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Nana76
3 months ago
12

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

e length and the diameter of this bar are doubled, the rate at which it will conduct heat between these reservoirs will be
Physics
1 answer:
serg [3.5K]3 months ago
7 0

Answer:

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

Explanation:

  1. 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.
  2. 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.
  3. 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.
  4. 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}.
  5. Rearranging offers us: Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1}).
  6. 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.
  7. It is important to point out that the temperatures in both the hot and cold reservoirs remain constant.
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Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

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As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

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C. At r = a

Following from equation 1:

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