A rod of length L with sides fully insulated | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The heat current $H$ through a rod of cross-sectional area $A$ is given by $H = -KA \frac{dT}{dx}$.Since the rod is insulated,$H$ is constant alon

Vedclass · Accepted Answer

$T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}}$

Vedclass · Answer

(A) The heat current $H$ through a rod of cross-sectional area $A$ is given by $H = -KA \frac{dT}{dx}$.Since the rod is insulated,$H$ is constant along the length.Substituting $K = \frac{\alpha}{T}$,we get $H = -\frac{\alpha A}{T} \frac{dT}{dx}$.Rearranging the terms: $\frac{H}{\alpha A} dx = -\frac{dT}{T}$.Integrating from $x=0$ to $x$ and $T=T_1$ to $T$:$\int_0^x \frac{H}{\alpha A} dx = -\int_{T_1}^T \frac{dT}{T}$.$\frac{H}{\alpha A} x = -(\ln T - \ln T_1) = \ln \frac{T_1}{T}$.At $x=L$,$T=T_2$,so $\frac{H}{\alpha A} L = \ln \frac{T_1}{T_2}$.Thus,$\frac{H}{\alpha A} = \frac{1}{L} \ln \frac{T_1}{T_2}$.Substituting this back: $\frac{x}{L} \ln \frac{T_1}{T_2} = \ln \frac{T_1}{T}$.$\ln \left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \ln \frac{T_1}{T}$.Taking the exponential of both sides: $\left( \frac{T_1}{T_2} \right)^{\frac{x}{L}} = \frac{T_1}{T}$.Therefore,$T = T_1 \left( \frac{T_2}{T_1} \right)^{\frac{x}{L}}$.

Similar Questions

Three rods made of the same material and having the same cross-sectional area but different lengths $10\, cm, 20\, cm$,and $30\, cm$ are joined as shown. The temperature of the junction is ......... $^oC$.

Three rods of the same dimensions have thermal conductivities $3K, 2K$, and $K$. They are arranged as shown in the figure. The temperature of the junction in the steady state is:

For a spherical shell with inner and outer radii $r_1$ and $r_2$ respectively,the formula for the heat flow rate is = ......... where $T_1$ and $T_2$ are the temperatures of the inner and outer surfaces of the shell $(T_1 > T_2)$. The thermal conductivity of the material of the shell is $k$.

Two plates of equal area are placed in series. The ratios of their thicknesses and thermal conductivities are both $2:3$. The temperature of the outer surface of one plate is $100 ^\circ C$ and that of the other is $0 ^\circ C$. The temperature of the common surface is ....... $^\circ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module