a steel rod of diameter $d_1 = 1.0$ cm and $l_1=25.0$ cm and an aluminum rod of diameter $d_2 = 1.2$ cm and…

Explanation:

Step1: Recall heat - conduction formula

The heat - conduction formula is $\Phi=kA\frac{\Delta T}{L}$, where $\Phi$ is the heat - conduction rate, $k$ is the thermal conductivity, $A$ is the cross - sectional area, $\Delta T$ is the temperature difference, and $L$ is the length of the rod. The cross - sectional area of a circular rod is $A = \frac{\pi d^{2}}{4}$.

Step2: Set up the heat - conduction equation for the composite rod

Since the heat - conduction rate through the steel part $\Phi_1$ and the aluminum part $\Phi_2$ is the same in the steady - state ($\Phi_1=\Phi_2$), we have $k_1A_1\frac{T_h - T}{L_1}=k_2A_2\frac{T - T_c}{L_2}$, where $T_h = 100^{\circ}C$ is the temperature of the hot end (boiling water), $T_c = 20^{\circ}C$ is the temperature of the cold end, $T$ is the temperature at the joint, $k_1$ and $k_2$ are the thermal conductivities of steel and aluminum respectively, $A_1=\frac{\pi d_1^{2}}{4}$ and $A_2=\frac{\pi d_2^{2}}{4}$ are the cross - sectional areas of steel and aluminum rods, and $L_1$ and $L_2$ are their lengths. Substituting $A_1$ and $A_2$ into the equation and simplifying, we get $k_1d_1^{2}\frac{T_h - T}{L_1}=k_2d_2^{2}\frac{T - T_c}{L_2}$. Expanding this gives $k_1d_1^{2}T_h-k_1d_1^{2}T=k_2d_2^{2}T - k_2d_2^{2}T_c$. Rearranging terms to solve for $T$: [ \begin{align*} T\left(k_1\frac{d_1^{2}}{L_1}+k_2\frac{d_2^{2}}{L_2}\right)&=k_1\frac{d_1^{2}}{L_1}T_h + k_2\frac{d_2^{2}}{L_2}T_c\ T&=\frac{k_1\frac{d_1^{2}}{L_1}T_h + k_2\frac{d_2^{2}}{L_2}T_c}{k_1\frac{d_1^{2}}{L_1}+k_2\frac{d_2^{2}}{L_2}} \end{align*} ] Given $k_1 = 14.0\ W/(m\cdot K)$, $d_1 = 1.0\ cm=0.01\ m$, $L_1 = 25.0\ cm = 0.25\ m$, $k_2 = 235\ W/(m\cdot K)$, $d_2 = 1.2\ cm=0.012\ m$, $L_2 = 18.0\ cm = 0.18\ m$, $T_h = 100^{\circ}C$, and $T_c = 20^{\circ}C$. [ \begin{align*} \frac{k_1d_1^{2}}{L_1}&=\frac{14\times(0.01)^{2}}{0.25}=5.6\times 10^{-3}\ W/(m\cdot K)\ \frac{k_2d_2^{2}}{L_2}&=\frac{235\times(0.012)^{2}}{0.18}=1.88\ W/(m\cdot K)\ \frac{k_1d_1^{2}}{L_1}T_h&=5.6\times 10^{-3}\times100 = 0.56\ W\ \frac{k_2d_2^{2}}{L_2}T_c&=1.88\times20 = 37.6\ W \end{align*} ] [ \begin{align*} T&=\frac{0.56 + 37.6}{5.6\times 10^{-3}+1.88}\ &=\frac{38.16}{1.8856}\ &\approx20.2^{\circ}C \end{align*} ] To find the heat - conduction rate $\Phi$, we can use either $\Phi = k_1A_1\frac{T_h - T}{L_1}$ or $\Phi = k_2A_2\frac{T - T_c}{L_2}$. Using $\Phi = k_1A_1\frac{T_h - T}{L_1}$, $A_1=\frac{\pi d_1^{2}}{4}=\frac{\pi\times(0.01)^{2}}{4}=7.85\times 10^{-5}\ m^{2}$ [ \begin{align*} \Phi&=14\times7.85\times 10^{-5}\times\frac{100 - 20.2}{0.25}\ &=14\times7.85\times 10^{-5}\times\frac{79.8}{0.25}\ &=14\times7.85\times 10^{-5}\times319.2\ &\approx0.35\ W \end{align*} ]

Answer:

The temperature at the joint is approximately $20.2^{\circ}C$ and the heat - conduction rate is approximately $0.35\ W$.