Two metal plates of same area and ratio of thickness $1 : 2$ and having ratio of thermal conductivities $1 : 2$ are in contact. If the free sides of the first plate is maintained at $- 20^{o}$ C and the free side of the other plate is maintained at $+ 40^{o}$ C ,the temperature of the common surface after attaining steady state is:

0^{o} C

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- 10^{o} C

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10^{o} C

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20^{o} C

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Solution

The correct option is B $10^{o} C$
From the fourier's law we know that heat transfer through a plane wall is given as:
$Q = \frac{K A Δ T}{l}$
Here, area are same for both the rod,
$Q = \frac{K 4 Δ T}{l}$
$Q_{1} = \frac{K_{1} (T_{1} - T)}{l_{1}}$
$Q_{2} = \frac{K_{2} (T - T_{2})}{l_{2}}$

On equating both $Q_{1}$ and $Q_{2}$ we get
$\frac{K_{1} (T_{1} - T)}{l_{1}} = \frac{K_{2} (T - T_{2})}{l_{2}}$
$(T_{1} - T) = (T - T_{2})$
$- 20 - T = T - 40$
$T = 10^{\circ} C$

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The correct option is B 10oCFrom the fourier's law we know that heat transfer through a plane wall is given as:Q=KAΔTlHere, area are same for both the rod,Q=K4ΔTlQ1=K1(T1−T)l1Q2=K2(T−T2)l2On equating both Q1 and Q2 we getK1(T1−T)l1=K2(T−T2)l2(T1−T)=(T−T2)−20−T=T−40T=10∘C