First slide

Conduction

Question

One end of a thermally insulated rod is kept at a temperature T₁ and the other at T₂. The rod is composed of two sections of lengths l₁ and l₂ and thermal conductivities k₁ and k₂ respectively. The temperature at the interface of the two sections is

Easy

A

$(k_{1} l_{2} T_{1} + k_{2} l_{1} T_{2}) / (k_{1} l_{2} + k_{2} l_{1})$
B

$(k_{1} l_{1} T_{1} + k_{2} l_{2} T_{2}) / (k_{1} l_{1} + k_{2} l_{2})$
C

$(k_{2} l_{2} T_{1} + k_{1} l_{1} T_{2}) / (k_{1} l_{1} + k_{2} l_{2})$
D

$(k_{2} l_{1} T_{1} + k_{1} l_{2} T_{2}) / (k_{2} l_{1} + k_{1} l_{2})$

Solution

At steady state, the rate of flow of heat through first slab = rate of flow of heat through second slab.
$\frac{k_{1} A [T_{1} - T]}{l_{1}} = \frac{k_{2} A [T - T_{2}]}{l_{2}} . . . . (1)$
where 7 = temperature of junction
From eq. (1)
$k_{1} l_{2} T_{1} - k_{1} l_{2} T = k_{2} l_{1} T - k_{2} l_{1} T_{2}$
or $k_{1} l_{2} T_{1} + k_{2} l_{1} T_{2} = k_{2} l_{1} T + k_{1} l_{2} T$
Solving we get
$T = \frac{k_{1} l_{2} T_{1} + k_{2} l_{1} T_{2}}{k_{1} l_{2} + k_{2} l_{1}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App