A wall has two layers A and B made of different materials. The thickness of both the layers is the same. The thermal conductivity of A and B are K A and K B such that K A = 3K B . The temperature across the wall is 20°C. In thermal equilibrium
Text Solution
Verified by ExpertsB
In series rate of flow of heat is same
$\Rightarrow \frac{K_A A (\theta_1 - \theta)}{l} = \frac{K_B A (\theta - \theta_2)}{l}$
⇒ ⇒ $3K_{\mathrm{B}}(\theta_{1} - \theta) = K_{\mathrm{B}}(\theta - \theta_{2})$
⇒ ⇒ $3(\theta_1 - \theta) = (\theta - \theta_2)$
⇒ ⇒ $3\theta_{1} - 3\theta = \theta - \theta_{2}$ ⇒ ⇒ $4\theta_{1} - 4\theta_{0} = \theta_{1} - \theta_{2}$
⇒ ⇒ $4(\theta_1 - \theta) = (\theta_1 - \theta_2)$
⇒ ⇒ $4(\theta_1 - \theta) = 20$ ⇒ ⇒ $(\theta_1 - \theta) = 5^\circ \mathrm{C}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems