Two identical long bars A and B made of diffe | vedclass

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(C) In the steady state, the heat conducted through the bar is lost to the surroundings through the surface of the bar by convection and radiation. Le

Vedclass · Accepted Answer

$\frac{k_A}{k_B} = \frac{l_A^2}{l_B^2}$

Vedclass · Answer

(C) In the steady state, the heat conducted through the bar is lost to the surroundings through the surface of the bar by convection and radiation. Let $P$ be the perimeter of the bars and $h$ be the heat transfer coefficient. The heat flow rate at a distance $x$ from the hot end is $q = -kA \frac{dT}{dx}$.The heat lost by the element $dx$ to the surroundings is $dQ = hP(T - T_0) dx$.In steady state, the heat balance equation is $-kA \frac{d^2T}{dx^2} = hP(T - T_0)$.Let $	heta = T - T_0$, then $\frac{d^2	heta}{dx^2} = \frac{hP}{kA} 	heta$. The solution is $	heta = 	heta_0 e^{-mx}$, where $m = \sqrt{\frac{hP}{kA}}$.The wax melts up to a length $l$ where the temperature $	heta$ reaches the melting point $	heta_m$. Thus, $	heta_m = 	heta_0 e^{-ml}$.Since $	heta_m, 	heta_0, h,$ and $P$ are the same for both bars, $ml$ must be constant. Therefore, $l \propto \frac{1}{\sqrt{m^2}} \propto \sqrt{k}$.Thus, $\frac{l_A}{l_B} = \sqrt{\frac{k_A}{k_B}}$, which implies $\frac{k_A}{k_B} = \frac{l_A^2}{l_B^2}$.

Two identical long bars $A$ and $B$ made of different materials are coated with wax and have one end immersed in a hot oil bath. When the steady state is reached, the lengths for which the wax melts are $l_A$ and $l_B$. If $k_A$ and $k_B$ are the thermal conductivities of the materials, then:

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