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In-s [12.5K]
3 years 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:
expeople1 [14]3 years ago
7 0

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

  1. Conduction could be described by the Law of Fourierin the form: Q=kA\frac{T_1-T_2}{L} where Q is the rate of heat transferred  by conduction, k is the thermal conductivity of the material, T_1 and T_2 are the temperatures of each heat deposit, A is the cross area to the flow of heat, and {L} is the distance that the flow of heat has to go.
  2. For the original cylinder the Fourier's law would be: kA_1\frac{T_1-T_2}{L_1}=25W, and if A_1=\frac{\pi D_{1}^{2}}{4}, then the expression would be:k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W where D_1 is the diameter of the original cylinder, and {L_1} is the length of the original cylinder.
  3. For the new cylinder, in the same fashion that for the first, Fourier's Law would be: Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2},where Q_2 is the heat rate in the second case, D_2 and {L_2 are the new diameter and length.
  4. But, D_2=2D_1 and L_2=2L_1, substituting in the expression for Q_2: Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}.
  5. Rearranging: Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1}).
  6. In the last declaration of  Q_2, it could be noted that the expressión inside the parenthesis is actually  Q_1, then:  Q_2=\frac{2^2}{2}(25W)=50W.
  7. <u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>
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