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(C) In steady state,the rate of heat flow through each rod is the same:$\left(\frac{dQ}{dt}\right)_1 = \left(\frac{dQ}{dt}\right)_2 = \left(\frac{dQ}{

Vedclass · Accepted Answer

$5/3$

Vedclass · Answer

(C) In steady state,the rate of heat flow through each rod is the same:$\left(\frac{dQ}{dt}\right)_1 = \left(\frac{dQ}{dt}\right)_2 = \left(\frac{dQ}{dt}\right)_3$Using the formula for heat conduction $\frac{dQ}{dt} = \frac{KA(T_H - T_L)}{L}$,we have:$\frac{2KA(3T - T_1)}{L} = \frac{KA(T_1 - T_2)}{L} = \frac{2KA(T_2 - T)}{L}$Simplifying by dividing by $KA/L$:$2(3T - T_1) = (T_1 - T_2) = 2(T_2 - T)$From the first equality:$6T - 2T_1 = T_1 - T_2 \Rightarrow 3T_1 - T_2 = 6T$  --- $(1)$From the second equality:$T_1 - T_2 = 2T_2 - 2T \Rightarrow T_1 - 3T_2 = -2T$ --- $(2)$Multiply equation $(2)$ by $3$:$3T_1 - 9T_2 = -6T$ --- $(3)$Subtracting $(3)$ from $(1)$:$(3T_1 - T_2) - (3T_1 - 9T_2) = 6T - (-6T)$$8T_2 = 12T \Rightarrow T_2 = 1.5T$Substitute $T_2$ into $(2)$:$T_1 - 3(1.5T) = -2T \Rightarrow T_1 - 4.5T = -2T \Rightarrow T_1 = 2.5T$Therefore,the ratio $T_1 / T_2 = 2.5T / 1.5T = 5/3$.

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