Answer: Option (C) is the correct answer.
Step-by-step explanation:
For heat conduction (Q), the general form is as follows.

where, q = heat generation
= density
= specific heat
Z = time
k = thermal conductivity
x = spatial direction (positional)
When k = constant
No heat is generated hence, q = 0
steady state conduction
= 0
Hence,
+ 0 =

= 0
= 0
On integrating both sides we get the following.
= 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.