Question

The two ends of a metal rod are maintained at temperature ${100^ \circ }C$ and ${110^ \circ }C$. The rate of heat flow in the rod is found to be $4.0J/s$. If the ends are maintained at temperatures ${200^ \circ }C$ and ${210^ \circ }C$, the rate of heat flow will be
$
  (A){\text{ }}8.0J/s \\
  (B){\text{ 4}}.0J/s \\
  (C){\text{ 44}}.0J/s \\
  (D){\text{ 16}}.8J/s \\
 $

Answer 1

Hints: In this question, the concept of the heat transfer is used, that is the rate of heat flow is directly proportional to the temperature difference should be applied to solve this problem.

Complete Step by step solution:
The rate of heat flow is given as the amount of heat transferred per unit of time in some material. It is usually measured in Joules per second.
In the given question, it is mentioned that the two ends of a metal rod are maintained at temperature ${100^ \circ }C$ and ${110^ \circ }C$.
So, the difference in the temperature is given as, $\Delta {T_1} = 110 - 100 = {10^ \circ }C$
Now, according to the question, the rate of heat flow is given by $\dfrac{{d{Q_1}}}{{dt}} = 4.0J/s$
Now in the second case of the question, it is given as the ends are maintained at temperatures ${200^ \circ }C$ and ${210^ \circ }C$.
So, the difference in the temperature in the second case is given as, $\Delta {T_2} = 210 - 200 = {10^ \circ }C$
As the rate of heat flow is directly proportional to the temperature difference and the difference in temperature in both the cases $\Delta {T_1},\Delta {T_2}$ is the same that is ${10^ \circ }C$. So, the same rate of heat will flow in the second case and will also be the same.
Thus, we can say that $\dfrac{{d{Q_2}}}{{dt}} = 4.0J/s$.

So $4.0J/s$ is the correct answer that is option $(B)$.

Note:Now we calculate the heat transfer by using the alternative method for the above question as follows:
Given: $
  \dfrac{{\Delta {Q_A}}}{{\Delta {Q_B}}} = \dfrac{{\Delta {T_A}}}{{\Delta {T_B}}} \\
   \Rightarrow \dfrac{4}{x} = \dfrac{{10}}{{10}} \\
   \Rightarrow x = 4J/s \\
 $
So $\Delta {T_A} = 110 - 100 = {10^ \circ }C$ and $\Delta {T_B} = 210 - 200 = {10^ \circ }C$
We know, the rate of heat flow is directly proportional to the temperature difference.
Hence $\Delta {Q_A} = \Delta {T_B} = {T_3} - {T_4} = {10^ \circ }C$
Now,
\[
  \dfrac{{\Delta {Q_A}}}{{\Delta {Q_B}}} = \dfrac{{\Delta {T_A}}}{{\Delta {T_B}}} \\
   \Rightarrow \dfrac{4}{x} = \dfrac{{10}}{{10}} \\
   \Rightarrow x = 4J/s \\
 \]