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Correct option is B)

The temperatures at the left of $A$ = $T_{a}$

The temperatures at the right of $B$ = $T_{b}$

The temperature at the junction = $T$

Heat flow is a constant in steady state:

$Q˙ =KAdxdT =constant$

Equating heat flow for both walls we get:

$2KAt−0T−T_{a} =KA2t−tT_{b}−T $

Which simplifies to: $3T=T_{b}+2T_{a}$

Also given that the temperature difference between the walls is $36_{∘}C$

$T_{b}−T_{a}=36$

Combining the two equation in $T,T_{a},T_{b}$ and eliminating $T_{b}$

We get $T−T_{a}=12_{∘}C$

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