A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th

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e length and the diameter of this bar are doubled, the rate at which it will conduct heat between these reservoirs will be

1 answer:

serg [3.5K] · Answer 1 · 2026-02-27T12:12:20+03:00

Answer:

The heat transfer rate using the new cylinder would amount to 50 W

Explanation:

Conduction is defined by Fourier's Law in this format: $Q=kA\frac{T_1-T_2}{L}$ , where $Q$ represents the conduction heat transfer rate, $k$ indicates the material's thermal conductivity, $T_1$ and $T_2$ denote the temperatures at the two heat sources, $A$ is the area through which heat flows, and ${L}$ is the distance the heat travels.
The expression for the original cylinder based on Fourier's law is: $kA_1\frac{T_1-T_2}{L_1}=25W$ , and should $A_1=\frac{\pi D_{1}^{2}}{4}$ hold, the formula becomes: $k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W$ , where $D_1$ refers to the diameter of the original cylinder, and ${L_1}$ signifies its length.
In the case of the new cylinder, applying Fourier's Law as with the first cylinder results in: $Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2}$ , where $Q_2$ signifies the heat transfer rate in this scenario, and $D_2$ and ${L_2$ are the new diameter and length.
Thus, $D_2=2D_1$ and $L_2=2L_1$ , replacing in the equation for $Q_2$ gives us: $Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}$ .
Rearranging offers us: $Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1})$ .
In the final statement of $Q_2$ , it can be observed that the term inside the brackets equals $Q_1$ , leading to: $Q_2=\frac{2^2}{2}(25W)=50W$ .
It is important to point out that the temperatures in both the hot and cold reservoirs remain constant.