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 the length and the diameter of this bar are doubled, the rate at which it
Will conduct heat between these reservoirs will be
A) 25 W
B) 12.5 W
C) 100 W
D) 200 W
E) 50 W

Question

Quizplus · Accepted Answer

The rate of heat conduction ($Q$) through a cylindrical bar is given by Fourier's law of heat conduction, $Q = \frac{kA(T_{hot} - T_{cold})}{L}$, where $k$ is the thermal conductivity, $A$ is the cross-sectional area, $T_{hot}$ and $T_{cold}$ are the temperatures of the hot and cold reservoirs, respectively, and $L$ is the length of the bar. Doubling both the diameter and the length of the bar quadruples the cross-sectional area ($A$) but also doubles the length ($L$). These changes result in the heat conduction rate being doubled ($2 \times 25 W = 50 W$).

A Solid Cylindrical Bar Conducts Heat at a Rate of 25