Two plates of thermal conductivities k_1 and | vedclass

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(C) When two slabs of the same thickness $l$ but different cross-sectional areas $A_1$ and $A_2$ are connected in parallel,the total rate of heat flow

Vedclass · Accepted Answer

$\frac{k_1 A_1 + k_2 A_2}{A_1 + A_2}$

Vedclass · Answer

(C) When two slabs of the same thickness $l$ but different cross-sectional areas $A_1$ and $A_2$ are connected in parallel,the total rate of heat flow $\frac{dQ}{dt}$ is the sum of the heat flow rates through each slab.The rate of heat flow through a slab is given by $\frac{dQ}{dt} = k A \frac{(T_1 - T_2)}{l}$.Since the slabs are in parallel,the total heat flow rate is $\frac{dQ}{dt} = \frac{dQ_1}{dt} + \frac{dQ_2}{dt}$.Substituting the expressions for each:$k (A_1 + A_2) \frac{(T_1 - T_2)}{l} = k_1 A_1 \frac{(T_1 - T_2)}{l} + k_2 A_2 \frac{(T_1 - T_2)}{l}$.Canceling the common terms $\frac{(T_1 - T_2)}{l}$ from both sides:$k (A_1 + A_2) = k_1 A_1 + k_2 A_2$.Therefore,the equivalent thermal conductivity $k$ is:$k = \frac{k_1 A_1 + k_2 A_2}{A_1 + A_2}$.

Two plates of thermal conductivities $k_1$ and $k_2$,cross-sectional areas $A_1$ and $A_2$,and the same thickness $l$ are joined as shown in the figure. The equivalent thermal conductivity $k$ of the combination is:

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