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 definitely result in lowering the temperature $θ$ of the interface?

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by increasing the thermal conductivity of inner layer

by increasing the thermal conductivity of outer layer

by increasing thickness of inner layer

by increasing thickness of outer layer

answer is A, D.

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Detailed Solution

Net resistance of two wall= $(\frac{l_{1}}{k_{1}} + \frac{l_{2}}{k_{2}}) \frac{1}{A}$
Temperature of interface = $θ = 1000 - (\frac{Δ T}{\frac{l_{1}}{k_{1}} + \frac{l_{2}}{k_{2}}}) \times \frac{l_{1}}{k_{1}} \dots . . (1)$

On increasing thermal conductivity of outer layer $k_{2}$ , temp of interface certainly decreases.
or $θ = (\frac{Δ T}{\frac{l_{1}}{k_{1}} + \frac{l_{2}}{k_{2}}}) \times \frac{l_{2}}{k_{2}} + 100 \dots \dots \dots . . (2)$

On increasing thickness of Inner layer $l_{1}$ temperature of interface decreases.