Cylindrical rod of copper of length $2 \mathrm{~m}$ and cross-sectional area $2 \mathrm{~cm}^2$ is insulated at its curved surface. The one end of rod is maintained in steam chamber and other is maintained in ice at $0^{\circ} \mathrm{C}$. The thermal conductivity of copper is $386…

Question

Cylindrical rod of copper of length $2 \mathrm{~m}$ and cross-sectional area $2 \mathrm{~cm}^2$ is insulated at its curved surface. The one end of rod is maintained in steam chamber and other is maintained in ice at $0^{\circ} \mathrm{C}$. The thermal conductivity of copper is $386 \mathrm{Js}^{-1} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ ). Find the temperature at a point which is at a distance of $120 \mathrm{~cm}$ from the colder end.

Aicharya · Accepted Answer

For cylindrical wire, $ \begin{aligned} l & =2 \mathrm{~m}, A=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2 \\ \theta_1 & =100^{\circ} \mathrm{C}, \theta_2=0^{\circ} \mathrm{C} \\ \theta & =? \end{aligned} $ At steady state, temperature gradient of rod remains constant.

$ \begin{array}{ll} \text { i.e., } & \frac{\theta_1-\theta}{x_1}=\frac{\theta-\theta_2}{x_2} \\ \Rightarrow & \frac{100^{\circ}-\theta}{80 \times 10^{-2}}=\frac{\theta-0^{\circ}}{120 \times 10^{-2}} \\ \Rightarrow & 100^{\circ}-\theta=\frac{80}{120} \times \theta \\ \Rightarrow & 100^{\circ}-\theta=\frac{2}{3} \theta \Rightarrow \frac{5 \theta}{3}=100^{\circ} \\ \Rightarrow & \theta=\frac{100^{\circ} \times 3}{5}=60^{\circ} \mathrm{C} \end{array} $

Cylindrical rod of copper of length $2 \mathrm{~m}$ and cross-sectional area $2 \mathrm{~cm}^2$ is insulated…

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