Two plates of the same area are placed in con | vedclass

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

Question

(C) Given that two plates have the same surface area $A$.Let $K_1, K_2$ be their thermal conductivities and $t_1, t_2$ be their thicknesses respective

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Given that two plates have the same surface area $A$.Let $K_1, K_2$ be their thermal conductivities and $t_1, t_2$ be their thicknesses respectively.According to the problem,the ratios are:$\frac{K_1}{K_2} = \frac{2}{3}$ and $\frac{t_1}{t_2} = \frac{2}{3}$.Since the plates are in series,the rate of heat flow through both plates must be the same in steady state:$\frac{dQ}{dt} = \frac{K_1 A (T_1 - T)}{t_1} = \frac{K_2 A (T - T_2)}{t_2}$Where $T$ is the temperature of the common surface.Substituting the given values $T_1 = 10^{\circ} C$ and $T_2 = 0^{\circ} C$:$\frac{K_1}{t_1} (10 - T) = \frac{K_2}{t_2} (T - 0)$$\frac{K_1}{K_2} \cdot \frac{t_2}{t_1} (10 - T) = T$Substituting the ratios $\frac{K_1}{K_2} = \frac{2}{3}$ and $\frac{t_2}{t_1} = \frac{3}{2}$:$\left( \frac{2}{3} \right) \cdot \left( \frac{3}{2} \right) (10 - T) = T$$1 \cdot (10 - T) = T$$10 - T = T$$2T = 10$$T = 5^{\circ} C$

Two plates of the same area are placed in contact. Their thicknesses as well as their thermal conductivities are in the ratio $2:3$. The outer surface of one plate is maintained at $10^{\circ} C$ and that of the other at $0^{\circ} C$. The temperature at the common surface is (in $^{\circ} C$)

Similar Questions

Why do some cooking vessels have a copper coating at the bottom?

The temperature difference between two sides of an iron plate,$1.8 \ cm$ thick is $9^{\circ} C$. Heat is transmitted through the plate at a rate of $10 \ kcal / (s \cdot m^2)$ at steady state. The thermal conductivity of iron is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module