Two slabs A and B of equal surface area are p | vedclass

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Question

(A) Let the thickness of slab $B$ be $d$ and its thermal conductivity be $K$. Then,for slab $A$,the thickness is $2d$ and thermal conductivity is $2K$

Vedclass · Accepted Answer

$62.5$

Vedclass · Answer

(A) Let the thickness of slab $B$ be $d$ and its thermal conductivity be $K$. Then,for slab $A$,the thickness is $2d$ and thermal conductivity is $2K$.Let the temperature at the contact surface be $T$.Since the slabs are in series,the rate of heat flow $H$ through both slabs must be equal:$H = \frac{K_A A (T_1 - T)}{d_A} = \frac{K_B A (T - T_2)}{d_B}$Substituting the given values:$\frac{2K \cdot A (100 - T)}{2d} = \frac{K \cdot A (T - 25)}{d}$$100 - T = T - 25$$2T = 125$$T = 62.5^{\circ} C$

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