Question

The two ends of a metal rod are maintained at temperatures $100^{\circ }C$ and $110^{\circ }C$. The rate of heat flow in the rod is found to be $4J{s}^{-1}$. If the ends are maintained at temperatures $200^{\circ }C$ and $210^{\circ }C$, the rate of heat flow will be

• A. $44.0J{s}^{-1}$
• B. $16.8J{s}^{-1}$
• C. $8.0J{s}^{-1}$
• D. $4.0J{s}^{-1}$

Answer 1

Answer: (D)

The two ends of a rod are maintained at temperatures $100^{\circ }C$ and $110^{\circ }C$.
Given, $\Delta T_1=110^{\circ }C-100^{\circ }C=10^{\circ }C$
$\frac{d{Q}_1}{dt}=4J{s}^{-1}$
$\Delta T_2=210-200=10^{\circ }C$
$\frac{d{Q}_2}{dt}=?$
As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both cases is same, i.e. $10^{\circ }C$.
So, the same rate of heat will flow in the second case.
Hence, $\frac{d{Q}_2}{dt}=4J{s}^{-1}$
So, if the ends of the rod are maintained at temperatures $200^{\circ }C$ and $210^{\circ }C$, then the rate of heat flow will remain same,
i.e. $4J{s}^{-1}$.