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(A) In steady state,the rate of heat flow through both plates is equal.$\frac{Q}{t} = \frac{K_1 A (	heta - 	heta_1)}{L_1} = \frac{K_2 A (	heta_2 -

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(A) In steady state,the rate of heat flow through both plates is equal.$\frac{Q}{t} = \frac{K_1 A (	heta - 	heta_1)}{L_1} = \frac{K_2 A (	heta_2 - 	heta)}{L_2}$Given: $	heta_1 = -20^{\circ}C$,$	heta_2 = 20^{\circ}C$,$L_1 = 2 \ cm$,$L_2 = 5 \ cm$,and $\frac{K_1}{K_2} = \frac{2}{5}$.Let $	heta$ be the temperature of the contact surface.Substituting the values into the equation:$\frac{K_1}{L_1} (	heta - (-20)) = \frac{K_2}{L_2} (20 - 	heta)$$\frac{K_1}{K_2} \cdot \frac{L_2}{L_1} = \frac{20 - 	heta}{	heta + 20}$$\frac{2}{5} \cdot \frac{5}{2} = \frac{20 - 	heta}{	heta + 20}$$1 = \frac{20 - 	heta}{	heta + 20}$$	heta + 20 = 20 - 	heta$$2	heta = 0$$	heta = 0^{\circ}C$.

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