Heat And Thermodynamics Ques 70

  1. Two metal cubes $A$ and $B$ of same size are arranged as shown in figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of $A$ and $B$ are $300 W / m^{\circ} C$ and $200 W / m{ }^{\circ} C$, respectively. After steady state is reached the temperature $T$ of the interface will be

(1996, 2M)

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Solution:

Formula:

Series And Parallel Combination Of Rod :

  1. Thermal resistance, $R=\frac{l}{K A}$.

i.e. $R \propto \frac{1}{K}, l$ and $A$ being the same for both the blocks

$\therefore \quad \frac{R _A}{R _B}=\frac{K _B}{K _A}=\frac{200}{300}=\frac{2}{3}$

Rate of flow of heat (thermal current) along both the cubes will be equal (in series). Therefore, temperature difference across both the cubes will be in the ratio of their thermal resistances. Hence,

$$ \begin{aligned} & \frac{(TD) _A}{(TD) _B}=\frac{R _A}{R _B}=\frac{2}{3} \quad \text { or } \quad \frac{100^{\circ} C-T}{T-0^{\circ} C}=\frac{2}{3} \\ & \Rightarrow \quad 300^{\circ} C-3 T=2 T \Rightarrow T=60^{\circ} C \end{aligned} $$



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