rod can be expressed using Fourier's law of heat conduction:

$$ Q = \frac{k \cdot A \cdot (\theta_1 - \theta_2)}{L} $$

where:

When the linear dimensions of the rod are doubled, both the length and the cross-sectional area will change. Let's look at each dimension:

If the original length is $$L$$, the new length will be $$2L$$.
If the original cross-sectional area is $$A$$, the new cross-sectional area will be $$4A$$ (since doubling the diameter of a circle quadruples its area).

Substituting the new dimensions into the formula, we get the new rate of heat flow, $$Q'$$:

$$ Q' = \frac{k \cdot 4A \cdot (\theta_1 - \theta_2)}{2L} $$

Simplifying the equation:

$$ Q' = \frac{4 \cdot k \cdot A \cdot (\theta_1 - \theta_2)}{2L} $$

We can further simplify this to:

$$ Q' = 2 \cdot \frac{k \cdot A \cdot (\theta_1 - \theta_2)}{L} $$

Since $$\frac{k \cdot A \cdot (\theta_1 - \theta_2)}{L}$$ is the original rate of heat flow $$Q$$, we have:

$$ Q' = 2Q $$

Therefore, the new rate of flow of heat is twice the original rate of flow. The correct answer is:

Option C $$2 \mathrm{Q}$$

Heat And Thermodynamics

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