Question 450: Thermodynamics & KTG

Question

Two materials having coefficients of thermal
conductivity '3K' and 'K' and thickness 'd' and
'3d', respectively, are joined to form a slab as
shown in the figure. The temperatures of the
outer surfaces are '$$	heta $$2' and '$$	heta $$1' respectively,
($$	heta $$2 > $$	heta $$1). The temperature at the interface is :-

Accepted Answer

Let interface temperature in steady state conduction is $	heta$, then assuming no heat loss through sides;
$$
\left(\begin{array}{l}
	ext { Rate of heat } \
	ext { flow through } \
	ext { first slab }
\end{array}ight)=\left(\begin{array}{l}
	ext { Rate of heat } \
	ext { flow through } \
	ext { second slab }
\end{array}ight)
$$
$$
\begin{aligned}
& \Rightarrow \frac{(3 K) A\left(	heta_2-	hetaight)}{d}=\frac{K A\left(	heta-	heta_1ight)}{3 d} \\
& \Rightarrow \quad 9\left(	heta_2-	hetaight)=	heta-	heta_1 \\
& \Rightarrow \quad 9 	heta_2+	heta_1=10 	heta \\
& \Rightarrow \quad 	heta=\frac{9}{10} 	heta_2+\frac{1}{10} 	heta_1
\end{aligned}
$$

Answer

$${{{	heta _1}} \over {10}} + {{9{	heta _2}} \over {10}}$$

Answer

$${{{	heta _2} + {	heta _1}} \over 2}$$

Answer

$${{{	heta _1}} \over {6}} + {{5{	heta _2}} \over {6}}$$

Answer

$${{{	heta _1}} \over {3}} + {{2{	heta _2}} \over {3}}$$

Thermodynamics & KTG

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