Two rods of same length and cross-section are | vedclass

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

Question

(C) At steady state,the rate of heat flow $(H)$ through both rods must be equal.The formula for the rate of heat flow is $H = \frac{KA(\Delta T)}{l}$.

Vedclass · Accepted Answer

$\frac{K_1 	heta_1 + K_2 	heta_2}{K_1 + K_2}$

Vedclass · Answer

(C) At steady state,the rate of heat flow $(H)$ through both rods must be equal.The formula for the rate of heat flow is $H = \frac{KA(\Delta T)}{l}$.Let $	heta$ be the temperature of the common junction. Since the rods are joined in series,the heat current flowing through the first rod equals the heat current flowing through the second rod:$\frac{K_1 A (	heta_1 - 	heta)}{l} = \frac{K_2 A (	heta - 	heta_2)}{l}$Since the lengths $(l)$ and cross-sectional areas $(A)$ are the same,they cancel out:$K_1(	heta_1 - 	heta) = K_2(	heta - 	heta_2)$$K_1 	heta_1 - K_1 	heta = K_2 	heta - K_2 	heta_2$$K_1 	heta_1 + K_2 	heta_2 = 	heta(K_1 + K_2)$$	heta = \frac{K_1 	heta_1 + K_2 	heta_2}{K_1 + K_2}$

Similar Questions

The thermal conductivity of a wire is $1.7 \ W \ m^{-1} \ K^{-1}$ and that of cement is $2.9 \ W \ m^{-1} \ K^{-1}$. The thickness of the cement insulation is $..... \ cm$. Here,the thickness of the wire is $20 \ cm$.

The area of the glass of a window of a room is $10\;m^2$ and thickness is $2\;mm$. The outer and inner temperatures are $40^{\circ}C$ and $20^{\circ}C$ respectively. The thermal conductivity of glass in the $MKS$ system is $0.2\;W/(m\cdot K)$. The heat flowing into the room per second will be:

There are two identical vessels filled with equal amounts of ice. The vessels are made of different metals. If the ice melts in the two vessels in $20$ and $35$ minutes respectively,the ratio of the coefficients of thermal conductivity of the two metals is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module