# Three building materials, plasterboard [k = 0.30 J/(s m Co)], brick [k = 0.60 J/(s m Co)], and wood [k = 0.10 J/(s m Co)], are sandwiched together as

Three building materials, plasterboard [k = 0.30 J/(s m Co)], brick [k = 0.60 J/(s m Co)], and wood [k = 0.10 J/(s m Co)], are sandwiched together as the drawing illustrates. The temperatures at the inside and outside surfaces are $$\displaystyle{25.1}^{\circ}$$ C and $$\displaystyle{0}^{\circ}$$ C, respectively. Each material has the same thickness and cross-sectional area. Find the temperature (a) at the plasterboard-brick interface and (b) at the brick-wood interface.

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I think here is the solution:

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Given that

thermal conductivity of plaster board, $$k_p=0.30\ J/(s\cdot m\cdot C)$$

thermal conductivity of brick, $$k_b=0.60\ J/(s\cdot m\cdot C)$$

thermal conductivity of wood, $$k_w=0.10\ J/(s\cdot m\cdot C)$$

The inside temperature, $$T_i=25.5^\circ C$$

The outside temperature, $$T_0=0^\circ\ C$$

a) The rate of heart transfer is the same for all three materials, so

$$\frac{Q}{t}=\frac{k_pA\triangle T_b}{L}=\frac{k_bA\triangle T_b}{L}=\frac{k_wA\triangle T_w}{L}$$

Let $$T_1$$ be the temperature at the plasterboard-brick interface, $$T_2$$  be the temperature at the brick-wood interface.

For plaster-board interface,

$$\frac{k_pA\triangle T_p}{L}=\frac{k_bA\triangle T_b}{L}$$

$$k_p(T_i-T_1)=k_b(T_1-T_2)$$

$$k_pT_i-k_pT_1=k_bT_1-k_bT_2$$

$$T_2=(\frac{k_p+k_b}{k_b})T_1-(\frac{k_p}{k_b})T_i$$

Hence, for brick-wood interface

$$\frac{k_bA\triangle T_b}{L}=\frac{k_wA\triangle T_w}{L}$$

$$k_b(T_1-T_2)=k_w(T_2-T_0)$$

$$k_bT_1=(k_w+k_b)T_2-k_wT_0$$

$$k_bT_1=(k_w+k_b)\{(\frac{k_p+k_b}{k_b})T_1-(\frac{k_p}{k_b}T_i)\}-k_wT_0$$

Therefore, the tempaerature at the plasterboard-brick interface is $$=19.83^\circ C$$