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(A) In steady-state heat conduction through rods in series,the rate of heat flow $\frac{dH}{dt}$ is the same for both rods.The formula for heat flow i

Vedclass · Accepted Answer

$K_1 > K_2$

Vedclass · Answer

(A) In steady-state heat conduction through rods in series,the rate of heat flow $\frac{dH}{dt}$ is the same for both rods.The formula for heat flow is $\frac{dH}{dt} = K A \left| \frac{dT}{dx} \right|$,where $K$ is the thermal conductivity,$A$ is the cross-sectional area,and $\left| \frac{dT}{dx} \right|$ is the temperature gradient.Since $\frac{dH}{dt}$ and $A$ are constant for both rods,we have $K_1 \left| \frac{dT}{dx} \right|_1 = K_2 \left| \frac{dT}{dx} \right|_2$.From the given graph,the slope of the temperature-distance line represents the magnitude of the temperature gradient $\left| \frac{dT}{dx} \right|$.It is clear from the graph that the slope for the second rod (from $L$ to $2L$) is greater than the slope for the first rod (from $0$ to $L$).Therefore,$\left| \frac{dT}{dx} \right|_2 > \left| \frac{dT}{dx} \right|_1$.Since $K_1 \left| \frac{dT}{dx} \right|_1 = K_2 \left| \frac{dT}{dx} \right|_2$,and $\left| \frac{dT}{dx} \right|_2 > \left| \frac{dT}{dx} \right|_1$,it must be that $K_1 > K_2$.

Two rods of the same length and same cross-sectional area are joined in series. The temperatures at the two ends are as shown in the figure. As we move along the rod,the temperature variation is as shown in the following graph.

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