Transmission Of Heat Question 84
Question: A wall has two layers A and B made of different materials. The thickness of both the layers is the same. The thermal conductivity of A and B are KA and KB such that KA = 3KB. The temperature across the wall is 20°C. In thermal equilibrium
[CPMT 1998]
Options:
A) The temperature difference across $ A=15{}^\circ C $
B) The temperature difference across $ A=5{}^\circ C $
C) The temperature difference across A is 10°C
D) The rate of transfer of heat through A is more than that through B.
Show Answer
Answer:
Correct Answer: B
Solution:
In series rate of flow of heat is same
Therefore $ \frac{K _{A}A({\theta _{1}}-\theta )}{l}=\frac{K _{B}A(\theta -{\theta _{2}})}{l} $
Therefore $ 3K _{B}({\theta _{1}}-\theta )=K _{B}(\theta -{\theta _{2}}) $
Therefore $ 3({\theta _{1}}-\theta )=(\theta -{\theta _{2}}) $
Therefore $ 3{\theta _{1}}-3\theta =\theta -{\theta _{2}} $
Therefore $ 4{\theta _{1}}-4\theta ={\theta _{1}}-{\theta _{2}} $
Therefore $ 4({\theta _{1}}-\theta )=({\theta _{1}}-{\theta _{2}}) $
Therefore $ 4({\theta _{1}}-\theta )=20 $
Therefore $ ({\theta _{1}}-\theta )=5{}^\circ C $
BETA
