Transmission Of Heat Question 97
Question: A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is $ 36^{o}C $ . Then the difference of temperature at the two surfaces of A will be
[IIT 1980; CPMT 1991; BHU 1997; MP PET 1996, 99; DPMT 2000]
Options:
A) $ 6^{o}C $
B) $ 12^{o}C $
C) $ 18^{o}C $
D) $ 24^{o}C $
Show Answer
Answer:
Correct Answer: B
Solution:
Suppose thickness of each wall is x then $ {{( \frac{Q}{t} )} _{combination}}={{( \frac{Q}{t} )} _{A}} $
Therefore $ \frac{K _{S}A({\theta _{1}}-{\theta _{2}})}{2x}=\frac{2KA({\theta _{1}}-\theta )}{x} $
$ \because $
$ K _{S}=\frac{2\times 2K\times K}{(2K+K)}=\frac{4}{3}K $ and $ ({\theta _{1}}-{\theta _{2}})=36{}^\circ $
Therefore $ \frac{\frac{4}{3}KA\times 36}{2x}=\frac{2KA({\theta _{1}}-\theta )}{x} $ Hence temperature difference across wall A is $ ({\theta _{1}}-\theta )=12^{o}C $
BETA
