One end of a thermally insulated rod is kept at a temperature $$T1$$ and the other at $$T2$$. The rod is composed of two sections of lengths $$l1$$ and $$l2$$ and thermal conductivities $$k1$$ and $$k2$$ respectively. The temperature at the interface of the two sections is

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Infinity Learn · Accepted Answer

At steady state, the rate of flow of heat through first slab $$=$$ rate of flow of heat through second slab.$$k1AT1$$−$$Tl1=k2AT$$−$$T2l2$$                ....(1)where$$ $$7$$ $$=$$ temperature of junction From eq. $$(1)         $$k1l2T1$$−$$k1l2T=k2l1T$$−$$k2l1T2or$$     $$k1l2T1+k2l1T2=k2l1T+k1l2TSolving$$ we $$getT=k1l2T1+k2l1T2k1l2+k2l1$$

One end of a thermally insulated rod is kept at a temperature T₁ and the other at T₂. The rod is composed of two sections of lengths l₁ and l₂ and thermal conductivities k₁ and k₂ respectively. The temperature at the interface of the two sections is

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One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of lengths l1 and l2 and thermal conductivities k1 and k2 respectively. The temperature at the interface of the two sections is

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One end of a thermally insulated rod is kept at a temperature T₁ and the other at T₂. The rod is composed of two sections of lengths l₁ and l₂ and thermal conductivities k₁ and k₂ respectively. The temperature at the interface of the two sections is