Question

An aluminium meter rod of area of cross section $4\;{\rm{c}}{{\rm{m}}^{\rm{2}}}$ with $K = 0.5\;{\rm{cal}} \cdot {{\rm{g}}^{ - 1}}{C^{ - 1}}$ is observed that at steady state $360\;{\rm{cal}}$ of heat flows per minute. The temperature gradient along the rod is
A $3\;^\circ {{\rm{C}} \mathord{\left/ {\vphantom {{\rm{C}} {{\rm{cm}}}}} \right. } {{\rm{cm}}}}$
B $6\;^\circ {{\rm{C}} \mathord{\left/ {\vphantom {{\rm{C}} {{\rm{cm}}}}} \right. } {{\rm{cm}}}}$
C $12\;^\circ {{\rm{C}} \mathord{\left/ {\vphantom {{\rm{C}} {{\rm{cm}}}}} \right. } {{\rm{cm}}}}$
D $20\;^\circ {{\rm{C}} \mathord{\left/ {\vphantom {{\rm{C}} {{\rm{cm}}}}} \right. } {{\rm{cm}}}}$

Answer 1

Thus question is based on the steady state process. WE have to know the concept of a steady state process. The steady state process occurs when there is a transfer of heat. The heat is transferred throughout the aluminum rod. The expression of state process we can calculate the temperature gradient. The temperature gradient can be written as $\dfrac{{dT}}{{dx}}$

Complete step by step answer:
Given: The cross sectional area $4\;{\rm{c}}{{\rm{m}}^{\rm{2}}}$ and the thermal conductivity is $K = 0.5\;{\rm{cal}} \cdot {{\rm{g}}^{ - 1}}{C^{ - 1}}$.

In aluminum rods, the heat flow remains the same throughout the pipe, as we have given that there is steady state heat flow per minute. A steady state flow process is the process in which conditions at all points in a rod remains constant with the change in time. The same mass flow rate will remain throughout the cross sectional area.
Now, we have to find the temperature gradient that is $\dfrac{{dt}}{{dx}}$
Now, we can use the expression of steady state heat flow we get,
$Q = kA\left( {\dfrac{{dT}}{{dx}}} \right)$
Now, substitute the values in above equation we get,
$\dfrac{{360\;{{{\rm{cal}}} \mathord{\left/ {\vphantom {{{\rm{cal}}} {{\rm{min}}}}} \right. } {{\rm{min}}}}}}{{60\;s}} \times \min = 0.5\;{\rm{cal}} \cdot {{\rm{g}}^{ - 1}}{C^{ - 1}} \times 4\;{\rm{c}}{{\rm{m}}^{\rm{2}}}\left( {\dfrac{{dT}}{{dx}}} \right)\\\ \implies \dfrac{{dT}}{{dx}} = \dfrac{{60}}{{20}}\\\ \implies \dfrac{{dT}}{{dx}} = 3\;^\circ {{\rm{C}} \mathord{\left/ {\vphantom {{\rm{C}} {{\rm{cm}}}}} \right. } {{\rm{cm}}}}\;$
Therefore, the temperature gradient is $3\;^\circ {{\rm{C}} \mathord{\left/ {\vphantom {{\rm{C}} {{\rm{cm}}}}} \right. } {{\rm{cm}}}}\;$.

Thus, the correct option is (A).

Note:
In this question, students must have the knowledge of steady state processes. In a steady state process, the parameters may have different values at different positions, but the value remains the same. In unsteady state heat conditions there is no system that exists initially. If the temperature of the system reaches the temperature of the heat source, then the system will be at steady state.