A) \[6{}^\circ C\]
B) \[12{}^\circ C\]
C) \[18{}^\circ C\]
D) \[24{}^\circ C\]
Correct Answer: B
Solution :
Heat flowing through the layers \[Q=\frac{KA\Delta \theta }{d}t\] Now, if Q, A, d, t are constant, then \[\Delta \,{{\theta }_{A}}{{K}_{A}}=\Delta \,{{\theta }_{B}}{{K}_{B}}\] \[({{\theta }_{1}}-\theta ).2K=(\theta -{{\theta }_{2}})K\] or \[2({{\theta }_{1}}-\theta )=(\theta -{{\theta }_{2}})\] ?.(i) \[\because \] \[{{\theta }_{1}}-{{\theta }_{2}}={{36}^{o}}C\] \[{{\theta }_{2}}={{\theta }_{1}}-36\] From Eq. (i) \[2({{\theta }_{1}}-\theta )=\theta -{{\theta }_{1}}+36\] or \[3({{\theta }_{1}}-\theta )=36\] or \[{{\theta }_{1}}-\theta =12\] So, temperature difference across A is\[{{12}^{o}}C\].You need to login to perform this action.
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