4 J/g/K PA = 1 g/cm LA = 10 cm T = 450 K LB KB = 5 W/m/K 4 J/g/K PB = 1 g/cm LB = 30 cm TOOR = 280 K LC kc = 1 W/m/K Cpc = 4 J/g/K Pc = 1 g/cm Lc = 10 cm R hR 80 W/m/K = b) Solve the above steady state 1-D problem, but now with heat generation within slab B with = 5x10 W/m. Recall, to solve a heat generation slab problem, you will need to write a system of THREE equations for T(x) or q"(x) in the domains: (1) left of the heat generating wall, (2) the heat generating wall component, and (3) right of the heat generating wall. You will then need to solve the system of equations by equating T(x) or q(x) at the interfaces of the three domains. For a heat generating wall, recall we write the solution w.r.t. a coordinate system at the center of the heat generating domain.