The temperature drop through a two layer furnace wall is `900^@C`. Each layer is of equal area of cross section. Which of the following actions will result in lowering the temperature `theta` of the interface?

Options

A. By increasing the thermal conductivity of outer layer .
B. By increasing the thermal conductivity of outer layer .
C. By increasing the thermal conductivity of outer layer .
D. By increasing the thermal conductivity of outer layer .

Answer

A, B

Explanation

H =rate of heat flow `=(900)/((l_(1))/((K_(i)A))+(l_(0))/(K_(0A)))`
Now, `1000 -theta = (Hl_(i))/(K_(i)A)or`
`theta=1000-[(900)/((l_(i))/(K_(i0A))+(l_(0))/(K_(0)A))](l_(i))/(K_(i)A)=100-(900)/(1+(l_(0)K)/(K_(0))(K_(i))/(l_(i)))`
Now we can see that `theta` can be decreased by increasing thermal conductivity of outer layer `(K_(0))` and thickness of inner layer `(I_(i))` .