Two rods with thermal conductivities K and 3K | vedclass

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

Question

(C) In a steady state,the rate of heat flow through both rods must be equal.Let $A$ be the cross-sectional area,$L_1 = 1 \ cm$,$L_2 = 2 \ cm$,$K_1 = K

Vedclass · Accepted Answer

$60^{\circ}C$

Vedclass · Answer

(C) In a steady state,the rate of heat flow through both rods must be equal.Let $A$ be the cross-sectional area,$L_1 = 1 \ cm$,$L_2 = 2 \ cm$,$K_1 = K$,$K_2 = 3K$,$T_1 = 0^{\circ}C$,$T_2 = 100^{\circ}C$,and $\phi$ be the interface temperature.The rate of heat flow is given by $H = \frac{KA(T_{high} - T_{low})}{L}$.For the first rod: $H_1 = \frac{K \cdot A \cdot (\phi - 0)}{1}$.For the second rod: $H_2 = \frac{3K \cdot A \cdot (100 - \phi)}{2}$.Since $H_1 = H_2$,we have:$K \cdot A \cdot \phi = \frac{3K \cdot A \cdot (100 - \phi)}{2}$$\phi = \frac{3(100 - \phi)}{2}$$2\phi = 300 - 3\phi$$5\phi = 300$$\phi = 60^{\circ}C$.

Similar Questions

On a cold morning,a metal surface will feel colder to touch than a wooden surface because

$A$ cylindrical rod has temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is $Q \ J s^{-1}$. All the linear dimensions of the rod are doubled while keeping the temperatures constant. What is the new rate of flow of heat?

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

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module