Question 514: Thermodynamics & KTG

Question

Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivites of these rods are: $\frac{\mathrm{L}_{\mathrm{A}}}{\mathrm{L}_{\mathrm{B}}}=\frac{1}{2}, \frac{\mathrm{r}_{\mathrm{A}}}{\mathrm{r}_{\mathrm{B}}}=2$ and $\frac{\mathrm{K}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{B}}}=\frac{1}{2}$. The free ends of rods A and B are maintained at 400 K , 200 K , respectively. The temperature of rods interface is ______________ $K$, when equilibrium is established.

Accepted Answer

$$\begin{aligned}
& \mathrm{R}_1=\frac{\ell_1}{\mathrm{~K}_1 \mathrm{~A}_1}, \mathrm{R}_2=\frac{\ell_2}{\mathrm{~K}_2 \mathrm{~A}_2} \
& \frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\Delta \mathrm{T}}{\mathrm{R}} \
& \left(\frac{\mathrm{dQ}}{\mathrm{dt}}ight)_1=\left(\frac{\mathrm{dQ}}{\mu}ight)_2 \
& \frac{400-\mathrm{T}}{\mathrm{R}_1}=\frac{\mathrm{T}-200}{\mathrm{R}_2} \
& \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left(\frac{\ell_1}{\ell_2}ight)\left(\frac{\mathrm{r}_2}{\mathrm{r}_1}ight)^2 	imes \frac{\mathrm{K}_2}{\mathrm{~K}_1} \
& =\frac{1}{2} 	imes\left(\frac{1}{2}ight)^2 	imes 2 \
& =\left(\frac{1}{4}ight)
\end{aligned}$$
$$\begin{aligned}
& \frac{400-\mathrm{T}}{\mathrm{~T}-200}=\frac{1}{4} \
& 1600-4 \mathrm{~T}=\mathrm{T}-200 \
& 5 \mathrm{~T}=1800 \
& \mathrm{~T}=360 \mathrm{~K}
\end{aligned}$$

Thermodynamics & KTG

Integer / Numeric Answer Type