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?
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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 $2Q$!
Step 1: Understanding the Question: Determine how the heat current through a conductor changes when all its linear dimensions are doubled. Step 2: Key Formula or Approach: Thermal resistance R_th = L/(KA). Heat current Q = Δθ/R_th. Scaling length by 2 and cross-sectional area by 2² = 4 changes the resistance by (2/4) = 1/2. Step 3: Detailed Explanation: Doubling all linear dimensions increases the conduction path length by a factor of 2, which alone would double the resistance. However, the cross-sectional area grows by 2² = 4, which reduces resistance fourfold. The net thermal resistance becomes (2/4) = ½ of the original. Since heat flow is inversely proportional to resistance, halving R_th doubles the heat current to 2Q. Step 4: Final Answer: The heat current doubles to 2Q.
