InterviewSolution
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InterviewSolution
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A wall has two layers A and B each made of different materials. The thickness of both the layers is the same. The thermal conductivity of A, `K_(A) = 3K_(B)`. The temperature different across the wall is `20^(@)C` in thermal equilibriumA. The temperature difference across A is `15^(@)C`.B. Rate of heat transfer across A is more than across BC. Rate of heat transfer across both is same.D. Temperature difference aross A is `5^(@)C` |
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Answer» Correct Answer - D `(Q)/(At) =(K(theta_(1)-theta_(2)))/(d)`= constant `:. K_(A)((theta_(1)=theta)/(d)) = K_(B)((theta-theta_(2))/(d))` `(K_A)/(K_B) = (theta-theta_(2))/(theta_(1)-theta) or 3 = (theta-theta_(2))/(theta_(1)=theta)` or, `3theta_(1)+theta_(2) = 4 theta` ...(1) Given `theta_(1)-theta_(2) = 4 theta` ..(2) Solving (1) and (2) we have, `theta-theta_(2)=15^(@)C` `:. theta_(1) - theta = theta_(1)-theta_(2)+theta_(2) - theta` `=(theta_(1)-theta_(2)) - (theta-theta_(2))` `=20^(@)C - 15^(@)C = 5^(@)C`. |
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