Answer: Option (C) is the correct answer.
Explanation:
For heat conduction (Q), the general form is as follows.
       [tex]\frac{d}{dx}(k\frac{dT}{dx}) + q = \rho C_{p}\frac{dT}{dZ}[/tex]
where, Â Â Â Â q = heat generation
        [tex]\rho[/tex] = density
         [tex]C_{p}[/tex] = specific heat
          Z = time
          k = thermal conductivity
          x = spatial direction (positional)
When k = constant
  No heat is generated hence, q = 0
  steady state conduction [tex]\frac{dT}{dx}[/tex] = 0
Hence, Â [tex]k \frac{d^{2}T}{dx^{2}}[/tex] + 0 = [tex]\rho C_{p} \times 0[/tex]
        [tex]k \frac{d^{2}T}{dx^{2}}[/tex] = 0
           [tex]\frac{d^{2}T}{dx^{2}}[/tex] = 0
On integrating both sides we get the following.
     [tex]\frac{dT}{dx} = C_{1}[/tex] = constant
Therefore, the temperature gradient along the specimen remains constant.
Hence, we can conclude that the statement in steady-state heat conduction  the temperature gradient along the specimen is constant, is true.
