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 25.1∘ C and 0∘ 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 · Accepted Answer

I think here is the solution:

Jeffrey Jordon · Accepted Answer

Given that  thermal conductivity of plaster board, kp=0.30 J/(s⋅m⋅C)  thermal conductivity of brick, kb=0.60 J/(s⋅m⋅C)  thermal conductivity of wood, kw=0.10 J/(s⋅m⋅C)  The inside temperature, Ti=25.5∘C  The outside temperature, T0=0∘ C  a) The rate of heart transfer is the same for all three materials, so  Qt=kpA△TbL=kbA△TbL=kwA△TwL  Let T1 be the temperature at the plasterboard-brick interface, T2  be the temperature at the brick-wood interface.  For plaster-board interface,  kpA△TpL=kbA△TbL  kp(Ti−T1)=kb(T1−T2)  kpTi−kpT1=kbT1−kbT2  T2=(kp+kbkb)T1−(kpkb)Ti  Hence, for brick-wood interface  kbA△TbL=kwA△TwL  kb(T1−T2)=kw(T2−T0)  kbT1=(kw+kb)T2−kwT0  kbT1=(kw+kb){(kp+kbkb)T1−(kpkbTi)}−kwT0  Therefore, the tempaerature at the plasterboard-brick interface is =19.83∘C

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

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