The thermal conductivity of a material in the CGS system is 0.4. In steady state, the rate of flow of heat $10\;{\text{cal/s - c}}{{\text{m}}^2}$, then the thermal gradient will be
A. 15 $ {^\circ C/cm}$
B. 25 ${^\circ C/cm}$
C. 50 ${^\circ C/cm}$
D. 75 ${^\circ C/cm}$

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Hint:The ability of a material to conduct heat is measured as thermal conductivity. The amount of heat flowing is directly proportional to the surface area, temperature difference, time period and inversely proportional to the thickness of the body. So, by using the formula of calculating the rate of flow of heat, we can find out the temperature gradient.

Complete step by step answer:
From the question, we know that the thermal conductivity of the material in CGS unit system is $k = 0.4\;{cal/^ \circ C} \cdot {\text{s}} \cdot {\text{cm}}$ and the rate of flow of heat per unit area, $\dfrac{{dQ}}{{Adt}} = 10cal/cm \cdot {s^2}$ .We know that the amount of rate of heat that flows across the body or material is expressed as,
$\dfrac{{dQ}}{{dt}} = - KA\dfrac{{dT}}{{dx}}$
Here, $A$ is the surface area of the material, $\dfrac{{dT}}{{dx}}$ is the temperature gradient, $T$ is the temperature and $x$ is the distance.
Rewrite the above equation, we have,
$\dfrac{{dQ}}{{Adt}} = - K\dfrac{{dT}}{{dx}}$ ,
Substitute the given values in the above equation, we get,
$
10 = - 0.4\dfrac{{dT}}{{dx}} \\
\Rightarrow \dfrac{{dT}}{{dx}} = - \dfrac{{10}}{{0.4}} \\
\therefore \dfrac{{dT}}{{dx}} = - 25^\circ C $
Here, a negative sign indicates that the fall in the temperature of the material with respect to distance.

Thus, the thermal gradient of the material is $25^\circ {\text{C/cm}}$ and option B is correct.

Note:In the formula of rate of flow of heat, the negative sign indicates the decrement of the temperature of the body along the positive x direction and $dt$ is expressed as${T_2} - {T_1}$ , where ${T_2}$ is final temperature and ${T_1}$ is initial temperature. The unit of thermal gradient in SI unit system is $J{s^{ - 1}}{m^{ - 1}}{K^{ - 1}}$ and in CI unit system is $cal \cdot {s^{ - 1}}c{m^{ - 1}}^\circ {C^{ - 1}}$ .