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(B) Let the thermal conductivity of layer $B$ be $k$ and that of layer $A$ be $2k$. Let the thickness of each layer be $x$ and the cross-sectional are

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$20$

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(B) Let the thermal conductivity of layer $B$ be $k$ and that of layer $A$ be $2k$. Let the thickness of each layer be $x$ and the cross-sectional area be $A$.In a steady state,the rate of heat flow $H$ through both layers in series is the same.$H = \frac{A(\Delta T)_A}{R_A} = \frac{A(\Delta T)_B}{R_B}$Since $R = \frac{x}{kA}$,we have $R_A = \frac{x}{(2k)A}$ and $R_B = \frac{x}{kA}$.Thus,$R_B = 2R_A$.The total temperature difference is $(\Delta T)_A + (\Delta T)_B = 60 \ K$.Since $H$ is constant,$(\Delta T)_A / R_A = (\Delta T)_B / R_B$,which implies $(\Delta T)_B = (\Delta T)_A 	imes (R_B / R_A) = 2(\Delta T)_A$.Substituting this into the total temperature difference equation:$(\Delta T)_A + 2(\Delta T)_A = 60 \ K$$3(\Delta T)_A = 60 \ K$$(\Delta T)_A = 20 \ K$.

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