Three rods each of length l and cross-section | vedclass

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(A) In a series combination of heat conductors,the rate of heat flow $(H)$ remains constant through each rod.$H = \frac{dQ}{dt} = \frac{kA(T_{high} -

Vedclass · Accepted Answer

$\frac{600}{7} {}^{\circ}C, \frac{400}{7} {}^{\circ}C$

Vedclass · Answer

(A) In a series combination of heat conductors,the rate of heat flow $(H)$ remains constant through each rod.$H = \frac{dQ}{dt} = \frac{kA(T_{high} - T_{low})}{l}$Since $A$ and $l$ are the same for all three rods,the heat current $H$ is proportional to $k \Delta T$.Let $H$ be the steady heat current. Then:$H = \frac{(2K)A(100 - T_1)}{l} = \frac{KA(T_1 - T_2)}{l} = \frac{(K/2)A(T_2 - 0)}{l}$Canceling $\frac{KA}{l}$ from all parts:$2(100 - T_1) = (T_1 - T_2) = 0.5 T_2$From the second and third parts:$T_1 - T_2 = 0.5 T_2 \Rightarrow T_1 = 1.5 T_2 = \frac{3}{2} T_2$From the first and second parts:$2(100 - T_1) = T_1 - T_2$$200 - 2T_1 = T_1 - T_2$$200 = 3T_1 - T_2$Substitute $T_1 = \frac{3}{2} T_2$ into the equation:$200 = 3(\frac{3}{2} T_2) - T_2$$200 = \frac{9}{2} T_2 - T_2 = \frac{7}{2} T_2$$T_2 = \frac{400}{7} {}^{\circ}C$Now,find $T_1$:$T_1 = \frac{3}{2} (\frac{400}{7}) = \frac{600}{7} {}^{\circ}C$Thus,the temperatures are $T_1 = \frac{600}{7} {}^{\circ}C$ and $T_2 = \frac{400}{7} {}^{\circ}C$.

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