A cylindrical bar that us well insulated around its sides connects hot and cold reservoirs and conducts heat at a rate of 10.0 J/s under steady state conditions.If all of its linear dimensions (diameter and length)are reduced by half,the rate at which it will now conduct heat between the same reservoirs is closest to
A) 80.0 J/s.
B) 20.0 J/s.
C) 5.00 J/s.
D) 2.50 J/s.
E) 1.25 J/s.

Question

Quizplus · Accepted Answer

The rate of heat conduction is given by Fourier's law: $Q = \frac{kA\Delta T}{L}$, where $Q$ is the heat transfer per unit time, $k$ is the thermal conductivity, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference, and $L$ is the length of the material. When both the diameter and length are reduced by half, the new area $A'$ becomes $\frac{1}{4}$ of the original (since area is proportional to the square of the diameter), and the new length $L'$ becomes $\frac{1}{2}$ of the original. Substituting these into the equation gives a new rate of heat conduction that is $\frac{1}{2}$ of the original, because the reduction in area (which decreases the rate) is partially offset by the reduction in length (which increases the rate). Therefore, the new rate is $10.0 \, \text{J/s} \times \frac{1}{2} = 5.00 \, \text{J/s}$.

A Cylindrical Bar That Us Well Insulated Around Its Sides