Respuesta :
Explanation:
As it is given that no heat is generating and the system is present in steady state.
As we know that,
  [tex]\frac{\partial^{2}T}{\partial x^{2}} + \frac{Q}{k}[/tex] = [tex]\frac{1}{\alpha} \frac{\partial T}{\partial t}[/tex]
So, according to the given conditions, [tex]\frac{Q}{k}[/tex] = 0, and [tex]\frac{1}{\alpha} \frac{\partial T}{\partial t}[/tex] = 0
So, Â Â [tex]\frac{\partial^{2}T}{\partial x^{2}}[/tex] = 0 , [tex]\frac{\partial T}{\partial x}[/tex] = [tex]C_{1}[/tex]
        T = [tex]C_{1}x + C_{2}[/tex]
Hence, boundary conditions will be as follows.
At x = 0, T = [tex]80^{o}C[/tex]
At x = L, T = [tex]40^{o}C[/tex]
Applying boundary condition is [tex]T_{0} = C_{2}[/tex]
Therefore, [tex]T_{0} = C_{2}[/tex] = [tex]80^{o}C[/tex]
    [tex]T_{L} = C_{1}x + 80[/tex]
        40 = [tex]C_{1} (100 \times 10^{-3}) + 80[/tex]
     [tex]\frac{-40}{100 \times 10^{-3}} = C_{1}[/tex]
         [tex]C_{1}[/tex] = -400
Hence, temperature distribution in a block will be as follows.
         T = -400x + 80
At center, x = [tex]50 \times 10^{-3}[/tex]
Therefore, Â T = [tex]-400 \times 50 \times 10^{-3} + 80[/tex]
           = [tex]60^{o}C[/tex]
Thus, we can conclude that the temperature of block at the center point is [tex]60^{o}C[/tex].
