Question

In steady-state heat conduction O A. the temperature at each position along the specimen rises slowly over time. O B. the thermal conductivity of the specimen decreases. ° C. the temperature gradient along the specimen is constant. D. changing the flowrate of cooling water against the cold side does not affect its temperature.

Answer 1

Answer: Option (C) is the correct answer.

Step-by-step explanation:

For heat conduction (Q), the general form is as follows.

`(d)/(dx)(k(dT)/(dx)) + q = \rho C_(p)(dT)/(dZ)`

where, q = heat generation

`\rho` = density

`C_(p)` = specific heat

Z = time

k = thermal conductivity

x = spatial direction (positional)

When k = constant

No heat is generated hence, q = 0

steady state conduction
`(dT)/(dx)` = 0

Hence,
`k (d^(2)T)/(dx^(2))` + 0 =
`\rho C_(p) * 0`

`k (d^(2)T)/(dx^(2))` = 0

`(d^(2)T)/(dx^(2))` = 0

On integrating both sides we get the following.

`(dT)/(dx) = C_(1)` = 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.