A wall is of two layers P and Q each made of different material. Both the layers have same thickness. The thermal conductivity of the material P is twice that of Q. Under thermal equilibrium, the temperature difference across the walls is $36^{o}$ C. What is the temperature difference across the layer P?

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Solution

$\frac{2 K (△ T_{1}) A}{l} = \frac{K (△ T_{2}) A}{l}$

$\frac{△ T_{1}}{△ T_{2}} = \frac{1}{2}$

$\frac{△ T_{1}}{36 - △ T_{2}} = \frac{1}{2} = △ T_{1} = 12^{o} C$

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A wall is of two layers P and Q each made of different material. Both the layers have same thickness. The thermal conductivity of the material P is twice that of Q. Under thermal equilibrium, the temperature difference across the walls is 36oC. What is the temperature difference across the layer P?

2K(△T1)Al=K(△T2)Al △T1△T2=12 △T136−△T2=12=△T1=12oC

$\frac{2 K (△ T_{1}) A}{l} = \frac{K (△ T_{2}) A}{l}$

$\frac{△ T_{1}}{△ T_{2}} = \frac{1}{2}$

$\frac{△ T_{1}}{36 - △ T_{2}} = \frac{1}{2} = △ T_{1} = 12^{o} C$