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(A) Let the thickness of the first plate be $x_1 = 2$ and the second plate be $x_2 = 3$. Let the temperature of the contact surface be $	heta$. The h

Vedclass · Accepted Answer

$(a) -5^{\circ}C, (b) 0^{\circ}C$

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(A) Let the thickness of the first plate be $x_1 = 2$ and the second plate be $x_2 = 3$. Let the temperature of the contact surface be $	heta$. The heat flow rate through both plates in series must be equal: $\frac{dQ}{dt} = \frac{K_1 A (T_1 - 	heta)}{x_1} = \frac{K_2 A (	heta - T_2)}{x_2}$.$(a)$ If they are of the same material,$K_1 = K_2 = K$. Given $T_1 = -25^{\circ}C$ and $T_2 = 25^{\circ}C$ (or vice versa,the setup implies a gradient between $-25^{\circ}C$ and $25^{\circ}C$). Using the steady state condition: $\frac{K A (	heta - (-25))}{2} = \frac{K A (25 - 	heta)}{3}$.$3(	heta + 25) = 2(25 - 	heta) \Rightarrow 3	heta + 75 = 50 - 2	heta \Rightarrow 5	heta = -25 \Rightarrow 	heta = -5^{\circ}C$.$(b)$ If thermal conductivities are in ratio $K_1 : K_2 = 2 : 3$,then $\frac{2 A (	heta + 25)}{2} = \frac{3 A (25 - 	heta)}{3}$.$	heta + 25 = 25 - 	heta \Rightarrow 2	heta = 0 \Rightarrow 	heta = 0^{\circ}C$.

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