question_answer

A partition wall has two layers of different materials A and B in contact with each other. They have the same thickness but the thermal conductivity of layer A is twice that of layer B. At steady state the temperature difference across the layer B is 50 K, then the corresponding difference across the layer A is

A) 50 K

B) 12.5 K

C) 25 K

D) 60 K

Correct Answer: C

Solution :

[c] Let T be temperature of the junction Here, \[{{\operatorname{K}}_{A}}=2{{K}_{B}},\] \[\operatorname{T}-{{T}_{B}}=50K\] At the steady state, \[{{H}_{A}}={{H}_{B}}\] \[\therefore \frac{{{K}_{A}}A\left( {{T}_{A}}-T \right)}{L}=\frac{{{K}_{B}}A\left( {{T}_{B}}-T \right)}{L}\] \[2{{K}_{B}}\left( {{T}_{A}}-T \right)={{K}_{B}}\left( T-{{T}_{B}} \right)\] \[{{T}_{A}}-T=\frac{T-{{T}_{B}}}{2}=\frac{50K}{2}=25K\]