Question

A cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q\ \text{J}\ \text{s}^{-1}$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. What is the new rate of flow of heat?

• A. $\frac{Q}{2}$
• B. $\frac{Q}{4}$
• C. $2Q$
• D. $\frac{3Q}{2}$

Hint:

For scaling transformations in conduction, think of it using thermal resistance: $R_{\text{th}} = \frac{L}{KA}$. Doubling all linear dimensions increases the length by 2, but increases the area by $2^2 = 4$. This causes the thermal resistance to drop by half ($R_{\text{th}}' = \frac{2}{4}R_{\text{th}} = \frac{1}{2}R_{\text{th}}$). Since heat current is inversely proportional to resistance ($Q = \frac{\Delta \theta}{R_{\text{th}}}$), halving the resistance automatically doubles the heat flow rate to $2Q$!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A solid cylindrical metal rod conducts heat between two fixed thermal reservoirs maintained at temperatures $\theta_1$ and $\theta_2$. Given an initial steady-state heat current $Q$, we need to calculate the new rate of heat flow when every single linear dimension of the rod (its length and radius) is exactly doubled.

Step 2: Key Formula or Approach:
According to Fourier's Law of Heat Conduction, the rate of heat transfer ($Q$) along a conductor of cross-sectional area $A$ and length $L$ is given by: $$Q = \frac{KA(\theta_1 - \theta_2)}{L}$$ Where $K$ is the thermal conductivity of the material. For a cylinder, the cross-sectional area is $A = \pi r^2$. Substituting this into the formula gives: $$Q = \frac{K(\pi r^2)(\theta_1 - \theta_2)}{L} \implies Q \propto \frac{r^2}{L}$$

Step 3: Detailed Explanation:
Let's analyze the initial configuration: $$Q_1 = Q \propto \frac{r^2}{L}$$ The problem states that all linear dimensions are doubled. This means: The new radius becomes $r' = 2r$ The new length becomes $L' = 2L$ Let's write down the proportional expression for the modified heat flow rate $Q_2$: $$Q_2 \propto \frac{(r')^2}{L'} \implies Q_2 \propto \frac{(2r)^2}{2L}$$ $$Q_2 \propto \frac{4r^2}{2L} = 2 \left(\frac{r^2}{L}\right)$$ Setting up a direct ratio comparison between $Q_2$ and $Q_1$: $$\frac{Q_2}{Q_1} = \frac{2 \left(\frac{r^2}{L}\right)}{\frac{r^2}{L}} = 2$$ $$Q_2 = 2Q_1 = 2Q$$ Therefore, the rate of heat conduction is exactly doubled.

Step 4: Final Answer:
The new rate of heat flow is $2Q$, which corresponds to option (C).