Transmission Of Heat Question 44
Question: Two walls of thicknesses d1 and d2 and thermal conductivities k1 and k2 are in contact. In the steady state, if the temperatures at the outer surfaces are $ T _{1} $ and $ T _{2} $ , the temperature at the common wall is
[MP PMT 1990; CBSE PMT 1999]
Options:
A) $ \frac{k _{1}T _{1}d _{2}+k _{2}T _{2}d _{1}}{k _{1}d _{2}+k _{2}d _{1}} $
B) $ \frac{k _{1}T _{1}+k _{2}d _{2}}{d _{1}+d _{2}} $
C) $ ( \frac{k _{1}d _{1}+k _{2}d _{2}}{T _{1}+T _{2}} )T _{1}T _{2} $
D) $ \frac{k _{1}d _{1}T _{1}+k _{2}d _{2}T _{2}}{k _{1}d _{1}+k _{2}d _{2}} $
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Answer:
Correct Answer: A
Solution:
In series both walls have same rate of heat flow.
Therefore $ \frac{dQ}{dt}=\frac{K _{1}A(T _{1}-\theta )}{d _{1}}=\frac{K _{2}A(\theta -T _{2})}{d _{2}} $
$ \Rightarrow K _{1}d _{2}(T _{1}-\theta )=K _{2}d _{1}(\theta -T _{2}) $
$ \Rightarrow \theta =\frac{K _{1}d _{2}T _{1}+K _{2}d _{1}T _{2}}{K _{1}d _{2}+K _{2}d _{1}} $
BETA
