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

Chardonnay Felix 2021-02-16 Answered
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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ottcomn
Answered 2021-02-18 Author has 22230 answers
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content_user
Answered 2021-09-09 Author has 10829 answers

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 \(=19.83^\circ C\)

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