Two plates of equal area are placed in contac | vedclass

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Question

(B) The rate of heat flow through the plates in series is given by:$\frac{Q}{t} = \frac{K_1 A (	heta_1 - 	heta)}{L_1} = \frac{K_2 A (	heta - 	heta

Vedclass · Accepted Answer

$-8.6$

Vedclass · Answer

(B) The rate of heat flow through the plates in series is given by:$\frac{Q}{t} = \frac{K_1 A (	heta_1 - 	heta)}{L_1} = \frac{K_2 A (	heta - 	heta_2)}{L_2}$Since the plates are of the same material,$K_1 = K_2 = K$. Also,the area $A$ is the same.Given: $	heta_1 = -20^{\circ}C$,$	heta_2 = 20^{\circ}C$,$L_1 = 2 \, cm$,$L_2 = 5 \, cm$.Substituting these values into the equation:$\frac{K A (-20 - 	heta)}{2} = \frac{K A (	heta - 20)}{5}$$\frac{-20 - 	heta}{2} = \frac{	heta - 20}{5}$$5(-20 - 	heta) = 2(	heta - 20)$$-100 - 5	heta = 2	heta - 40$$7	heta = -60$$	heta = -\frac{60}{7} \approx -8.6^{\circ}C$Thus,the temperature of the contact surface is $-8.6^{\circ}C$.

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