Question

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures${T_1}$ and ${T_2}$ $\left( {{T_1} > {T_2}} \right)$. The rate of heat transfer$\dfrac{{dQ}}{{dt}}$, through the rod in a Steady State, is given by
$\left( A \right).{\text{ }}\dfrac{{dQ}}{{dt}} = \dfrac{{KL\left( {{T_1} - {T_2}} \right)}}{A}$
$\left( B \right).{\text{ }}\dfrac{{dQ}}{{dt}} = \dfrac{{KL\left( {{T_1} - {T_2}} \right)}}{{LA}}$
$\left( C \right).{\text{ }}\dfrac{{dQ}}{{dt}} = KLA\left( {{T_1} - {T_2}} \right)$
$\left( D \right).{\text{ }}\dfrac{{dQ}}{{dt}} = \dfrac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$

Answer 1

Hint
Temperature is one of the physical properties of matter. The temperature of the matter defines the state of it. Also, there are alternative temperature-related properties of matter like thermal conductivity, thermal diffusivity, and so on. Steady-state in natural philosophy deals with notably no modification in energy and mass of the system into account with relevance time. Technically, up to the marked volume, there's no accumulation of any mass and energy.

Complete step by step answer
In the given question there is a rod having length L and their cross-section area is given as A. Also, it is given that there is the temperature difference which is represented by T that is $T_1-T_2$.
Since the rod is in steady-state condition and as we all know Steady State is that state wherever the temperature of every rod becomes constant.
During this state, no heat is absorbed by the rod. The warmth that reaches any crosswise is transmitted to the future, except that a tiny low a part of the warmth is lost to the encircling from the perimeters of convection and radiation. And this state of the rod is termed Steady State.
$ \Rightarrow \dfrac{{dQ}}{{dt}} = \dfrac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$
Where $\dfrac{{dQ}}{{dt}}$is the conduction through the rod,
$K$ is the Coefficient of thermal conductivity,
$A$ is the Cross-sectional area and
${T_1}{\text{ and }}{{\text{T}}_2}$ is the temperature difference.
So from the condition of Steady State, we can say that the rate of heat flowing from one face to another face in time t is given by
$ \Rightarrow \dfrac{{dQ}}{{dt}} = \dfrac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$
Therefore, Option (D) will be the correct answer.

Note
Steady-state conductivity is the kind of conductivity that happens once the temperature difference(s) driving the conductivity is constant, so (after a stabilization time), the spatial distribution of temperatures (temperature field) within the conducting object does not modify to any extent further. Thus, all partial derivatives of temperature regarding space could either be zero or have nonzero values, however, all derivatives of temperature at any purpose regarding time are uniformly zero. In steady-state conductivity, the quantity of heat coming into any region of an object is capable of the quantity of heat coming out (if this were not thus, the temperature would be rising or falling, as thermal energy was cornered in a very region).