A brass rod with a length of $30.0 \mathrm{~cm}$ is placed side by side with an aluminum rod with a length of $30.0 \mathrm{~cm}$, and this system is placed between a hot temperature of $100^{\circ} \mathrm{C}$ 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 brass rod has a cross-sectional area of $20.0 \mathrm{~cm}^{2}$, and the aluminum rod has a cross-sectional area of 30.0 $\mathrm{cm}^{2}$. 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 each rod can be calculated using the formula $Q/t = kA(T_h - T_c)/L$. For the brass rod, $Q/t = (100 \, \mathrm{W/m^{\circ}C})(0.002 \, \mathrm{m^2})(90 \, \mathrm{^{\circ}C})/(0.3 \, \mathrm{m}) = 60 \, \mathrm{W}$. For the aluminum rod, $Q/t = (230 \, \mathrm{W/m^{\circ}C})(0.003 \, \mathrm{m^2})(90 \, \mathrm{^{\circ}C})/(0.3 \, \mathrm{m}) = 207 \, \mathrm{W}$. The total rate of heat flow is the sum of the two, $60 \, \mathrm{W} + 207 \, \mathrm{W} = 267 \, \mathrm{W}$.

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A Brass Rod with a Length Of $30.0 \mathrm{~cm}$ Is Placed Side by Side with an Aluminum Rod with Placed