A brass rod with a length of $20.0 \mathrm{~cm}$ is placed side by side with an aluminum rod with a length of $20.0 \mathrm{~cm}$, and this system is placed between a hot temperature of $150^{\circ} \mathrm{C}$ and a cold temperature of $-10.0^{\circ} \mathrm{C}$. The thermal conductivities of the brass and the aluminum are $100 \mathrm{~W} / \mathrm{m}{ }^{\circ} \mathrm{C}$ and 230 $\mathrm{W} / \mathrm{m}^{\circ} \mathrm{C}$, respectively. The rods have the same cross-sectional area of $20.0 \mathrm{~cm}^{2}$. What is the rate of heat flow from the hot temperature to the cold temperature?

Question

Quizplus · Accepted Answer

The rate of heat flow ($Q/t$) through a material is given by the formula $Q/t = kA(T_{hot} - T_{cold})/d$, where $k$ is the thermal conductivity, $A$ is the cross-sectional area, $T_{hot}$ and $T_{cold}$ are the temperatures at the hot and cold ends, respectively, and $d$ is the thickness or length of the material. For both rods placed side by side, the effective thermal conductivity ($k_{eff}$) is the harmonic mean of their thermal conductivities because they are in parallel: $1/k_{eff} = (1/k_{brass} + 1/k_{aluminum})/2$. However, since they have the same cross-sectional area and are subjected to the same temperature difference, we can calculate the total heat flow by adding their individual heat flows. The total heat flow is $Q/t = (k_{brass}A(T_{hot} - T_{cold})/d) + (k_{aluminum}A(T_{hot} - T_{cold})/d)$. Substituting the given values, $A = 20.0 \times 10^{-4} \mathrm{m}^2$, $T_{hot} - T_{cold} = 160^{\circ}C$, $d = 0.2 \mathrm{m}$, $k_{brass} = 100 \mathrm{W/m}^{\circ}C$, and $k_{aluminum} = 230 \mathrm{W/m}^{\circ}C$, we find the total rate of heat flow to be $528 \mathrm{W}$.

A Brass Rod with a Length Of 20.0 cm20.0 \mathrm{~cm}20.0 cm Is Placed Side by Side with an Aluminum Rod with Placed

A Brass Rod with a Length Of $20.0 \mathrm{~cm}$ Is Placed Side by Side with an Aluminum Rod with Placed