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Two plates of equal areas are placed in contact wih each other. Their thickness are 2.0 cm and 5.0 cm respectively. The temperature of the external surface of the first plate is `-20^(@)C` and that of the external surface of the second plate is `20^(@)C`. What will be the temperature of the current surface if the plates (i) are of the same material, (ii) have thermal conductivities in the ratio `2:5`.

Text Solution

Verified by Experts

Rate of flow of heat in the plates is `Q/t = (K_(1)A(theta_(1)-theta))/(L_(1)) = (K_(2)A(theta-theta_(2)))/(L_(2)) ….(i)`
(i) Here `theta_(1) = -20^(@)C, theta_(2)=20^(@)C,` `L_(1)=2cm =0.02m, L_(2)=5cm = 0.05m and K_(1)=K_(2)=K`
`therefore` equation (i) becomes `(KA(-20-theta))/(0.02)= (KA(theta-20))/(0.05)`
`therefore 5(-20-theta)= 2(theta-20) implies -100-5theta= 2theta-40 implies 7theta= -60 implies theta=-8.6^(@)C`
(ii) `K_(1)/K_(2)=2/5` or `K_(1)=2/5K_(2)`
` therefore "from equation" (i) (2//5K_(2)A(-20-theta))/0.02= (K_(2)A(theta-20))/(0.05) -20-theta= theta-20` or `-2theta=0`
`therefore theta=0^(@)C`
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