Two rectangular blocks A and B of different m | vedclass

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(A) In the steady state,the rate of heat flow through block $A$ must be equal to the rate of heat flow through block $B$.Let $K_1$ and $K_2$ be the th

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) In the steady state,the rate of heat flow through block $A$ must be equal to the rate of heat flow through block $B$.Let $K_1$ and $K_2$ be the thermal conductivities of blocks $A$ and $B$ respectively,$L$ be their length,and $A$ be their cross-sectional area.Given: $\frac{K_1}{K_2} = \frac{1}{3}$,so $K_2 = 3K_1$.Let $	heta$ be the temperature of the junction.The rate of heat flow is given by $H = \frac{KA(\Delta T)}{L}$.Equating the heat flow rates: $\frac{K_1 A (100 - 	heta)}{L} = \frac{K_2 A (	heta - 0)}{L}$.Since $A$ and $L$ are the same for both,we have $K_1(100 - 	heta) = K_2	heta$.Substituting $K_2 = 3K_1$: $K_1(100 - 	heta) = 3K_1	heta$.$100 - 	heta = 3	heta$.$100 = 4	heta$.$	heta = 25^{\circ}C$.

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